100 200 100 200 100 0.8

# Reply to Paris’s Comments on Exactification of Stirling’s Approximation for
the Logarithm of the Gamma Function

Victor Kowalenko ARC Centre of Excellence for Mathematics
and Statistics of Complex Systems
Department of Mathematics and Statistics
The University of Melbourne
Victoria 3010, Australia<EMAIL_ADDRESS>
###### Abstract.

In a recent paper [1] Paris has made several comments concerning the author’s
recent work on the exactification of Stirling’s approximation for the
logarithm of the gamma function, $\ln\Gamma(z)$. Despite acknowledging that
the calculations in Ref. [2] are basically correct, he claims that there is no
need for the concept of regularisation when determining values of
$\ln\Gamma(z)$ from its complete asymptotic expansion. Here it is shown that
he has already applied the concept at the beginning of the analysis in Ref.
[1]. Next he claims that the definition used in Ref. [2] for the Stokes
multiplier in the subdominant part of a complete expansion, which is
responsible for demonstrating that the Stokes phenomenon is discontinuous
rather than a smooth transition, is not correct. It is shown that the Stokes
multiplier used in Ref. [2] is entirely consistent with Berry’s description of
the conventional view of the Stokes phenomenon in his famous work where he
developed the smoothing view [15]. Furthermore, it is pointed out that Paris
has not substantiated the smoothing view by at least reproducing or improving
the results in Table 7 of Ref. [2], which not only confirm the jump
discontinuous nature of the Stokes phenomenon, but also provide accurate
values of $\ln\Gamma(z)$ to 30 figures, regardless of the size of the variable
or whether the truncation is optimal or not. Finally, the issue of
computational expediency of MB-regularised values over Borel-summed values is
discussed since Paris is critical of the Borel-summed forms being used in
computations.

Keywords: Asymptotic series, Asymptotic form, Borel summation, Cauchy
integral, Complete asymptotic expansion, Conditional convergence, Conventional
view, Discontinuity, Divergence, Exactification, Gamma function, Mellin-Barnes
regularisation, Regularisation, Remainder, Smoothing, Stokes line, Stokes
multiplier, Stokes phenomenon, Stokes sector, Stirling’s approximation

2010 Mathematics Subject Classification: 30B10, 30B30, 30E15, 30E20, 34E05,
34E15, 40A05, 40G10, 40G99, 41A60

##  1\. Introduction

Recently Paris [1] has made several claims concerning a recent work [2] which
deals with the exactification of Stirling’s approximation for the first time
ever since its discovery almost three centuries ago. Exactification means that
a complete asymptotic expansion for a function/integral has undergone
regularisation so that it is able to provide exact values of the original
function for all values, including arguments or phases, of the power variable.
Rather than dealing with the gamma function, $\Gamma(z)$, the work
concentrated on Stirling’s approximation being applied to the more complex and
multivalued logarithm of the gamma function, viz. $\ln\Gamma(z)$. The purpose
of this article is to refute the more important claims made by Paris, whilst
at the same time elucidating the reader on fundamental aspects not presented
in Ref. [2] in order that they may gain a better understanding of it.

Before we can address Paris’s claims specifically, we need to present
background or preliminary material that will allow the reader to decide on
their own as to who is correct. Ref. [2] begins with a derivation of the
complete form of Stirling’s approximation for $\ln\Gamma(z)$, which is given
as Eq. (12). This result is

(1) $\displaystyle\ln\Gamma(z)-\Bigl{(}z-\frac{1}{2}\Bigr{)}\ln
z+z-\frac{1}{2}\,\ln(2\pi)\equiv
2z\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(2k-1)}\;\frac{\Gamma(2k)\,\zeta(2k)}{(2\pi
z)^{2k}}\;.$

Usually, the asymptotic series on the rhs of the above statement, which is
denoted by $S(z)$ in Ref. [2], is neglected and hence, the remaining terms on
the lhs become the most familiar form of Stirling’s approximation. On the few
occasions that the series on the rhs is included, only the first few terms are
usually given, while the remaining terms are neglected. That is, the
asymptotic series when displayed is often truncated at a few terms. The
remainder is often denoted by $+\cdots$ or by the appearance of the Landau
gauge symbol, viz. $O()$. When this occurs, it is known as a standard
Poincar$\acute{\rm e}$ asymptotic expansion. However, in order to obtain exact
values of $\ln\Gamma(z)$ for any value of $z$, one requires not only all the
terms in the above series, but also exponentially subdominant exponential
terms that are generally discarded in standard Poincar$\acute{\rm e}$
asymptotics. Such terms are said to lie beyond all orders of a standard
asymptotic expansion [3]. The analysis resulting in (1) was not intended to be
overly original, but to ensure that there were no exponentially subdominant
terms in Stirling’s approximation so that exactification could be achieved. In
short, the derivation of the above statement was complete.

The above statement possesses an equivalence symbol instead of an equals sign.
This is because the infinite series on the rhs is divergent in certain sectors
of the complex plane and conditionally convergent for the remaining sectors.
In the latter case the equivalence symbol can be replaced by the more
restrictive equals sign, but this is unnecessary since much of the analysis in
Ref. [2] is carried out with the equivalence symbol. In those sectors where
the rhs is divergent, the finite values obtained on the lhs cannot possibly
equal the values obtained from the infinite series on the rhs. Therefore, the
lhs is equivalent to the rhs, but it also means that the divergence must be
tamed. The origin of the divergence is due to an impropriety in the method
used to obtain the asymptotic series in the first place. That is, all
asymptotic methods are improper as described in great detail in Refs. [4] and
[5]. Moreover, the divergence results in an infinity in the remainder of the
asymptotic series. As a consequence, for the rhs of the above statement or
equivalence to agree with the finite lhs, the infinity in the remainder must
be removed in an appropriate manner, enabling the series to become summable.
This process is known as regularisation. Therefore, provided there is no
truncation or the remainder has not been neglected, all asymptotic series must
be regularised in order to yield a finite value. If they are truncated, then
one is dealing with an approximation, which depending upon the value of $|z|$
can be extremely accurate, but it can never equal the finite value on the lhs.
For other values the approximation can be highly inaccurate, which means that
one has ventured outside the range of applicability of the expansion. Hence,
exactification means that there are no ranges of applicability. That is, large
or small values of the power variable are no longer necessary. Thus,
exactification can only be achieved with complete asymptotic expansions with
all component asympototic series undergoing regularisation.

Once the above equivalence is derived, Ref. [2] proceeds to regularise $S(z)$
in order to obtain exact values of $\ln\Gamma(z)$ for all values, including
arguments, of $z$. To accomplish this, the Dirichlet series form for the
Riemann zeta function is introduced and the order of the summations in the
resulting equivalence are interchanged. As a consequence, the asymptotic
series on the rhs or $S(z)$ can be expressed as an infinite sum of Type I
terminants. Terminants were first introduced by Dingle in his remarkable, but
grossly underappreciated, book on asymptotic expansions [6] because he noticed
that many of the late terms in the asymptotic expansions of the special
functions of mathematical physics could be approximated by them. Basically, he
surmised that by studying the behaviour of the two types of terminants, one
could come up with very accurate approximations for the special functions of
mathematical physics. It is still an open question whether this can be
achieved mainly because although the late terms become more accurate as an
approximation to the asymptotic expansion, the truncated part may already be
too large as discussed in Ref. [7]. Nevertheless, if a Type I terminant can be
regularised, then it follows that the rhs of the above equivalence can be
regularised by introducing the regularised value for each terminant in $S(z)$.
This is basically the approach taken in the exactification of Stirling’s
approximation. The only problem with this approach is that because an infinite
sum is involved, the regularised value of $S(z)$ may also become infinite,
which is why the analysis in Ref. [2] represents a departure from the analysis
of asymptotic expansions with only one terminant as carried out in Ref. [4].
Thus, the theory in the latter reference, which advances the concepts and
methods in Dingles’s book, was applied to $S(z)$. Once this was done, the
resulting infinite series could be regularised as explained later in this
article, thereby enabling exact values of $\ln\Gamma(z)$ to be calculated from
its asymptotic forms.

As explained in Ref. [4], there are at present two methods that are used
widely for regularising divergent series: (1) Borel summation and (2) Mellin-
Barnes (MB) regularisation. A third technique based on the Euler-Maclaurin
summation formula is discussed in Ref. [8], but it has yet to be studied in
detail. Moreover, logarithmically divergent series need to be treated with
special care [9, 10], although from Lemma 2.2 in Ref. [2], one can see that
the logarithmic power series can be regularised by using the regularised value
of the geometric series, which also represents a crucial step in Borel
summation. Since there is a plethora of techniques that can be used evaluate
an integral or solve a differential equation, there is more than likely to be
a host of methods of regularising divergent series or asymptotic expansions,
although the exactification of Stirling’s approximation can be accomplished by
Borel summation and MB regularisation. Thus, there is no need to consider
other methods of regularisation, which may not be the case for other
asymptotic problems.

The first example of exactification of a complete asymptotic expansion
occurred in 1993 when T. Taucher and I carried out an extensive numerical
investigation of the complete asymptotic expansion for the exponential series
$S_{3}(a)=\sum_{n=1}^{\infty}\exp(-an^{3})$, which was given by

$\displaystyle S_{3}(a)$ $\displaystyle\equiv
2\sum_{k=0}^{\infty}\frac{(-1)^{k+1}a^{2k+1}}{(2\pi)^{6k+4}}\;\frac{\Gamma(6k+4)}{\Gamma(2k+2)}\;\zeta(6k+4)+\frac{2\sqrt{\pi}}{\Gamma(\frac{1}{6})\Gamma(\frac{5}{6})}\sum_{n=1}^{\infty}\frac{e^{-\sqrt{2z}}}{(6\pi
na)^{1/4}}$ (2)
$\displaystyle\times\;\;\sum_{k=0}^{\infty}\frac{\Gamma(k+1/6)}{(4\sqrt{z})^{k}}\frac{\Gamma(k+5/6)}{\Gamma(k+1/2)}\;\cos\Bigl{(}\sqrt{2z}\,-\frac{\pi}{8}-\frac{3k\pi}{4}\Bigr{)}.$

In the above equivalence $z\\!=\\!(2n\pi/3)^{3}a^{-1}$, while $\zeta(s)$
represents the Riemann zeta function. This presentation differs from the
original derivation in that the equivalence symbol has replaced an equals
sign. However, since the numerical investigation in Ref. [11] was concerned
with positive real values of $a$, where both series are deemed to be
conditionally convergent, one could use an equals sign in the analysis. In
standard Poincar$\acute{\rm e}$ asymptotics the above statement is expressed
as

(3) $\displaystyle
S_{3}(a)\sim-\frac{2\zeta(4)}{(2\pi)^{4}}\;\frac{\Gamma(4)}{\Gamma(2)}\;a+\frac{2\zeta(10)}{(2\pi)^{10}}\,\frac{\Gamma(10)}{\Gamma(4)}\;a^{3}+O(a^{5})\;\;.$

It would also be called a small $a$-expansion truncated at the third order.
The second series in Equivalence (2) is referred to as being exponentially
subdominant because the factor of $\exp(-\sqrt{2z})$ vanishes as $a\to 0$.
These terms are the ones that are said to lie beyond all orders of the first
or dominant asymptotic series [3]. Nevertheless, all these terms are required
if one wishes to obtain exact values of $S_{3}(a)$ from Equivalence (2).

Both series in Equivalence (2) diverge when $\pi/2<|{\rm arg}\,z|<\pi$, but
not in the positive half of the complex plane. Despite the fact that they are
conditionally convergent there, the coefficients diverge rapidly for values of
$a$ greater than unity, which makes the standard asymptotic form given by
Approximation (3) an inaccurate approximation when it is truncated. Moreover,
even though both asymptotic series do not possess an infinity in their
remainder, they can still undergo regularisation except that now there is no
need to remove an infinity. Initially, our investigation concentrated on the
Borel summation of both series, but it was found that the forms obtained via
this method were not amenable to fast computation. So, we devised an
alternative method of regularisation, which cast the remainders of both series
in the form of MB integrals. This enabled us to exploit the rapid exponential
decay of the gamma function in the integrands along the imaginary axis.
Consequently, we were able to obtain exact values of both series for values of
$a$ ranging from 0.01 to 10 to incredible accuracy, in some cases as high as
65 figures. All the results from the MB-regularisation of Equivalence (2) were
eventually presented in Secs. 7 and 8 of Ref. [11] together with a comparison
of the first step in hyperasymptotics as described in Refs. [13]-[14]. The
latter approach was found to be far more inferior, especially for the larger
values of $a$. Despite this remarkable achievement in that for first time ever
the exact values of a function had been obtained from its complete asymptotic
expansion, the study has either gone largely unnoticed or even worse, been
badly misunderstood.

As discussed in the preface of Ref. [11], in order to be able to extend the
analysis to the complex plane, it was necessary that a complete understanding
of the Stokes phenomenon would be needed, which entailed not only determining
where the lines of discontinuity occurred, but also the size of the
subdominant jumps in the asymptotic expansion. A major accomplishment towards
this goal occurred with the publication of Ref. [4], but it concentrated only
on asymptotic series that could be Borel-summed, namely both types of
generalised terminants. In order to verify these results, the MB-regularised
forms were also derived. Nevertheless, the advent of Ref. [4] meant that it
was now possible to consider the exactification of Stirling’s approximation,
which as mentioned previously, involves an infinite number of terminants.
Thus, Ref. [2] represents the next stage in the development of improved
methods for handling asymptotic expansions over the past two decades.

##  2\. Regularisation

Now that the preliminary or background information has been presented, we can
now address the comments made by Paris in Ref. [1]. The first major issue in
these comments concerns Stokes smoothing or rather the Berry smoothing of the
Stokes phenomenon since it is clear from Stokes’s 1858 publication [16] that
he believed that discontinuities arose in what he described as the arbitrary
constants of asymptotic expansions when they were extended over the complex
plane. The second claim made by Paris is that there is no need for
regularisation to appear in the analysis of asymptotic expansions. We shall
discuss the second claim first because it will be helpful when addressing the
first claim.

Paris [1] expresses the main asymptotic series in the Stirling approximation
as

(4)
$\displaystyle\Omega(z)=\sum_{k=1}^{N-1}\frac{B_{2k}}{2k(2k-1)}+R_{N}(z)\;\;,$

where the remainder is given by

(5) $\displaystyle
R_{N}(z)=\sum_{k=1}^{\infty}\frac{1}{k}\Bigl{(}e^{2piikz}T_{\nu}(2\pi
ikz)-e^{-2\pi ikz}T_{\nu}(-2\pi ikz)\Bigl{)}\;\;,$

$T_{\nu}(z)$ is defined in terms of the incomplete gamma function as

(6) $\displaystyle T_{\nu}(z)=\frac{e^{\pi i\nu}}{2\pi
i}\,\Gamma(\nu)\Gamma(1-\nu,z)\;\;,$

and $\nu=2N-1$. As stated in Ref. [1], this result was first derived in Paris
and Wood [17] via an MB integral approach. It was derived by a totally
different approach based on Borel summation in Ref. [2], where it appears as
Eq. (83) and is expressed as

$\displaystyle R_{N}^{SS}(z)$ $\displaystyle=\frac{\Gamma(2N-1)}{2\pi
i}\sum_{n=1}^{\infty}\frac{1}{n}\Bigl{(}e^{-2\pi nzi}\,\Gamma(2-2N,-2\pi nzi)$
(7) $\displaystyle-\;\;e^{2\pi nzi}\,\Gamma(2-2N,2\pi nzi)\Bigr{)}\;.$

In fact, the awkward presentation in Ref. [17] makes it difficult to see that
both Eqs. (5) and (7) are identical. Nevertheless, this reference will be
inserted in Ref. [2]. It should be noted that Ref. [17] was never used in Ref.
[2] because from the outset Equivalence (1) was accompanied by the $\sim$
symbol instead of the equivalence symbol. That is, it already began by
employing standard Poincar$\acute{\rm e}$ asymptotics, which set it on a
collision course with Ref. [2] as will become evident here.

Ref. [1] begins with a major inconsistency between (1.1) and (1.3). The first
statement is written as a complete asymptotic expansion, where the $\sim$
symbol implies divergence, but then the finite terms are represented by the
following statement:

(8) $\displaystyle\Omega(z):=\log\Gamma(z)-(z-1/2)\log z+z-(1/2)\log
2\pi\;\;.$

At no stage does Paris discuss the meaning of the := symbol, which looks like
another version of the equivalence symbol. In fact, this symbol does not
appear throughout the remainder of Ref. [1]. Nevertheless, let us accept the
premise that $\Omega(z)$ is finite according to Eq. (8) and that as a result
it always has a finite remainder for any value of $N$. This means in turn that
it does not need to be regularised as claimed by Paris.

Where Paris has missed the need for regularisation is in the introduction of
his so-called terminant function $T(z)$ in Eq. (5). As explained in Ch. 21 of
Dingle’s book, terminants are divergent asymptotic series whose coefficients
are given by the gamma function. For example, Dingle shows by Borel summation
that a Type I terminant can be expressed as

(9)
$\displaystyle\sum_{k=n}^{\infty}\frac{\Gamma(k+\alpha+1)}{(-x)^{k}}=\frac{\Gamma(n+\alpha+1)}{(-x)^{n}}\;\Lambda_{n+\alpha}(x)\;\;,\quad|{\rm
arg}\,x|<\pi\;\;,$

where

(10)
$\displaystyle\Lambda_{s}(x)=\frac{1}{\Gamma(n+1)}\int_{0}^{\infty}\frac{t^{s}\,e^{-t}}{1+t/x}\;dt\;\;,$

The integral in Eq. (10) can be written in terms of the incomplete gamma
function by using No. 3.383(10) in Ref. [18]. Then one finds that the above
result reduces to

(11)
$\displaystyle\sum_{k=n}^{\infty}\frac{\Gamma(k+\alpha+1)}{(-x)^{k}}=(-1)^{n}x^{\alpha-1}e^{x}\,\Gamma(n+\alpha)\,\Gamma(1-n-\alpha,x)\;\;,$

where, in addition to the condition on ${\rm arg}\,x$, $\Re\,(n+\alpha)>0$.

Eq. (11) also gives the impression that one does not need regularisation as
Paris claims for his terminant function. Unfortunately, nothing could be
further from the truth. When $|{\rm arg}\,x|>\pi/2$, all we have to do is is
consider $\Re\,(-x)>>\Im\,(-x)$ for the terminant, in which case the positive
real part of $(-x)^{k}$ dominates the negative value produced by the imaginary
part of $(-x)^{k}$. Consequently, all the terms in the terminant will be
positive and thus, the series will become divergent. That is, an infinity
exists in the remainder. Yet the rhs of Eq. (11) is finite. In fact, the
“sleight of hand” in Eq. (11) has occurred as a result of Borel summation.

Basically, Borel summation consists of a few steps. The first is that the
gamma function on the lhs is replaced by its integral representation and then
the order of the summation and integration are interchanged. When this is
done, one arrives at

(12)
$\displaystyle\sum_{k=n}^{\infty}\frac{\Gamma(k+\alpha+1)}{(-x)^{k}}=\Bigl{(}\frac{-1}{x}\Bigl{)}^{n}\int_{0}^{\infty}e^{-t}\,t^{n+\alpha}\sum_{k=0}^{\infty}\Bigl{(}-\frac{t}{x}\Bigr{)}^{k}\;dt\;\;.$

Hence, the terminant has been expressed in terms of the geometric series. If
we replace the geometric series by $1/(1+t/x)$, then we obtain Eq. (11).
However, a major problem arises because according to p. 19 of Ref. [19], the
geometric series is absolutely convergent for $|t/x|\\!<\\!1$ and divergent
for $|t/x|\\!\geq\\!1$. Since $t$ ranges from zero to infinity, this
substitution cannot be made for all values of $t$. In short, “Eq.” (11) is
simply not valid.

The situation is even more complex because the geometric series is not always
divergent outside the its radius of absolute convergence. To see this more
clearly, let us replace $-t/x$ in the geometric series by $z$ for the time
being. Then we can use the material in Ch. 4 of Ref. [4] or in Ref. [5], where
the series is expressed as

(13)
$\displaystyle\sum_{k=0}^{\infty}z^{k}=\sum_{k=0}^{\infty}\Gamma(k+1)\,\frac{z^{k}}{k!}=\lim_{p\to\infty}\sum_{k=0}^{\infty}\frac{z^{k}}{k!}\int_{0}^{p}e^{-t}\,t^{k}\;dt\;\;.$

Since the integral in Eq. (13) is finite, technically, we can interchange the
order of the summation and integration. In reality, an impropriety has
occurred when we do this, which will be discussed in more detail shortly. For
now, interchanging the summation and integration yields

$\displaystyle\sum_{k=0}^{\infty}z^{k}=$
$\displaystyle\lim_{p\to\infty}\int_{0}^{p}e^{-t}\sum_{k=0}^{\infty}\frac{(zt)^{k}}{k!}\;dt=\lim_{p\to\infty}\int_{0}^{p}e^{-t(1-z)}\;dt$
(14)
$\displaystyle=\;\;\lim_{p\to\infty}\left[\frac{e^{-p(1-z)}}{z-1}+\frac{1}{1-z}\right].$

For $\Re\,z\\!<\\!1$, the first term of the last member of the above equation
vanishes and the geometric series yields a finite value of $1/(1-z)$.
Therefore, we see that the same value is obtained for the series when
$\Re\,z\\!<\\!1$ as when $z$ lies in the circle of absolute convergence given
by $|z|\\!<\\!1$ or the unit disk. According to the definition on p. 18 of
Ref. [19], this means that the series is conditionally convergent for
$\Re\,z\\!<\\!1$ and $|z|\\!>\\!1$. For $\Re\,z\\!>\\!1$, however, the first
term in the last member of Eq. (14) is infinite. Since regularisation is the
process of removing the infinity in the remainder of a divergent series, we
remove or neglect the first term of the last member of Eq. (14). As a
consequence, we are left with a finite part that equals $1/(1-z)$, which is
known as the regularised value. Hence, for all complex values of $z$ except
$\Re\,z\\!=\\!1$, we have

(15) $\displaystyle\sum_{k=0}^{\infty}z^{k}\begin{cases}\equiv
1/(1-z)\;\;,&\;\;\Re\,z>1\;\;,\\\ =1/(1-z)\;\;,&\;\;\Re\,z<1.\end{cases}$

At the barrier of $\Re\,z\\!=\\!1$, the situation appears to be unclear. For
$z\\!=\\!1$ the last member of Eq. (14) vanishes, which is consistent with
removing the infinity from $1/(1-z)$. For other values of $\Re\,z\\!=\\!1$,
the last member of Eq. (14) is clearly undefined. This is to be expected as
this vertical line represents the border between the domains of convergence
and divergence for the geometric series. Nevertheless, because the finite
value remains the same to the right and to the left of the barrier at
$\Re\,z\\!=\\!1$ and in keeping with the fact that regularisation is
effectively the removal of the first term on the rhs of Eq. (14), we take
$1/(1-z)$ to be the finite or regularised value when $\Re\,z\\!=\\!1$. Hence,
Equivalence (15) becomes

(16) $\displaystyle\sum_{k=0}^{\infty}z^{k}\begin{cases}\equiv
1/(1-z)\;\;,&\;\;\Re\,z\geq 1\;\;,\\\ =1/(1-z)\;\;,&\;\;\Re\,z<1.\end{cases}$

Frequently, the values of $z$ over which an asymptotic series is convergent
and those over which it is divergent are not known. So, in these cases it is
better to replace the equals sign by the less stringent equivalence symbol on
the understanding that we may be dealing with a series that is absolutely or
conditionally convergent for some values of $z$. As a result, the above
statement can be succinctly expressed as

(17)
$\displaystyle\sum_{k=N}^{\infty}z^{k}=z^{N}\sum_{k=0}^{\infty}z^{k}\equiv\frac{z^{N}}{1-z}\;\;.$

Such a statement is no longer an equation, but an equivalence statement or
equivalence for short. Moreover, it can be extended to the binomial series as
explained in Ch. 4 of Ref. [4]. It should also be noted that the above result
is only applicable when the form for the regularised value of the divergent
series is identical to the form of the limiting value of the absolutely
convergent series. Later in Ch. 4 of Ref. [4] the regularisation of the
hypergeometric series ${}_{2}F_{1}(a+1,b+1;a+b+2-x;1)$ is discussed, which is
found to possess a the regularised value for $\Re\,x>0$ that is different from
the limit when the series is convergent for $\Re\,x<0$. Hence, in this
instance the two forms must be kept separate from each other as in (16).

Now let us see what happens when we replace $z$ by $-t/x$. According to the
above analysis, the geometric series in Eq. (12) is divergent whenever
$\Re\,(-t/x)>1$. As $t$ ranges from zero to infinity there will always be
values where it is divergent provided $\Re\,x<0$. For these values of $t$ and
$x$ we require the regularised value of the geometric series because the
series is outside the barrier of convergence. Rather than determine the exact
regions where the convergent and divergent regions apply, we simply introduce
Equivalence (17) into Eq. (12). Then we obtain

(18)
$\displaystyle\sum_{k=n}^{\infty}\frac{\Gamma(k+\alpha+1)}{(-x)^{k}}\equiv\Bigl{(}\frac{-1}{x}\Bigl{)}^{n}\int_{0}^{\infty}\frac{t^{n+\alpha}\,e^{-t}}{(1+t/x)}\;dt\;\;,$

for $|{\rm arg}\,x|\neq\pi$. In other words, we have obtained Dingle’s result
given by Eq. (9) here, although with one major difference. That is, we now
have an equivalence symbol instead of an equals sign, which informs us that
the series may be divergent for certain values of $x$. Even when it is
convergent, the series may only be conditionally convergent, which is equally
troublesome to treat. Therefore, we have seen that regularisation is crucial
for handling terminants contrary to Paris’s comments [1].

It should also be noted that if the steps in Borel summation are reversed,
then an asymptotic expansion is obtained. That is, if we wish to obtain an
asymptotic series from the convergent integral on the rhs of “Eq.” (11), then
all we need to do is expand the denominator as a power series. This method of
deriving an asymptotic series is known as the method of expanding most of the
exponential and is discussed on p. 113 of Ref. [6]. Like all asymptotic
methods it has an impropriety, which arises from using a power series
expansion outside its radius of absolute convergence. The divergence in a
series expansion obtained by employing a standard technique such as Laplace’s
method or the method of steepest descent is therefore an indication that
something improper has occurred. Regularisation represents the necessary
corrective measure for rendering the values of the original function. Hence,
this is why it is necessary that the complete Stirling approximation must be
regularised to yield the exact values of $\ln\Gamma(z)$ in Ref. [2].

Since we have seen that terminants can become divergent, the question now
becomes: Why does Paris not need it in his analysis in Ref. [1]? The answer to
this question is that he has not used Borel summation to obtain the Borel-
summed forms given in Ref. [2]. Rather he has used MB integrals in his
analysis. In fact, unbeknownst to him he has invoked MB-regularisation, which
is evident when one examines his “Eq”. (2.1), but before we can do this, the
reader needs to understand what MB regularisation is. So, let us consider the
general series given by

(19) $\displaystyle S_{I}(N,z)=\sum_{k=N}^{\infty}f(k)(-z)^{k}\;\;,$

where $N$ is referred to as the truncation parameter.

Basically, MB regularisation applies whenever the function $f(k)$ in Eq. (19)
possesses the following properties: (1) as $L\\!\to\\!\infty$,
$|f(s)|\\!=\\!O\bigl{(}\exp(-\epsilon_{1}L)\bigr{)}$ for $s\\!=\\!c\\!+\\!iL$,
and $|f(s)|\\!=\\!O\bigl{(}\exp(-\epsilon_{2}L)\bigr{)}$ for
$s\\!=\\!c\\!-\\!iL$, where $\epsilon_{1},\epsilon_{2}\\!>\\!0$, (2)
$-\pi\\!<\\!\theta\\!=\\!{\rm arg}\,z\\!<\\!\pi$, (3) there exists a real
number $c$ such that the poles of $\Gamma(N-s)$ lie to the right of the line
$N\\!-\\!1\\!<\\!c=\Re\,s\\!<\\!N$ in the complex plane and that the poles of
$f(s)\Gamma(s+1-N)$ to the left of it and (4)
$z^{s}f(s)\Gamma(s+1-N)\Gamma(N-s)$ is single-valued to the right of the line.

Consider the following integral along the imaginary axis given by

(20) $\displaystyle
I=(-1)^{N}\int_{c-i\infty}^{c+i\infty}z^{s}f(s)\Gamma(1+s-N)\Gamma(N-s)\;ds\;\;,$

where the cut-off $c$ is given above. Eq. (20) is basically an MB integral
passing through $c$ when the line contour intersects the real axis. The first
two properties of $f(s)$ are required for ensuring that the modulus of the
integrand in the above integral decays exponentially at the endpoints. That
is,

(21) $\displaystyle\left|\frac{z^{s}\,f(s)}{e^{-i\pi s}-e^{i\pi
s}}\right|\stackrel{{\scriptstyle s=c\pm iL}}{{\sim}}\,|z|^{c}e^{\mp
L\theta}e^{-\pi L}\,|f(c\pm iL)|\;\;.$

In particular, the upper limit given by $s\\!=\\!c\\!+\\!i\infty$ decays
exponentially for all values of $\theta$ in the principal branch provided
$f(c\\!+\\!iL)|\\!=\\!O(\exp(-\epsilon_{1}L))$ as $L\\!\to\\!\infty$ and
$\epsilon_{1}\\!>\\!0$. Similarly, the lower limit decays exponentially
provided $|f(c\\!-\\!iL)|=O(\exp(-\epsilon_{2}L))$ as $L\\!\to\\!\infty$ and
$\epsilon_{2}\\!>\\!0$. Furthermore, the integrand of the above integral is
single-valued because ${\rm arg}\,z$ has been confined to the principal branch
of the complex plane from the second property. Since the integrand is single-
valued to the right of the line ${\rm Re}\,s=c$, we can close the line contour
integral to the right and apply Cauchy’s residue theorem. Hence, we obtain

$\displaystyle\int_{c-i\infty}^{c+i\infty}z^{s}f(s)\Gamma(1+s-N)\Gamma(N-s)\;ds+\int_{C}z^{s}f(s)\Gamma(1+s-N)\Gamma(N-s)\;ds$
(22) $\displaystyle=\;2\pi i\sum{\rm
Res}\left\\{z^{s}f(s)\Gamma(1+s-N)\Gamma(N-s)\right\\}\;\;,$

where the contour $C$ represents the great arc contour integral closing the
limits of the MB integral, while $\sum{\rm Res}\\{f(s)\\}$ denotes the sum of
the residues of $f(s)$ in the region bounded by the line $\Re\,s=c$ and the
arc contour.

In accordance with the third property of $f(s)$ only the poles of
$\Gamma(N-s)$ lie in the region to the right of the line contour. Therefore,
the residues in Eq. (22) are the simple poles of the gamma function for all
positive integers greater than or equal to $N$. Hence, Eq. (22) reduces to

$\displaystyle\int_{c-i\infty}^{c+i\infty}z^{s}f(s)\Gamma(1+s-N)\Gamma(N-s)\;ds$
$\displaystyle+\int_{C}z^{s}f(s)\Gamma(1+s-N)\Gamma(N-s)\;ds$ (23)
$\displaystyle=\;\;2\pi i(-1)^{N}\sum_{k=N}^{\infty}(-z)^{k}f(k)\;\;.$

It should be noted that for the case when the series on the rhs of Eq. (23) is
convergent, we have two cases; either the MB integral on the lhs and the
integral along the great arc are both convergent or they are both divergent.
Whichever of the two cases applies depends on the magnitude of ${\rm arg}\,z$.
For example, consider the MB integral for the exponential function, which is

(24)
$\displaystyle\exp(-z)=\sum_{k=0}^{\infty}\frac{(-z)^{k}}{k!}=\frac{1}{2\pi
i}\int_{\scriptstyle{c-i\infty}\atop c={\rm
Re}\,s>0}^{c+i\infty}\Gamma(s)z^{-s}\;ds\;\;.$

The above equation appears on p. 261 of Ref. [21] except that $s$ has been
replaced by $-s$. According to Copson, this result is valid only for $|{\rm
arg}\,z|<\pi/2$. When evaluating the MB integral numerically by using the
NIntegrate routine in Mathematica [22], one obtains values of $\exp(-z)$ very
quickly for $|{\rm arg}\,z|\\!<\\!\pi/4$, but as ${\rm arg}$ approaches
$\pi/2$, the evaluation becomes very slow, indicating that convergence
problems are arising. For $|{\rm arg}\,z|\\!>\\!\pi/2$, the MB integral no
longer converges. This means that we have encountered the second case where
the integral along the great arc is also divergent. Hence, we have seen that
the integral along the great arc does not always vanish even when the series
on the rhs converges.

For the case where the complex series is divergent, Eq. (24) can be written as

$\displaystyle\sum_{k=N}^{\infty}(-z)^{k}f(k)-\frac{(-1)^{N}}{2\pi
i}\int_{C}z^{s}f(s)\Gamma(1+s-N)\Gamma(N-s)\;ds$ (25)
$\displaystyle=\;\frac{(-1)^{N}}{2\pi i}\int_{c-i\infty\atop{N-1<c={\rm
Re}\,s<N}}^{c+i\infty}z^{s}f(s)\Gamma(1+s-N)\Gamma(N-s)\;ds\;\;.$

If the MB integral on the rhs of Eq. (25) is defined, i.e., it yields a
definite value, then the integral along the great arc must also be divergent
or else the divergence in the series cannot be removed. By removing the
integral in this situation, we are effectively regularising the series. As
noted in Ref. [11], this is somewhat analogous to evaluating the Hadamard
finite or regularised part of a divergent integral [23]. So, let us consider
the evaluation of the contour integral along the great arc, which can be
expressed as

(26) $\displaystyle I_{{\text{a}rc}}=\lim_{L\to\infty}i\pi
L\int_{-\pi/2}^{\pi/2}d\gamma\;\frac{f(Le^{i\gamma})\,e^{i\gamma}}{\sin(\pi
Le^{i\gamma})}\;\exp(Le^{i\gamma}(\ln|z|+i\theta))\;\;.$

In obtaining this result we have used the reflection formula for the gamma
function, which is

(27) $\displaystyle\Gamma(1+s-N)\Gamma(N-s)=\frac{(-1)^{N+1}\pi}{\sin(\pi
s)}\;\;.$

The magnitude of the integral in Eq. (26) is bounded by

$\displaystyle|I_{{\text{a}rc}}|\leq 2\pi$
$\displaystyle\lim_{L\to\infty}L\int_{-\pi/2}^{\pi/2}d\gamma\;|f(Le^{i\gamma})|\,\exp\left(L\ln|z|\cos\gamma\right)$
(28)
$\displaystyle\times\;\;\exp\left(-L\pi|\sin\gamma|-L\theta\sin\gamma\right)\;\;.$

The above integral can be split into two integrals, one between $(-\pi/2,0)$
and the other between $(0,\pi/2)$. Then the Jordan inequality can be applied.
At the lower limits of both integrals there is no contribution from the
$\sin\gamma$ terms in the exponential leaving only the terms involving
$\cos\gamma$, which are bounded by unity in the integrals. Therefore, the
conditions under which the contour along the great arc vanishes are
$\epsilon_{1}\\!>\\!\ln|z|$ and $\epsilon_{2}\\!>\\!\ln|z|$. For
$\epsilon_{1}\\!=\\!\ln|z|$ and $\epsilon_{2}\\!=\\!\ln|z|$, the integral can
vanish but this will depend upon the algebraic behaviour of $f(s)$. For
$\epsilon_{1}\\!<\\!\ln|z|$ or $\epsilon_{2}\\!<\\!\ln|z|$, $I_{{\text{a}rc}}$
will yield infinity. In this case we simply remove the infinity in accordance
with the process of regularisation. Therefore, irrespective of whether the
contour integral along the great arc yields infinity or vanishes, we arrive at
the following equivalence statement:

$\displaystyle S_{I}(N,z)=\sum_{k=N}^{\infty}$
$\displaystyle\;f(k)\,(-z)^{k}\equiv\frac{(-1)^{N}}{2\pi
i}\int_{\scriptstyle{c-i\infty}\atop{N-1<c=\Re s<N}}^{c+i\infty}z^{s}f(s)$
(29) $\displaystyle\times\;\;\Gamma(1+s-N)\,\Gamma(N-s)\;ds\;\;.$

We can simplify the above result by introducing again the reflection formula
for the gamma function or Eq. (27). Then we arrive at

(30) $\displaystyle
S_{I}(N,z)=\sum_{k=N}^{\infty}f(k)(-z)^{k}\equiv\int_{c-i\infty\atop{N-1<c=\Re\,s<N}}^{c+i\infty}\frac{z^{s}\,f(s)}{e^{-i\pi
s}-e^{i\pi s}}\;ds\;\;.$

The conditions on $z^{s}f(s)$ as $s\\!\to\\!c\pm i\infty$ are required to
ensure that the integral in Eq. (30) is convergent. They have been included
more for the sake of completeness rather than utility because the above result
has to be integrated numerically in order to obtain a definite value. If the
integral on the rhs Eq. (30) is divergent, then numerical integration will
fail.

Now we let $f(k)\\!=\\!|B_{2k}|\,z/(2k(2k-1)$ and $z\\!=\\!1/z^{2}$ in
$S_{I}(N,z)$ as given by Eq. (19). In order to determine the MB integral
involving this $f(k)$, we have to continue the Bernoulli numbers analytically
to complex values. This is done by expressing the Bernoulli numbers in terms
of integer values of the Riemann zeta function and analytically continuing the
latter to complex values. From No. 9.616 of Ref. [18], we have

(31) $\displaystyle
B_{2k}=\frac{(-1)^{k-1}\,(2k)!}{2^{2k-1}\,\pi^{2k}}\;\zeta(2k)\;\;.$

Hence, $f(k)=\Gamma(2k-1)\zeta(2k)/2(2\pi)^{2k}z$. Then according to Eq. (25),
we obtain

$\displaystyle\sum_{k=N}^{\infty}\frac{B_{2k}}{2k(2k-1)z^{2k-1}}$
$\displaystyle=\frac{1}{2\pi i}\int_{c-i\infty\atop{{\rm Max}[1/2,N-1]<c={\rm
Re}\,s<N}}^{c+i\infty}\frac{1}{(2\pi
z)^{2s-1}}\;\frac{\Gamma(2s-1)\zeta(2s)}{\sin(s\pi)}\;ds$ (32)
$\displaystyle+\;\;\frac{1}{2\pi i}\int_{C}\frac{1}{(2\pi
z)^{2s-1}}\;\frac{\Gamma(2s-1)\zeta(2s)}{\sin(s\pi)}\;ds\;\;.$

Note that the lower limit has been adjusted. This is required in order that
the singularities of the integrand other than those for $\Gamma(N-s)$ lie to
the left of the line contour. Specifically, the line contour needs to lie to
the right of the pole at $s=1/2$ in $\Gamma(2s-1)$, while the truncation
parameter $N$ must be greater than zero. We now make a change of variable by
putting $s=(s^{{}^{\prime}}+1)/2$ in the MB integral on the rhs of the above
equation. This yields

$\displaystyle\sum_{k=N}^{\infty}\frac{B_{2k}}{2k(2k-1)z^{2k-1}}$
$\displaystyle=\frac{1}{4\pi
i}\int_{c^{{}^{\prime}}-i\infty}^{c^{{}^{\prime}}+i\infty}\frac{1}{(2\pi
z)^{s^{{}^{\prime}}}}\;\frac{\Gamma(s^{{}^{\prime}})\zeta(s^{{}^{\prime}}+1)}{\sin((s^{{}^{\prime}}+1)\pi/2)}\;ds^{{}^{\prime}}$
(33) $\displaystyle+\;\;\frac{1}{2\pi i}\int_{C}\frac{1}{(2\pi
z)^{2s-1}}\;\frac{\Gamma(2s-1)\zeta(2s)}{\sin(s\pi)}\;ds\;\;,$

where ${\rm Max}[0,2N-3]<c^{{}^{\prime}}=\Re\,s^{{}^{\prime}}<2N-1$. By
introducing the reflection formula for the Riemann zeta function as given on
p. 269 of Ref. [19] or No. 9.535(3) of Ref. [18], which is

(34) $\displaystyle 2^{1-s}\,\Gamma(s)\,\zeta(s)\,\cos\Bigl{(}\frac{\pi
s}{2}\Bigr{)}=\pi^{s}\,\zeta(1-s)\;\;,$

we obtain

$\displaystyle\sum_{k=N}^{\infty}\frac{B_{2k}}{2k(2k-1)z^{2k-1}}$
$\displaystyle=-\frac{1}{2\pi i}\int_{c-i\infty\atop{{\rm Max}[0,2N-3]<c={\rm
Re}\,s<2N-1}}^{c+i\infty}\;\frac{z^{-s}\,\zeta(-s)}{\sin(s\pi)}\;ds$ (35)
$\displaystyle+\;\;\frac{1}{2\pi i}\int_{C}\frac{1}{(2\pi
z)^{2s-1}}\;\frac{\Gamma(2s-1)\zeta(2s)}{\sin(s\pi)}\;ds\;\;,$

where the prime superscripts have been dropped. For $N\\!=\\!1$, Eq. (35)
reduces to

$\displaystyle\sum_{k=1}^{\infty}\frac{B_{2k}}{2k(2k-1)z^{2k-1}}$
$\displaystyle=-\frac{1}{2\pi i}\int_{c-i\infty\atop{{\rm Max}[0<c={\rm
Re}\,s<1}}^{c+i\infty}\;\frac{z^{-s}\,\zeta(-s)}{\sin(s\pi)}\;ds$ (36)
$\displaystyle+\;\;\frac{1}{2\pi i}\int_{C}\frac{1}{(2\pi
z)^{2s-1}}\;\frac{\Gamma(2s-1)\zeta(2s)}{\sin(s\pi)}\;ds\;\;.$

This is identical to the result with which Paris begins his analysis in Ref.
[1] except for the contour integral along the great arc. The problem is that
the contour along the great arc does not always vanish and is divergent for
certain values of $z$. When this occurs, it must be reflected in the series on
the lhs becoming divergent. So let us examine the series on the lhs.

By introducing Eq. (31) into the lhs of Eq. (32) and replacing the Riemann
zeta function by its Dirichlet series form, we find that

(37) $\displaystyle
L(N,z)=\sum_{k=N}^{\infty}\frac{B_{2k}}{2k(2k-1)z^{2k-1}}=z\sum_{n=1}^{\infty}\sum_{k=N}^{\infty}\frac{(-1)^{k}\,\Gamma(2k-1)}{(2\pi
n)^{2k}}\;\;.$

The inner series looks familiar indeed. Except for the factor of 2 inside the
gamma function, it is basically a Type I terminant, which we studied earlier.
In Ref. [4] such terminants are referred to as generalised terminants because
of the factor of 2. Specifically, a Type I generalised terminant is defined as

(38) $\displaystyle
S^{I}_{p,q}\Bigl{(}N,z^{\beta}\Bigr{)}=\sum_{k=N}^{\infty}\Gamma(pk+q)(-z^{\beta})^{k}\;\;.$

Therefore, the inner series in Eq. (37) can be represented by
$S^{I}_{2,-1}(N,-(1/2\pi n)^{2})$. Let us now introduce the integral
representation for the gamma function into Eq. (38) and interchange the order
summation and integration. As presented on p. 156 of Ref. [4], we find that

(39) $\displaystyle
S^{I}_{p,q}\Bigl{(}N,z^{\beta}\Bigr{)}=\int_{0}^{\infty}e^{-t}\,t^{q-1}\sum_{k=N}^{\infty}\Bigl{(}-z^{\beta}t^{p}\Bigr{)}^{k}\;\;.$

For the above integral to be always convergent, we require that
$\Re\,(pN+q)>0$. Nevertheless, we have arrived at the geometric series, which
has already been analysed here. In fact, from Equivalence (16) we know that

(40)
$\displaystyle\sum_{k=N}^{\infty}\Bigl{(}-z^{\beta}t^{p}\Bigr{)}^{k}\begin{cases}\equiv\left(-z^{\beta}t^{p}\right)^{N}/(1+z^{\beta}t^{p})\;\;,&\;\;\Re\,\left(-z^{\beta}t^{p}\right)\geq
1\;\;,\\\
=\left(-z^{\beta}t^{p}\right)^{N}/(1+z^{\beta}t^{p})\;\;,&\;\;\Re\,\left(-z^{\beta}t^{p}\right)<1.\end{cases}$

For the asymptotic series of the $\ln\Gamma(z)$ or $L(N,z)$, it has already
been stated that $\beta=2$ and $p=2$. Therefore, the series is divergent or
possesses an infinity, whenever $\Re\,(-z^{2}t^{2})>1$. As $t$ ranges from
zero to infinity, this means that integral will yield an infinite value when
$\Re\,z^{2}<0$. So there is no doubt that the integral along the great arc
must also be infinite for these values of $z$ or else Eq. (35) is nonsense.
Hence, MB regularisation involves the removal of the contour along the great
arc, which can either be zero or infinity. Consequently, Eq. (36) can be
written as

(41)
$\displaystyle\sum_{k=1}^{\infty}\frac{B_{2k}}{2k(2k-1)z^{2k-1}}\equiv-\frac{1}{2\pi
i}\int_{c-i\infty\atop{{\rm Max}[0<c={\rm
Re}\,s<1}}^{c+i\infty}\;\frac{z^{-s}\,\zeta(-s)}{\sin(s\pi)}\;ds\;\;.$

So, we have seen that the result with which Paris began his analysis in Ref.
[1] has already been regularised. More importantly, it is simply false to say
that there is no need for regularisation to appear in an asymptotic series as
Paris claims in Ref. [1]. All asymptotic series possess regions/sectors where
they are either divergent or conditionally convergent because the method used
to determine them is improper as explained in Ref. [4]. Regularisation is
necessary for correcting the divergence resulting from the application of an
asymptotic method. On the other hand, we have seen that an asymptotic series
is not divergent everywhere. It has regions where it is convergent, although
such series are conditionally convergent rather than possessing a region of
absolute convergence like the geometric and binomial series. Furthermore, the
above statement is not complete because the values of ${\rm arg}\,z$ over
which it is valid have not been stated. This leads to the concept of an
asymptotic form, which is discussed in the following section.

##  3\. The Stokes Phenomenon and Smoothing

There seems to be a line of thinking amongst the proponents of smoothing of
the Stokes phenomenon that the interpretation of the Stokes phenomenon given
in Refs. [2] and [4] is not the same as theirs and that as a consequence,
although the calculations in these references are correct as stated by Paris
[1], we are not talking about the same phenomenon. This section aims to refute
such a suggestion.

Before discussing the specific comments made by Paris concerning smoothing,
let us consider the following paragraphs from Berry’s famous paper on the
asymptotic smoothing of Stokes’s discontinuities [15]. There it is written:

The conventional view (Stokes 19864) is powerfully (and unconventionally)
argued by Dingle in I. It asserts that the change in $S$ is discontinuous and
localized at the Stokes line: on one side, $S$ takes a value, $S_{-}$ say; on
the other, $S=S_{-}+1$; on the line itself, $S=S_{-}+1/2$. For the example (3)
the intuition behind this view is illustrated by figure 1, which shows how the
steepest-descent contours of the integral (${\rm Im}\,t^{2}={\rm Im}\,Z^{2}$)
change discontinuously across the Stokes line, suddenly bringing in the
subdominant contribution from the stationary point at $t=0$ ($S_{-}=0$ in this
case). It is worth repeating …..

From the context it is clear that Stokes is referring to asymptotic series
interpreted by truncation near their least term. My aim here is to dispel
Stokes’s mist and show that his discontinuity is an artefact of poor
resolution: with the appropriate magnification, $S$ changes smoothly.

From the above paragraphs it is obvious that we are not just talking about a
minor difference in interpretation, but a completely different view or
understanding of the Stokes phenomenon with major ramifications to
asymptotics. Moreover, Berry is aiming to demonstrate that since Stokes and
Dingle are the main proponents of the conventional view, both of them are
wrong. In fact, Paris refers to the dramatic change in the field of
asymptotics as a result of Berry’s paper in Ch. 6 of his book [24]. As we
shall see, this view is very disturbing because if it is true, then it would
be impossible to obtain exact values of a function from its complete
asymptotic expansion as has been accomplished in Refs. [2], [4], [9]-[11] and
[20].

For the benefit of the reader, the smoothing effect takes place near what are
known today as Stokes lines. These are lines where an asymptotic series
attains maximal dominance over all other terms in the asymptotic expansion,
which appears as Rule B in Ch. 1 of Dingle’s book [6]. Basically, this means
that Stokes lines or rays are determined by those phases in which the terms in
an asymptotic series are homogeneous in phase and all of the same sign (Rule
A). The regions or sectors bordered by Stokes lines are referred to as Stokes
sectors. In terms of the Type I terminant given in “Eqs.” (9) and (11), the
Stokes lines occur at ${\rm arg}\,x=(2k+1)\pi$, where $k$ is an integer
because all the terms in the asymptotic series are all positive real numbers
with no complex or imaginary parts at all. Since they occur virtually outside
the principal branch, Dingle does not study the Stokes phenomenon in relation
to Type I terminants in Ch. 22 of Ref. [6]. Instead, he applies the
conventional view to Type II terminants, which are identical to Type I
terminants except the factor of $(-1)^{k}$ is missing. For these series the
Stokes lines occur at ${\rm arg}\,x=2k\pi$ (including zero), well within the
principal branch of the complex plane. In actual fact, this is a shortcoming
in his great book because often an asymptotic series is in powers of
$x^{\beta}$, where $\beta>1$. Then the conditions for the Stokes lines become
either ${\rm arg}\,z=(2k+1)\pi/\beta$ for a generalised Type I terminant or
${\rm arg}\,z=2k\pi/\beta$ for a generalised Type II terminant. Hence, there
can be many Stokes lines lying in the principal branch, which is usually taken
to be $(-\pi,\pi]$. Because of this, all phases and sectors are studied for
both types of generalised terminants in Ref. [4].

Not long after Berry’s paper appeared, Olver [25] claimed that he had
“rigorously” proven that the smoothing did occur near a Stokes line and that
Stokes had got it wrong after all. Basically, Olver re-expressed Berry’s form
for the Stokes multiplier of the subdominant asymptotic solution of the one-
dimensional Helmholtz equation, viz.

(42) $\displaystyle
S_{n}(F)=\frac{1}{2}-\frac{i}{2\pi}\;P\\!\int_{0}^{\infty}dt\;\frac{t^{n-\beta}}{1-t}\,e^{F(1-t)}\;\;,$

as

(43) $\displaystyle S_{n}=-\frac{1}{2\pi
i}\int_{-1}^{\infty}\frac{d\tau}{\tau}\,e^{-A(\tau-\ln(1+\tau))}\,(1+\tau)^{\mu}\;e^{-iB\tau}\;\;,$

where $2z\\!=\\!-A\\!-\\!iB$, $n\\!-\\!1\\!=\\!A\\!+\\!\mu$, $A$ is large,
real and positive, $n$ is an integer and $B$ and $\mu$ are real. From
Laplace’s method, which is a variant of the method of expanding most of the
exponential mentioned earlier, one expects as $A\\!\to\\!\infty$ that the
major contribution to the above integral comes from the neighbourhood around
$\tau\\!=\\!0$. Hence, Olver expands the first exponential in the above
integral in powers of $\tau$ and extends the lower limit of the integral to
$-\infty$. In so doing, he eventually obtains

$\displaystyle S_{n}$
$\displaystyle\overset{\sim}{=}\frac{1}{2}+\frac{1}{2}\;{\rm
erf}\Bigl{(}\frac{B}{\sqrt{2A}}\Bigr{)}+i\;\frac{e^{-B^{2}/2A}}{\sqrt{2\pi
A}}\left(\mu+\frac{1}{3}-\frac{B^{2}}{3A}+\Bigl{(}\frac{1}{A}-\frac{B^{2}}{A^{2}}\Bigr{)}\right.$
(44)
$\displaystyle\left.\times\;\;\Bigl{(}\frac{\mu}{3}-\frac{\mu^{2}}{2}+\frac{\mu^{3}}{6}\Bigr{)}+\cdots\right)-\frac{B}{2A}\;\frac{e^{-B^{2}/2A}}{\sqrt{2\pi
A}}\Bigl{(}\mu-\mu^{2}+\cdots\Bigr{)}\;\;.$

Hence, we see the emergence of the error function appearing in the Stokes
multiplier. It represents the leading order term of complicated asymptotic
expansion, which at best is conditionally convergent, but whose terms diverge
in magnitude eventually. This is what is referred to as rigorous mathematics.
All that has happened here is that more than the first few orders have been
obtained via standard Poincar$\acute{\rm e}$ asymptotics with the remaining
terms denoted by + $\cdots$. On the other hand, not one asymptotic result has
been truncated in Ref. [2]. The only instance of truncation was to a
convergent series in order to expedite the calculations and even then, it was
after $10^{5}$ terms in the series. For a more detailed description of the
terms missing in the above result, the reader should consult Ch. 6 and the
appendix in Ref. [4]. So, it appears that Berry’s smoothing of the Stokes
phenomenon is an “artefact of poor resolution” rather than the other way
around with Stokes mist.

What analysts such as Berry, Olver and Paris fail to realise is that Dingle
came up with an explanation of the Stokes phenomenon when discussing Borel
summation of Type II terminants in Ch. 22 of his book. There he finds that

(45)
$\displaystyle\sum_{k=n}^{\infty}\frac{\Gamma(k+\alpha+1)}{x^{k}}\equiv\frac{\Gamma(n+\alpha+1)}{x^{n}}\;{\bar{\Lambda}}_{n+\alpha}(-x)\;\;,\quad{\rm
arg}\,x=0,$

where

(46)
$\displaystyle{\bar{\Lambda}}_{s}(-x)=\frac{1}{\Gamma(s+1)}\;P\int_{0}^{\infty}\frac{t^{s}\,e^{-t}}{1-t/x}\;dt\;\;,$

and $P$ denotes that the Cauchy principal value must be taken. In the above we
have replaced the equals sign in Dingle’s “Eq.” (16) by an equivalence symbol
in accordance with the discussion on regularisation in the previous section.
The above result is actually an asymptotic form because it indicates the
values of $x$ for which it is valid. Moreover, the result applies only to a
Stokes line since according to the conventional view there is a jump
discontinuity as $x$ moves either above or below the positive real axis. So,
what is this jump discontinuity? Well, if we move to the next section in
Dingle’s book, then we see the jump discontinuities emerge due to semi-
circular residues of the integral in Eq. (46). So, basically we are
interpreting the integral as a Cauchy integral with a line contour along the
positive real axis, but as $x$ moves either above or below the positive axis,
we have to take into account the semi-residues of the Cauchy integral. Dingle
states for ${\rm arg}\,x>0$, the semi-residue is evaluated in an anticlockwise
direction, while for ${\rm arg}\,x<0$, it is taken in a clockwise direction.
As a consequence, he finds that

(47) $\displaystyle{\bar{\Lambda}}_{s}(-x)=\begin{cases}\Lambda_{s}(-x)+i\pi
x^{s+1}e^{-x}/\Gamma(s+1)\;,\quad 0<{\rm
arg}\,x<2\pi\;,\cr\Lambda_{s}(-x)-i\pi
x^{s+1}e^{-x}/\Gamma(s+1)\;,\quad-2\pi<{\rm arg}\,x<0\;.\end{cases}$

In the above results $\Lambda_{s}(-x)$ is given by Eq. (10). In addition, the
above results are virtually identical to the Plemelj relations discussed on p.
414 of Ref. [26].

We can summarise Dingle’s results by expressing them as

(48)
$\displaystyle\sum_{k=n}^{\infty}\frac{\Gamma(k+\alpha+1)}{x^{k}}\equiv\frac{\Gamma(n+\alpha+1)}{x^{n}}\;\Lambda_{n+\alpha}(-x)+\frac{2\pi
i\,x^{s+1}\,e^{-x}}{\Gamma(n+\alpha+1)}\;S\;\;,$

where we take the Cauchy principal value when evaluating
$\Lambda_{n+\alpha}(-x)$ for ${\rm arg}\,x=0$ and the factor $S$ multiplying
the full residue is given by

(49) $\displaystyle S=\begin{cases}1/2\;\;,\quad 0<{\rm arg}\,x<2\pi\;\;,\cr
0\;\;,\quad{\rm arg}\,x=0\;\;,\cr-1/2\;\;,\quad-2\pi<{\rm
arg}\,x<0\;\;.\end{cases}$

This indeed looks familiar. It is in fact the conventional view of the Stokes
multiplier as described by Berry at the beginning of this section with
$S_{-}=-1/2$. Magic! So, Dingle has provided us with the key to the Stokes
phenomenon; we simply interpret the integrals obtained by Borel summation as
Cauchy integrals and then investigate their singular behaviour. Moreover, from
these results we see that it is necessary that the values of ${\rm arg}\,z$
need to be specified for an asymptotic series. The combination of the
asymptotic expansion with the values of ${\rm arg}\,z$ over which it is valid
gives rise to asymptotic forms.

As indicated earlier, the asymptotic series in the Stirling’s approximation
for $\ln\Gamma(z)$ involves Type I terminants, which are not discussed in Ch.
22 of Dingle’s book [6]. For this situation we require the material in Ch.
10.1 of Ref. [4], which studies Borel summation of generalised Type I
terminants given by Eq. (38). In fact, Eq. (38) is extended by introducing a
factor of $\exp(-2li\pi)$ with $z^{\beta}$. Thus, Borel summation of a
generalised Type I terminant yields

(50) $\displaystyle
S^{I}_{p,q}\\!\left(N,z^{\beta}e^{-2li\pi}\right)\equiv(-1)^{N}p^{-1}z^{\beta(N-1)}\int_{C}\frac{s^{N+q/p-1}\,e^{-s^{1/p}}}{s-(-z^{-\beta}e^{2li\pi})}\;ds\;\;.$

If we let

(51) $\displaystyle f(s)=z^{\beta(N-1)}s^{N+q/p-1}\exp(-s^{1/p})/p\;\;,$

then the rhs in the above result can be regarded as a Cauchy integral whose
contour $C$ is the line contour along the positive real axis. Furthermore, it
possesses a singularity at $s=-z^{-\beta}$. Although $\exp(2li\pi)$ is equal
to unity for all values of $l$, it was found in Ch. 7 of Ref. [4] to have an
effect on the MB-regularised value of $S^{I}_{p,q}(N,z^{\beta}e^{-2li\pi})$.
In particular, the difference between the MB-regularised values of
$S^{I}_{p,q}\\!\left(N,z^{\beta}e^{-2li\pi}\right)$ and
$S^{I}_{p,q}\\!\left(N,z^{\beta}e^{-2(l\\!-\\!1)i\pi}\right)$ was found to be
given by

$\displaystyle\Delta
S^{I}_{l}(N,z^{\beta})=S^{I}_{p,q}\bigl{(}N,z^{\beta}e^{-2li\pi}\bigr{)}-S^{I}_{p,q}\bigl{(}N,z^{\beta}e^{-2(l-1)i\pi}\bigr{)}$
(52) $\displaystyle\equiv\int_{c-i\infty\atop{{\rm
Max}[N-1,-q/p]<c=\Re\,s<N}}^{c+i\infty}\\!\\!\\!\\!z^{\beta
s}\;\Gamma(ps+q)\,e^{-(2l-1)i\pi s}\;ds\;\;,$

where the lower bound on the offset $c$ is adjusted to ensure that the poles
for the MB integral remain to the right of $-q/p$, if it should be greater
than $N\\!-\\!1$, in accordance with the conditions on $f(k)$ in Eq. (19). By
making the change of variable, $y\\!=\\!ps\\!+\\!q$, we find that Equivalence
(52) becomes

(53) $\displaystyle\Delta S^{I}_{l}(N,z^{\beta})\equiv
p^{-1}(z^{-\beta}\,e^{(2l-1)i\pi})^{q/p}\int_{c-i\infty\atop{c=\Re\,s>-q/p}}^{c+i\infty}\\!\\!\\!\\!\left(z^{-\beta}e^{(2l-1)i\pi}\right)^{\\!-y/p}\Gamma(y)\;dy$

The MB integral can be regarded as the inverse Mellin transform of $\exp(-x)$
with $x\\!=\\!z^{-\beta/p}\exp((2l-1)i\pi/p)$ in Eq. (24). Consequently, the
above equivalence reduces to

(54) $\displaystyle\Delta S^{I}_{l}(N,z^{\beta})\equiv\frac{2\pi
i}{p}\;z^{-\beta
q/p}\,e^{(2l-1)iq\pi/p}\exp\\!\left(-z^{-\beta/p}e^{(2l-1)i\pi/p}\right)\;\;.$

From p. 412 of Ref. [26] we know that the Cauchy integral on the rhs of
Equivalence (50) develops jump discontinuities as $-z^{-\beta}e^{2il\pi}$
moves across the line contour. This means that while $z^{-\beta}e^{2li\pi}$ is
located in a branch of the complex plane, say $(2j\\!-\\!1)\pi\\!<\\!{\rm
arg}\,(z^{-\beta}e^{2li\pi})\\!<\\!(2j\\!+\\!1)\pi$, the regularised value is
given by the Cauchy integral, but once $z^{-\beta}e^{2li\pi}$ moves outside of
this branch, it acquires extra terms or else the regularised value would be
the same for all Stokes sectors. Therefore, the regularised value of
$S^{I}_{p,q}(N,z^{\beta}e^{-2li\pi})$ cannot be represented solely by a Cauchy
integral. As a result, another problem emerges. We need to determine the
specific Stokes sector over which the Cauchy integral is the sole contribution
to the regularised value. This appears to be arbitrary, much like selecting a
principal branch in the complex plane. So, a primary Stokes sector must be
nominated. This is taken to be the $j\\!=\\!l\\!=\\!0$ branch of the complex
plane. Consequently, the Cauchy integral on the rhs of Equivalence (50)
represents the regularised value only for $-\pi/\beta\\!<\\!{\rm
arg}\,z\\!<\\!\pi/\beta$. That is, the $j\\!=\\!l\\!=\\!0$ branch reduces to
the principal branch of the complex plane when $\beta\\!=\\!1$, thereby
yielding the regularised value of the first type of terminant as given on p.
406 of Dingle’s book [6]. Hence, for $l\\!=\\!0$ we arrive at

(55) $\displaystyle
S^{I}_{p,q}\\!\left(N,z^{\beta}\right)\equiv(-1)^{N}p^{-1}z^{\beta(N-1)}\int_{C}\frac{s^{N+q/p-1}\,e^{-s^{1/p}}}{s-(-z^{-\beta})}\;ds\;\;,$

which is only valid for $-\pi/\beta\\!<\\!{\rm arg}\,z\\!<\\!\pi/\beta$.

We are now in a position to determine the jump discontinuities that apply over
the other or secondary Stokes sectors in the complex plane. First, we note
that the Cauchy integral in Equivalence (50) is singular whenever ${\rm
arg}\,z^{-\beta}\\!=\\!(2j\\!+\\!1)\pi$, and $j$ is an integer. As
$z^{-\beta}\exp((2l\\!-\\!1)i\pi)$ crosses from the primary sector Stokes
sector to the adjacent sector or $j\\!=\\!l\\!+\\!1$ branch of the complex
plane, $-z^{-\beta}e^{2li\pi}$ moves from below the line contour or positive
real axis to above the axis. During this transition the Cauchy integral
becomes undefined when $-z^{-\beta}e^{2li\pi}$ is situated on the positive
real axis. To evaluate the residue, let us consider an infinitesimal circle
around the pole at $s\\!=\\!-z^{-\beta}\exp(-2li\pi)$. At this stage, we shall
not be concerned with whether we are considering a complete rotation or a
semi-circular rotation around the pole. Nor will we be concerned with the
direction of the indentation. Hence, we shall assume that the infinitesimal
indentation begins at an angle, $\gamma_{1}$ in the complex plane, and ends at
another angle, $\gamma_{2}$. Then we find that the contribution from the pole
from the Cauchy integral in Equivalence (50) is given by

$\displaystyle I^{I}=ip^{-1}(-1)^{N}z^{\beta(N-1)}\lim_{\epsilon\to
0}\int_{\gamma_{1}}^{\gamma_{2}}\left(z^{-\beta}e^{i(2l-1)\pi}\right)^{\\!N+q/p-1}$
(56)
$\displaystyle\left.\times\;\;\exp\\!\left(-z^{-\beta/p}e^{i(2l-1)\pi/p}\right)\;d\gamma=i(-1)^{N}\Delta\gamma\,f(s)\right|_{s=z^{-\beta}\exp((2l-1)i\pi)}\;\;,$

where $\Delta\gamma\\!=\\!\gamma_{2}\\!-\\!\gamma_{1}$. Therefore, we have
seeen that $f(s)$ or Eq. (51) with $s\\!=\\!-z^{-\beta}\exp((2l-1)i\pi)$
represents the residue of the Cauchy integral. Moreover, if
$\Delta\gamma\\!=\\!2\pi$, which corresponds to a complete rotation in a
clockwise direction, then we see that the residue of the Cauchy integral given
by Equivalence (50) is identical to the difference of the regularised values
for $S^{I}_{p,q}(N,z^{\beta}e^{-2li\pi})$ and
$S^{I}_{p,q}(N,z^{\beta}e^{-2(l-1)i\pi})$ or Equivalence (54). Therefore, the
above result confirms the remarkable insight made by Dingle on p. 412 of Ref.
[6] that the jump discontinuity due to crossing Stokes sectors is dependent
upon the singular behaviour of the Cauchy integral that emerges from the
introduction of the regularised value of the geometric series during Borel
summation.

By putting $l\\!=\\!1$ in Equivalence (54), we arrive at

(57) $\displaystyle S^{I}_{p,q}\\!$
$\displaystyle\left(N,z^{\beta}e^{-2i\pi}\right)-S^{I}_{p,q}\\!\left(N,z^{\beta}\right)\equiv\frac{2\pi
i}{p}\;z^{-\beta
q/p}e^{iq\pi/p}\exp\\!\left(-z^{-\beta/p}e^{i\pi/p}\right)\;\;.$

Since the above equivalence is valid only for $-\pi/\beta\\!<\\!{\rm
arg}\,z\\!<\\!\pi/\beta$, we can replace $S^{I}_{p,q}(N,z^{\beta})$ by its
regularised value since the Cauchy integral is also valid over this sector.
Hence, Equivalence (57) becomes

$\displaystyle S^{I}_{p,q}\\!\left(N,z^{\beta}e^{-2i\pi}\right)$
$\displaystyle\equiv(-1)^{N}p^{-1}\,z^{\beta(N-1)}\int_{C}\frac{s^{N+q/p-1}\;e^{-s^{1/p}}}{s-(-z^{-\beta})}\;ds$
(58) $\displaystyle+\;\;\frac{2\pi i}{p}\;z^{-\beta
q/p}e^{iq\pi/p}\exp\\!\left(-z^{-\beta/p}e^{i\pi/p}\right)\;\;.$

Now we replace $z\exp(-2i\pi/\beta)$ by $z_{*}$, which yields

$\displaystyle S^{I}_{p,q}\\!\left(N,z_{*}^{\beta}\right)$
$\displaystyle\equiv(-1)^{N}p^{-1}\,z^{\beta(N-1)}\int_{C}\frac{s^{N+q/p-1}\;e^{-s^{1/p}}}{s+z^{-\beta}}\;ds$
(59) $\displaystyle+\;\;\frac{2\pi i}{p}\;z^{-\beta
q/p}e^{iq\pi/p}\exp\\!\left(-z^{-\beta/p}e^{i\pi/p}\right)\;\;,$

where $-3\pi/\beta\\!<\\!{\rm arg}\,z_{*}\\!<\\!-\pi/\beta$ and
$-\pi/\beta\\!<\\!{\rm arg}\,z\\!<\\!\pi/\beta$. The terms on the rhs of
Equivalence (59) have been left in terms of $z$ to emphasise the fact that
when they are evaluated, they are done so in the primary Stokes sector. That
is, the regularised values of the series on the lhs of Equivalence (59) for
values of $z_{*}$ lying in the sector of $(-3\pi/\beta,-\pi/\beta)$ are
determined by evaluating the terms on the rhs for the corresponding values of
$z_{*}$ lying in the primary Stokes sector. This anomaly arises from the fact
that if software packages such as Mathematica are used to carry out
calculations of the rhs of the above equivalence in determining regularised
value of the Type I generalised terminant outside the primary sector, then
they will only do so for values of the complex variable lying in the principal
branch of the complex plane. That is, in order to obtain regularised value
outside the primary sector we need to evaluate forms where the complex
variable lies inside it.

To make Equivalence (59) appear less awkward, we replace $z_{*}$ and $z$,
respectively by $z$ and $z_{1}$, where the latter is defined as
$z_{1}\\!=\\!z\exp(2i\pi/\beta)$. Then the above equivalence can be re-written
as

$\displaystyle S^{I}_{p,q}\\!\left(N,z^{\beta}\right)$
$\displaystyle\equiv(-1)^{N}p^{-1}\,z_{1}^{\beta(N-1)}\int_{C}\frac{s^{N+q/p-1}\;e^{s^{1/p}}}{s-(-z_{1}^{-\beta})}\;ds$
(60) $\displaystyle+\;\;\frac{2\pi i}{p}\;z_{1}^{-\beta
q/p}e^{iq\pi/p}\exp\\!\left(-z_{1}^{-\beta/p}e^{i\pi/p}\right)\;\;,$

where $-3\pi/\beta\\!<\\!{\rm arg}\,z\\!<\\!-\pi/\beta$. It is this result
that is used to derive the regularised value of $S^{I}_{p,q}(N,z^{\beta})$ for
any Stokes sector as a result of the continuous rotation of $z^{\beta}$ in the
complex plane.

So far, we have only been concerned with Stokes sectors, but the behaviour on
the Stokes lines, where the singularity of the Cauchy integral is situated, is
also important. This issue has been completely overlooked by Paris’s analysis
in Ref. [1]. To obtain the Borel-summed regularised value of
$S^{I}_{p,q}(N,z^{\beta})$ on the Stokes line given by ${\rm
arg}\,z\\!=\\!-\pi/\beta$, we need to invoke Rule 8a presented in Ch. 3 of
Ref. [4]. The rules presented in this reference are virtually a re-expression
of those appearing in Ch. 1 of Dingle’s book [6]. Rule 8a states that the
asymptotic form along a Stokes line is the average of the asymptotic forms in
the adjoining Stokes sectors except that the principal value must be
evaluated. Although there is no statement about the Cauchy principal value in
Dingle’s rules, “Eq.” (46) demonstrates that it is implied. Therefore, the
Borel-summed regularised value of a Type I generalised terminant for ${\rm
arg}\,z\\!=\\!-\pi/\beta$ is given by

$\displaystyle S^{I}_{p,q}\\!\left(N,z\right)$
$\displaystyle\equiv-p^{-1}\,|z|^{\beta(N-1)}P\int_{0}^{\infty}\frac{t^{N+q/p-1}\;e^{-t^{1/p}}}{t-|z|^{-\beta}}\;dt$
(61) $\displaystyle+\;\;\frac{\pi i}{p}\;|z|^{-\beta
q/p}\exp\\!\left(-|z|^{-\beta/p}\right)\;\;.$

From this result we see that the regularised value is composed of the
regularised value in the primary Stokes sector except that now the principal
value of the Cauchy integral must be evaluated and only the semi-residue
contribution is taken in a clockwise direction ($\Delta\gamma\\!=\\!\pi$) of
the $l\\!=\\!1$ version of Eq. (60).

If we put $l\\!=\\!2$ in Equivalence (52), then with the aid of Equivalence
(54) we find that

(62) $\displaystyle S^{I}_{p,q}\\!\left(N,z^{\beta}e^{-4i\pi}\right)$
$\displaystyle-S^{I}_{p,q}\\!\left(N,z^{\beta}e^{-2i\pi}\right)\equiv\frac{2\pi
i}{p}\;z^{-\beta
q/p}e^{3iq\pi/p}\exp\\!\left(-z^{-\beta/p}e^{3iq\pi/p}\right)\;\;.$

We can replace $S^{I}_{p,q}(N,z^{\beta}\exp(-2i\pi))$ by introducing
Equivalence (58) into the above result. This yields

$\displaystyle
S^{I}_{p,q}\\!\left(N,z^{\beta}e^{-4i\pi}\right)\equiv(-1)^{N}p^{-1}\,z^{\beta(N-1)}\int_{C}\frac{s^{N+q/p-1}\;e^{-s^{1/p}}}{s-(-z^{-\beta})}\;ds+\frac{2\pi
i}{p}\;z^{-\beta q/p}$ (63) $\displaystyle\times
e^{3iq\pi/p}\exp\\!\left(-z^{-\beta/p}e^{3iq\pi/p}\right)+\frac{2\pi
i}{p}\;z^{-\beta q/p}e^{iq\pi/p}\exp\\!\left(-z^{-\beta/p}e^{i\pi/p}\right).$

Next we replace $z\exp(-4i\pi/\beta)$ by $z$ on the lhs, while on the rhs $z$
is replaced by $z_{2}$, the latter now being equal to $z\exp(4i\pi/\beta)$.
Hence, we arrive at

$\displaystyle S^{I}_{p,q}\\!\left(N,z^{\beta}\right)$
$\displaystyle\;\equiv(-1)^{N}p^{-1}\,z_{2}^{\beta(N-1)}\int_{C}\frac{s^{N+q/p-1}\;e^{-s^{1/p}}}{s+z_{2}^{-\beta}}\;ds+\frac{2\pi
i}{p}\;z_{2}^{-\beta q/p}$ (64)
$\displaystyle\times\sum_{j=1}^{2}e^{i(2j-1)q\pi/p}\exp\\!\left(-z_{2}^{-\beta/p}e^{i(2j-1)\pi/p}\right)\;\;,$

where $-(2(2)\\!+\\!1)\pi/\beta\\!<\\!{\rm
arg}\,z\\!<\\!-(2(2)\\!-\\!1)\pi/\beta$ or $-5\pi/\beta\\!<\\!{\rm
arg}\,z\\!<\\!-3\pi/\beta$. When $z$ lies on the Stokes line that borders the
$l\\!=\\!1$ and $l\\!=\\!2$ sectors, i.e. where ${\rm
arg}\,z\\!=\\!-3\pi/\beta$, all we need to do is average Equivalences (63) and
(64) and take the Cauchy principal value of the resulting contour integral.
Before we can average the two equivalences we must replace $z_{1}$ in
Equivalence (63) by $z\exp(2i\pi/\beta)$ and $z_{2}$ in Equivalence (64) by
$z\exp(4i\pi/\beta)$ so that $z$ is the same variable in both equivalences.
Then taking the average of the two modified equivalences and setting ${\rm
arg}\,z$ equal to $-3\pi/\beta$, we find that the regularised value of the
Type I generalised terminant is given by

$\displaystyle S^{I}_{p,q}\\!\left(N,z^{\beta}\right)$
$\displaystyle\;\equiv-p^{-1}\,|z|^{\beta(N-1)}P\int_{0}^{\infty}\frac{t^{N+q/p-1}\;e^{-t^{1/p}}}{t-|z|^{-\beta}}\;dt+\frac{2\pi
i}{p}\;|z|^{-\beta q/p}e^{2iq\pi/p}$ (65)
$\displaystyle\times\;\;\exp\\!\left(-|z|^{-\beta/p}e^{2i\pi/p}\right)+\frac{\pi
i}{p}\,|z|^{-\beta q/p}\exp\\!\left(-|z|^{-\beta/p}\right)\;\;.$

A pattern is now emerging that will allow us to determine the regularised
value of a Type I generalised terminant for the Stokes sectors due to
clockwise rotations of $z$. We simply replace the upper limit 2 by $M$ in the
sum on the rhs of Equivalence (64). As a consequence, we find that the
regularised value of $S^{I}_{p,q}(N,z^{\beta})$ for
$-(2M\\!+\\!1)\pi/\beta\\!<\\!{\rm arg}\,z\\!<\\!-(2M\\!-\\!1)\pi/\beta$ is
given by

$\displaystyle S^{I}_{p,q}(N,z^{\beta})$
$\displaystyle\;\equiv(-1)^{N}z_{M}^{\beta(N-1)}p^{-1}\int_{0}^{\infty}\frac{t^{N+q/p-1}\,e^{-t^{1/p}}}{t+z_{M}^{-\beta}}\;dt+\frac{2\pi
i}{p}\;z_{M}^{-\beta q/p}$ (66)
$\displaystyle\times\;\;\sum_{j=1}^{M}\,e^{i(2j-1)q\pi/p}\,\exp\\!\left(-z_{M}^{-\beta/p}e^{i(2j-1)\pi/p}\right)\;\;,$

where $z_{M}\\!=\\!z\exp(2Mi\pi/\beta)$. This equivalence, which appears as
(32) in the beginning of the proof to Thm. 2.1 in Ref. [2], represents the
base or platform for the exactification of Stirling’s approximation for
positive values of ${\rm arg}\,z$. Note that negative values of ${\rm arg}\,z$
correspond to positive values of ${\rm arg}\,z$ in Stirling’s approximation
since the asymptotic series in the latter, viz. $S(z)$, is in inverse powers
of $z$. On the Stokes line given by ${\rm
arg}\,z\\!=\\!-(2M\\!+\\!1)\pi/\beta$, one takes the average of the
regularised values for the two adjacent Stokes sectors bordered by the line
and replaces the Cauchy integral by its principal value. After a little
algebra one finds that the regularised value of the Type I generalised
terminant on the Stokes line simplifies to

$\displaystyle
S^{I}_{p,q}(N,z^{\beta})\equiv-|z|^{\beta(N-1)}p^{-1}\;P\int_{0}^{\infty}\frac{t^{N+q/p-1}\,e^{-t^{1/p}}}{t-|z|^{-\beta}}\;dt+\frac{2\pi
i}{p}\;|z|^{-\beta q/p}$ (67)
$\displaystyle\times\;\;\sum_{j=1}^{M}e^{2ijq\pi/p}\exp\\!\left(-|z|^{-\beta/p}e^{2ij\pi/p}\right)+\frac{\pi
i}{p}\;|z|^{-\beta q/p}\,\exp\\!\left(-|z|^{-\beta/p}\right)\;\;.$

This equivalence appears as (45) in the proof of Thm. 2.1 in Ref. [2].

For negative values of ${\rm arg}\,z^{-\beta}$ in the complex plane, the
rotations of $z^{-\beta}$ are clockwise. To obtain the regularised value of a
generalised Type I terminant for this case, we put $l\\!=\\!0$ in Equivalences
(52)-(54), which yields

$\displaystyle S^{I}_{p,q}\\!$
$\displaystyle\left(N,z^{\beta}e^{2i\pi}\right)-S^{I}_{p,q}\\!\left(N,z^{\beta}\right)\equiv-\frac{2\pi
i}{p}\;z^{-\beta q/p}$ (68) $\displaystyle\hskip
28.45274pt\times\;\;e^{-iq\pi/p}\exp\\!\left(-z^{-\beta/p}e^{-i\pi/p}\right)\;\;.$

Hence, we see that the regularised value for the lower Stokes sector given by
$S^{I}_{p,q}(N,z^{\beta})$ is related to the regularised value for the Stokes
sector immediately above, viz. $S^{I}_{p,q}(N,z^{\beta}$ $\exp(2i\pi))$, plus
the residue contribution of the Cauchy integral taken in a clockwise
direction. In fact, the regularised values of all lower Stokes sectors are
related to the regularised values for the Stokes sectors immediately above
them plus the residue contributions of the Cauchy integral taken in a
clockwise direction. Again, this is consistent with the conventional view of
the Stokes phenomenon. Furthermore, the above result represents the complex
conjugate of Equivalence (57). Since Equivalence (68) is only valid over the
primary Stokes sector, i.e. for $-\pi/\beta\\!<\\!{\rm
arg}\,z\\!<\\!\pi/\beta$, we can introduce the regularised value of
$S^{I}_{p,q}(N,z^{\beta})$ from Equivalence (55) into it. Then by replacing
$z\exp(2i\pi/\beta)$ in the resulting equivalence with $z$, one obtains

$\displaystyle S^{I}_{p,q}\\!\left(N,z\right)$
$\displaystyle\equiv(-1)^{N}p^{-1}\,z_{-1}^{\beta(N-1)}\int_{C}ds\;\frac{s^{N+q/p-1}}{s-(-z_{-1}^{-\beta})}\;e^{-s^{1/p}}$
(69) $\displaystyle-\;\;\frac{2\pi i}{p}\;z_{-1}^{-\beta
q/p}\,e^{-iq\pi/p}\exp\\!\left(-z_{-1}^{-\beta/p}e^{-i\pi/p}\right)\;\;,$

where $\pi/\beta\\!<\\!{\rm arg}\,z\\!<\\!3\pi/\beta$ and
$z_{-1}\\!=\\!z\exp(-2i\pi/\beta)$.

For the Stokes line of ${\rm arg}\,z\\!=\\!\pi/\beta$, we first re-write the
rhs of Equivalence (69) in terms of $z$. Then we average the resulting
equivalence with Equivalence (55). Next we take the Cauchy principal value of
the resulting integral in accordance with Rule 8a in Ch. 3 of Ref. [4].
Finally, we replace $z$ by $|z|\exp(i\pi/\beta)$. Thus, we arrive at the
Borel-summed regularised value of a Type I generalised terminant for ${\rm
arg}\,z\\!=\\!\pi/\beta$, which is

$\displaystyle S^{I}_{p,q}\\!\left(N,z\right)$
$\displaystyle\equiv-p^{-1}\,|z|^{\beta(N-1)}P\int_{0}^{\infty}\frac{t^{N+q/p-1}\;e^{-t^{1/p}}}{t-|z|^{-\beta}}\;dt$
(70) $\displaystyle-\;\;\frac{\pi i}{p}\;|z|^{-\beta
q/p}\exp\\!\left(-|z|^{-\beta/p}\right)\;\;.$

When $l\\!=\\!-1$, the combination of Equivalences (52) and (54) yields

$\displaystyle S^{I}_{p,q}\\!$
$\displaystyle\left(N,z^{\beta}e^{4i\pi}\right)-S^{I}_{p,q}\\!\left(N,z^{\beta}e^{2i\pi}\right)\equiv-\frac{2\pi
i}{p}\;z^{-\beta q/p}$ (71) $\displaystyle\hskip
28.45274pt\times\;\;e^{-3iq\pi/p}\exp\\!\left(-z^{-\beta/p}e^{-3i\pi/p}\right)\;\;.$

We can express $S^{I}_{p,q}(N,z^{\beta}\exp(2i\pi))$ in terms of
$S^{I}_{p,q}(N,z^{\beta})$ by putting $l\\!=\\!0$ in Equivalences (52) and
(54). This gives

$\displaystyle S^{I}_{p,q}\\!\left(N,z^{\beta}e^{4i\pi}\right)\equiv
S^{I}_{p,q}\\!\left(N,z^{\beta}\right)-\frac{2\pi i}{p}\;z^{-\beta q/p}$ (72)
$\displaystyle\times\;\;\sum_{j=1}^{2}e^{-i(2j-1)q\pi/p}\exp\\!\left(-z^{-\beta/p}e^{-i(2j-1)\pi/p}\right)\;\;,$

where $(2(2)\\!-\\!1)\pi/\beta\\!<\\!{\rm
arg}\,z\\!<\\!(2(2)\\!+\\!1)\pi/\beta$. Setting $z$ equal to
$z\exp(-4i\pi/\beta)$ and introducing the appropriate version of Equivalence
(50), one eventually obtains

$\displaystyle S^{I}_{p,q}\\!\left(N,z^{\beta}\right)$
$\displaystyle\;\equiv(-1)^{N}p^{-1}\,z_{-2}^{\beta(N-1)}\int_{C}\frac{s^{N+q/p-1}\;e^{-s^{1/p}}}{s+z_{-2}^{-\beta}}\;ds-\frac{2\pi
i}{p}\;z_{-2}^{-\beta q/p}$ (73)
$\displaystyle\times\sum_{j=1}^{2}e^{-i(2j-1)q\pi/p}\exp\\!\left(-z_{-2}^{-\beta/p}e^{-i(2j-1)q\pi/p}\right)\;\;,$

which, as expected, is the complex conjugate of Equivalence (64). For ${\rm
arg}\,z\\!=\\!3\pi/\beta$, we first express the rhs’s of Equivalences (69) and
(73) in terms of $z$ rather than $z_{-1}$ and $z_{-2}$. Then we average the
resulting equivalences and evaluate the Cauchy principal value of the
resulting contour integral. Alternatively, we can take the complex conjugate
of Equivalence (65). Hence, we find that

$\displaystyle
S^{I}_{p,q}\\!\left(N,z^{\beta}\right)\equiv-p^{-1}\,|z|^{\beta(N-1)}P\int_{0}^{\infty}\frac{t^{N+q/p-1}\;e^{-t^{1/p}}}{t-|z|^{-\beta}}\;dt-\frac{2\pi
i}{p}\;|z|^{-\beta q/p}$ (74)
$\displaystyle\times\;\;e^{-2iq\pi/p}\exp\\!\left(-|z|^{-\beta/p}e^{-2i\pi/p}\right)-\frac{\pi
i}{p}\,|z|^{-\beta q/p}\exp\\!\left(-|z|^{-\beta/p}\right)\;\;.$

From the above we see that a similar pattern is emerging as in the case of the
anti-clockwise rotations of $z^{-\beta}$. Therefore, by extending Equivalences
(73) and (74) we are able to derive general forms for the Borel-summed
regularised values of a Type I generalised terminant for any Stokes sector or
line, where ${\rm arg}\,z\\!>\\!0$. In particular, the generalisation of
Equivalence (73) to $(2M\\!-\\!1)\pi/\beta\\!<\\!{\rm
arg}\,z\\!<\\!(2M\\!+\\!1)\pi/\beta$, where $M\\!>\\!0$, can be carried out
simply by replacing 2 with $M$. Thus, we arrive at

$\displaystyle
S^{I}_{p,q}(N,z^{\beta})\equiv(-1)^{N}z_{-M}^{\beta(N-1)}\,p^{-1}\int_{0}^{\infty}\frac{t^{N+q/p-1}\,e^{-t^{1/p}}}{t+z_{-M}^{-\beta}}\;dt-\frac{2\pi
i}{p}\;z_{-M}^{-\beta q/p}$ (75)
$\displaystyle\;\;\times\;\;\sum_{j=1}^{M}e^{-i(2j-1)q\pi/p}\,\exp\\!\left(-z_{-M}^{-\beta/p}e^{-i(2j-1)\pi/p}\right)\;\;.$

In the above result $z_{-M}\\!=\\!z\exp(-2Mi\pi/\beta)$. Hence, we see that
Equivalence (75) represents the complex conjugate of Equivalence (66). This is
essentially the equivalence given as (41) in the proof of Thm. 2.1 of Ref.
[2]. For the Stokes line where ${\rm arg}\,z\\!=\\!(2M\\!+\\!1)\pi/\beta$, the
generalisation of Equivalence (74) yields

$\displaystyle S^{I}_{p,q}$
$\displaystyle(N,z^{\beta})\equiv-|z|^{\beta(N-1)}\,p^{-1}P\int_{0}^{\infty}\frac{t^{N+q/p-1}\,e^{-t^{1/p}}}{t-|z|^{-\beta}}\;dt-\frac{2\pi
i}{p}\;|z|^{-\beta q/p}\sum_{j=1}^{M}e^{-2ijq\pi/p}$ (76)
$\displaystyle\;\;\times\;\;\exp\\!\left(-|z|^{-\beta/p}e^{-2ij\pi/p}\right)-\frac{\pi
i}{p}\;|z|^{-\beta q/p}\,\exp\\!\left(-|z|^{-\beta/p}\right)\;\;,$

which represents the complex conjugate of Equivalence (67). This equivalence
appears as the lower-signed version of (45) in the proof of Thm. 2.1 of Ref.
[2].

At no stage in the preceding analysis has there been a requirement to
introduce the concept of smoothing in the vicinity of a Stokes line. To
observe the behaviour near a Stokes line, let us consider the vicinity of the
line where ${\rm arg}\,z=-(2M+1)\pi/\beta$. For $-(2M+1)\pi/\beta<{\rm
arg}\,z<-(2M-1)\pi/\beta$, Equivalence (66) yields

$\displaystyle S^{I}_{p,q}(N,z^{\beta})$
$\displaystyle\;\equiv(-1)^{N}z^{\beta(N-1)}p^{-1}\int_{0}^{\infty}\frac{t^{N+q/p-1}\,e^{-t^{1/p}}}{t+z^{-\beta}}\;dt+\frac{2\pi
i}{p}\;z^{-\beta q/p}$ (77)
$\displaystyle\times\;\;\sum_{j=0}^{M-1}\,e^{(2j+1)iq\pi/p}\,\exp\\!\left(-z^{-\beta/p}e^{(2j+1)i\pi/p}\right)\;\;,$

while for the next lower Stokes sector or $-(2M+3)\pi/\beta<{\rm
arg}\,z<-(2M+1)\pi/\beta$, it gives

$\displaystyle S^{I}_{p,q}(N,z^{\beta})$
$\displaystyle\;\equiv(-1)^{N}z^{\beta(N-1)}p^{-1}\int_{0}^{\infty}\frac{t^{N+q/p-1}\,e^{-t^{1/p}}}{t+z^{-\beta}}\;dt+\frac{2\pi
i}{p}\;z^{-\beta q/p}$ (78)
$\displaystyle\times\;\;\sum_{j=0}^{M}\,e^{(2j+1)iq\pi/p}\,\exp\\!\left(-z^{-\beta/p}e^{(2j+1)i\pi/p}\right)\;\;.$

For ${\rm arg}\,z=-(2M+1)\pi/\beta$, Equivalence (67) yields

$\displaystyle S^{I}_{p,q}(N,z^{\beta})$
$\displaystyle\equiv(-1)^{N}z^{\beta(N-1)}p^{-1}\;P\int_{0}^{\infty}\frac{t^{N+q/p-1}\,e^{-t^{1/p}}}{t+z^{-\beta}}\;dt+\frac{2\pi
i}{p}\;|z|^{-\beta q/p}$
$\displaystyle\times\;\;\sum_{j=0}^{M-1}e^{-(2j+1)iq\pi/p}\exp\\!\left(-z^{-\beta/p}e^{-(2j+1)i\pi/p}\right)+\frac{\pi
i}{p}\;z^{-\beta q/p}\,e^{-(2M+1)iq\pi/p)}\,$ (79)
$\displaystyle\times\;\;\exp\\!\left(-z^{-\beta/p}e^{-(2M+1)i\pi/p}\right)\;\;.$

If we let

$\displaystyle F(z)$
$\displaystyle=(-1)^{N}z^{\beta(N-1)}p^{-1}\int_{0}^{\infty}\frac{t^{N+q/p-1}\,e^{-t^{1/p}}}{t+z^{-\beta}}\;dt+\frac{2\pi
i}{p}\;z^{-\beta q/p}$ (80)
$\displaystyle\times\;\;\sum_{j=0}^{M-1}\,e^{(2j+1)iq\pi/p}\,\exp\\!\left(-z^{-\beta/p}e^{(2j+1)i\pi/p}\right)\;\;,$

where it is understood that the Cauchy principal of integral is evaluated on
the Stokes line, then Equivalences (77)-(79) can be expressed as

(81) $\displaystyle S^{I}_{p,q}(N,z^{\beta})\equiv F(z)+2\pi iSz^{-\beta
q/p}\,e^{-(2M+1)iq\pi/p)}\,\exp\\!\left(-z^{-\beta/p}e^{-(2M+1)i\pi/p}\right)\;\;,$

with the multiplier $S$ of the subdominant exponential term being given by

(82) $\displaystyle S=\begin{cases}0\;\;,\quad-(2M+1)\pi/\beta<{\rm
arg}\,z<-(2M-1)\pi/\beta\;\;,\cr 1/2\;\;,\quad{\rm
arg}\,z=-(2M+1)\pi/\beta\;\;,\cr 1\;\;,\quad-(2M+3)\pi/\beta<{\rm
arg}\,z<-(2M+1)\pi/\beta\;\;.\end{cases}$

This is simply the conventional view as described by Berry in Ref. [15] and
presented at the beginning of this section. In this instance it is clear that
$S_{-}=0$ for a Type I generalised terminant when the primary Stokes sector is
given by $|{\rm arg}\,z|<\pi/\beta$.

Now let us turn our attention to the asymptotic series $S(z)$ appearing in the
complete version of Stirling’s “approximation” for $\ln\Gamma(z)$. This series
in its entirety, i.e. without any truncation whatsoever, is found in Ref. [2]
to be

(83) $\displaystyle
S(z)=z\sum_{k=1}^{\infty}\frac{(-1)^{k}}{(2z)^{2k}}\,\Gamma(2k-1)\,c_{k}(1)\;,$

where the cosecant polynomials for unity [8] can be expressed in terms of the
Bernoulli numbers as

(84) $\displaystyle c_{k}(1)=\frac{(-1)^{k}}{(2k)!}\;2^{2k}\,B_{2k}\;.$

Later in Ref. [2] the truncation parameter $N$ is introduced with the
Dirichlet series form for the Riemann zeta function. Hence, the series becomes

(85) $\displaystyle
S(z)=z\sum_{k=1}^{N-1}\frac{(-1)^{k}}{(2z)^{2k}}\,\Gamma(2k-1)\,c_{k}(1)-2z\sum_{n=1}^{\infty}S^{I}_{2,-1}\left(N,(1/2n\pi
z)^{2}\right)\;.$

The first term on the rhs of Eq. (85) is basically Paris’s $\Omega(z)$, while
the second term is the frequently neglected remainder that is either divergent
or conditionally convergent. Moreover, we see that the second term represents
an infinite sum of Type I generalised terminants where $p\\!=\\!2$,
$q\\!=\\!v-1$, $z\\!=\\!1/2n\pi z$ and $\beta\\!=\\!2$. By introducing these
values into Equivalence (77), one obtains

$\displaystyle S^{I}_{2,-1}\left(N,(1/2n\pi z)^{2}\right)$
$\displaystyle\equiv\frac{(-1)^{N}}{(2n\pi
z)^{2N-2}}\int_{0}^{\infty}\frac{y^{2N-2}\,e^{-y}}{y^{2}+4n^{2}\pi^{2}z^{2}}\;dy-\frac{1}{2nz}$
(86)
$\displaystyle\times\;\;\sum_{j=1}^{M}(-1)^{M-j}\,\exp\Bigl{(}-2(-1)^{M-j}ni\pi
z\Bigr{)}\;.$

Consequently, the asymptotic series for $\ln\Gamma(z)$ becomes

$\displaystyle S(z)$ $\displaystyle\equiv
z\sum_{k=1}^{N-1}\frac{(-1)^{k}}{(2z)^{2k}}\,\Gamma(2k-1)\,c_{k}(1)-2\,\Bigl{(}-\frac{1}{4\pi^{2}z^{2}}\Bigr{)}^{N}\,z\sum_{n=1}^{\infty}\frac{1}{n^{2N-2}}$
(87) $\displaystyle\times\;\;\int_{0}^{\infty}\frac{e^{-y}\,y^{2N-2}}{((y/2\pi
z)^{2}+n^{2})}\;dy+\sum_{n=1}^{\infty}\frac{1}{n}\sum_{j=1}^{M}(-1)^{M-j}\exp\left(2(-1)^{M-j}\,ni\pi
z\right)\;.$

Equivalences (86) and (87) appear respectively as (33) and (34) in Ref. [2].
For those Stokes lines where ${\rm arg}\,(1/2n\pi z)=-(M+1/2)\pi$ or
$\theta={\rm arg}\,z=(M+1/2)\pi$, the introduction of Equivalence (79) into
Eq. (85) yields

$\displaystyle S(z)$ $\displaystyle\equiv
z\sum_{k=1}^{N-1}\frac{(-1)^{k}}{(2z)^{2k}}\,\Gamma(2k-1)\,c_{k}(1)+2\Bigl{(}\frac{1}{4\pi^{2}|z|^{2}}\Bigr{)}^{N-1}z\;\sum_{n=1}^{\infty}\frac{1}{n^{2N-2}}\;$
$\displaystyle\times\;\;P\int_{0}^{\infty}\frac{e^{-y}\,y^{2N-2}}{y^{2}-4n^{2}\pi^{2}|z|^{2}}\;dy-
ie^{i\theta}\sum_{n=1}^{\infty}\frac{1}{n}\sum_{j=1}^{M}(-1)^{j}\exp\left(2(-1)^{j}\,n\pi|z|\right)$
(88)
$\displaystyle-\;\;ie^{i\theta}\sum_{n=1}^{\infty}\frac{1}{2n}\exp\left(-2n\pi|z|\right)\;.$

Since each generalised Type I terminant can be expressed in terms of a Stokes
multiplier accompanied by an exponential term that is subdominant in the
vicinity of a Stokes line, it follows that $S(z)$ can be expressed in terms of
such a multiplier and an infinite series of subdominant exponential terms as
given by the last term on the rhs of the above equivalence. If we let

$\displaystyle G(z)$
$\displaystyle=z\sum_{k=1}^{N-1}\frac{(-1)^{k}}{(2z)^{2k}}\,\Gamma(2k-1)\,c_{k}(1)-2\,\Bigl{(}-\frac{1}{4\pi^{2}z^{2}}\Bigr{)}^{N}\,z\sum_{n=1}^{\infty}\frac{1}{n^{2N-2}}$
(89) $\displaystyle\times\;\;\int_{0}^{\infty}\frac{e^{-y}\,y^{2N-2}}{((y/2\pi
z)^{2}+n^{2})}\;dy+\sum_{n=1}^{\infty}\frac{1}{n}\sum_{j=1}^{M}(-1)^{M-j}\exp\left(2(-1)^{M-j}\,ni\pi
z\right)\;,$

where it is understood that the Cauchy principal value is evaluated at the
Stokes line, then $S(z)$ can be written as

(90) $\displaystyle S(z)\equiv
G(z)+(-1)^{M}S\sum_{n=1}^{\infty}\frac{e^{2(-1)^{M}ni\pi z}}{n}\;\;,$

with the Stokes multiplier given by Eq. (82). The above result is only
partially regularised because the series on the rhs, which arises from the
infinite number of singularities situated on the Stokes line, can become
divergent. However, the series can be regularised by using Lemma 2.2 in Ref.
[2], which proves that

(91)
$\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}\;z^{k}\begin{cases}\equiv\ln(1+z)\quad,&\quad\Re\,z\leq-1\quad,\cr=\ln(1+z)\quad,&\quad\Re\,z>-1\quad.\end{cases}$

Then Equivalence (90) becomes

(92) $\displaystyle S(z)\equiv G(z)+(-1)^{M+1}S\ln\Bigl{(}1-e^{2(-1)^{M}\pi
iz}\Bigr{)}\;\;.$

For the Stokes line where $M\\!=\\!0$, viz. ${\rm arg}\,z\\!=\\!\pi/2$, the
above result reduces to

(93) $\displaystyle S(z)\equiv G(z)-\frac{1}{2}\,\ln\Bigl{(}1-e^{2\pi
iz}\Bigr{)}\;\;.$

The second term on the rhs of both the above equivalences is referred to as
the Stokes discontinuity term, denoted by $SD^{+}(z)$, in Ref. [2] and appears
as Eq. (78) there. It is subsequently used in the numerical study of the
Stokes phenomenon in Sec. 3. It is also the expression which Paris claims is
not the correct interpretation of the Stokes phenomenon. Yes, it is not the
interpretation according to the smoothing view espoused by Berry and Olver,
but it is, nonetheless, totally consistent with the conventional view of the
Stokes phenomenon as stated in Ref. [2].

To summarise the preceding analysis, we have seen that the subdominant
exponential term in the Borel-summed regularised value of a generalised
terminant emerges as a result of a singularity being situated on a Stokes
line. In the case of the asymptotic series $S(z)$ for $\ln\Gamma(z)$, there is
an infinite number of generalised terminants with coinciding Stokes lines. As
a consequence, each Stokes line for $S(z)$ possesses an infinite number of
singularities, albeit located at different positions. This produces an
infinite number of subdominant exponential terms, each accompanied by the same
Stokes multiplier. The multiplier is entirely consistent with the conventional
view of the Stokes phenomenon. The interesting property of the resulting sum
of the residues due to the singularities is that it need not necessarily be
absolutely convergent and thus, may require regularisation. It is for this
reason that the exactification of Stirling’s approximation for $\ln\Gamma(z)$
is a challenging problem in asymptotics, which is discussed in the
introduction of Ref. [2]. Despite this, no one can be certain that the above
analysis is indeed correct until an effective numerical study has been carried
out. After all, we have seen that Olver claimed that his derivation of a
smoothed Stokes multiplier, viz. “Eq.” (39), was based on rigorous
mathematics. Yet he did not provide an indication as to just how accurate his
result is and for what range of values it is valid. In actual fact, the notion
of a proof does not really apply here because as we shall soon see, we are
talking about encountering discontinuities arising from singularities in
Cauchy integrals and introducing an approach or method to handle them as they
occur.

If anyone possesses an incorrect interpretation of the Stokes multiplier, then
it is surely Paris himself, who seems to believe that the Stokes multiplier
should only multiply the leading subdominant exponential in a complete
asymptotic expansion. Neither Stokes nor Dingle were concerned with the
leading exponential in a complete asymptotic expansion, although it must be
said that they did not study situations where a Stokes line possesses an
infinite number of singularities. In addition, the Berry/Olver derivation of
Stokes smoothing presented earlier does not isolate the leading exponential in
the subdominant part of an asymptotic solution at the expense of the other
terms. Despite this, it is not incorrect to isolate or factor out the leading
exponential from all the other subdominant exponential terms and then to refer
to all the remaining terms as a multiplier, even though it is not what Dingle
and Berry originally had in mind.

The problem with Paris’s approach like that of Berry and Olver is that in
order to come up with improved results on standard Poincar$\acute{{\rm e}}$
asymptotics is that they have been forced to truncate at some stage in their
analyses. As a result, we see the standard asymptotic constructs such as the
$\sim$ symbol and $+\cdots$, appear in Paris’s results, which are given by
(3.1) to (3.4) in Ref. [1]. Thus, Paris’s version of the Stokes multiplier is
given as

(94) $\displaystyle S(\theta)\sim\frac{1}{2}+\frac{1}{2}\,{\rm
erf}\Bigl{(}c\left(\theta+\pi/2\right)\sqrt{\pi|z|}\,\Bigr{)}-\frac{C_{0}e^{-2\pi\gamma|z|}}{2\pi\sqrt{z}}\;i\;\;,$

where $\gamma=1+i\,\exp(i\theta)$, $C_{0}=B_{0}\exp(-2\pi
i\omega|z|)+\exp(i\omega\nu)/(1+\exp(-i\omega))$,

(95) $\displaystyle
B_{0}=\frac{e^{-i\omega\alpha}}{1-e^{-i\omega}}+\frac{1}{c(\theta+\pi/2)}\;i\;\;,$

$\alpha=2N_{0}-1-2\pi|z|$, $\nu=2N_{0}-1$, $N_{0}$ is the optimal point of
truncation and

(96) $\displaystyle
c(\theta+\pi/2)=\omega+\frac{1}{6}\,i\omega^{2}-\frac{1}{36}\,\omega^{3}+\frac{1}{270}\,i\omega^{4}+\cdots\;\;.$

Consequently, the remainder is expressed as

(97) $\displaystyle R_{N_{0}}(z)\approx e^{2\pi iz}\,T_{\nu}(2\pi iz)-e^{-2\pi
iz}T_{\nu}(-2\pi iz)\sim e^{2\pi iz}S(\theta)\;\;.$

In essence, Paris’s “refined version” of $S(\theta)$ is no different from the
result given by (82) in Ref. [2] except that there is now an extra imaginary
term. This term was neglected in Ref. [2] as an unnecessary complication since
it is not possible to obtain hyperasymptotic values of the Stokes multiplier,
e.g. to 30 figures, near the Stokes line at ${\rm arg}\,z\\!=\\!\pi/2$,
anyway. In fact, the imaginary part raises another problem because imaginary
parts for the multiplier simply do not appear in the conventional view of the
Stokes phenomenon. So, if the smoothing view is correct, then it means that
extra terms need to be included in the conventional view in order to obtain
exact values of $\ln\Gamma(z)$ from its asymptotic forms. Note that since this
analysis employs symbols such as $\sim$ and $+\cdots$, it results in the usual
drawbacks associated with standard Poincar$\acute{\rm e}$ asymptotics, viz.
vagueness and limited range of applicability. As discussed in the prologue to
Ref. [6], these two drawbacks are responsible for giving the discipline a bad
name. On the other hand, none of these symbols has been introduced in the
derivation of the Borel-summed and MB-regularised asymptotic forms given in
Ref. [2].

The whole point about the numerical study in Ref. [2] is to determine those
values of ${\rm arg}\,z$ near the Stokes line, where Paris’s form for the
Stokes multiplier deviates the most from the conventional view. Specifically,
these values are determined by plotting the real part of Approximation (94) as
in Figs. 1 of Refs. [1] and [2]. According to Berry, Olver and Paris, one
should not obtain exact values of $\ln\Gamma(z)$ for these values of ${\rm
arg}\,z$ via the conventional view of the Stokes phenomenon. As can be seen
from the figures, the smoothed Stokes multiplier as given by Approximation
(94) is closer to 1/2 than being close to 0 or 1 as dictated by the
conventional view. Hence, it stands to reason that if Stokes smoothing is
correct, then the conventional view cannot possibly give accurate values of
$\ln\Gamma(z)$ in the vicinity of the Stokes line at ${\rm
arg}\,z\\!=\\!\pi/2$.

Table 1 here or Table 7 in Ref. [2] presents the results obtained from
programming the Borel-summed regularised results for $\ln\Gamma(z)$ as a
Mathematica module [22], which appear as Program 2 in the appendix of Ref.
[2]. Specifically, the module calculates $\ln\Gamma(z)$ using the following
asymptotic forms:

(98)
$\displaystyle\ln\Gamma(z)=\begin{cases}F(z)+TS_{N}(z)+R^{SS}_{N}(z)+SD^{SS,U}_{1}(z)\;,\quad&\pi/2<\theta\leq\pi\;,\cr
F(z)+TS_{N}(z)+R^{SS}_{N}(z)\;,\quad&-\pi/2<\theta<\pi/2\;.\cr\end{cases}$

In the above equation $F(z)$ represents the standard form of Stirling’s
approximation, i.e.,

(99) $\displaystyle F(z)=\Bigl{(}z-\frac{1}{2}\Bigr{)}\ln
z-z+\frac{1}{2}\,\ln(2\pi)\;,$

and the truncated series denoted by $TS_{N}(z)$ is given by

(100) $\displaystyle
TS_{N}(z)=z\sum_{k=1}^{N-1}\frac{(-1)^{k}}{(2z)^{2k}}\;\Gamma(2k-1)\,c_{k}(1)\;.$

Eq. (100) is basically Paris’s $\Omega(z)$. In addition, the Borel-summed
remainder term denoted by $R^{SS}_{N}(z)$ was evaluated via

(101) $\displaystyle R^{SS}_{N}(z)$
$\displaystyle=\frac{2\,(-1)^{N+1}\,z}{(2\pi
z)^{2N-2}}\sum_{n=1}^{\infty}\frac{1}{n^{2N-2}}\int_{0}^{\infty}dy\;\frac{y^{2N-2}\,e^{-y}}{\left(y^{2}+4\pi^{2}n^{2}z^{2}\right)}\;,$

while the final term $SD^{SS,U}_{1}(z)$ is the Stokes discontinuity term given
by Eq. (93). That is,

(102) $\displaystyle SD^{SS}_{1}(z)=-\ln\Bigl{(}1-e^{2\pi zi}\Bigr{)}\;.$

$\delta$ | Method | Value
---|---|---
1/10 | LogGamma[z] |  -5.1085546405054331385771175 - 2.43504864133618239587613036$\,i$
| $SD^{SS,U}_{1}(z)$ |  0.0000000146924137960847328 + 0.00000000724920978735477097$\,i$
| Top |  -5.1085546405054331385771175 - 2.43504864133618239587613036$\,i$
-1/10 | LogGamma[z] |  -3.1156770612855851062960250 + 0.79152717486178700663566144$\,i$
| Bottom |  -3.1156770612855851062960250 + 0.79152717486178700663566144$\,i$
1/100 | LogGamma[z] |  -4.4448078360199294879676721 - 0.68426539470619315579497619$\,i$
| $SD^{SS,U}_{1}(z)$ |  0.0000000054543808883397577 - 0.00000000366845661861183983$\,i$
| Top |  -4.4448078360199294879676721 - 0.68426539470619315579497619$\,i$
-1/100 | LogGamma[z] |  -4.2360547825638102221663061 - 0.35681003461125834209091866$\,i$
| Bottom |  -4.2360547825638102221663061 - 0.35681003461125834209091866$\,i$
1/1000 | LogGamma[z] |  -4.3531757575591613140088085 - 0.53385166100905755261595669$\,i$
| $SD^{SS,U}_{1}(z)$ |  0.0000000065016016472424544 - 0.00000000038545945628149871$\,i$
| Top |  -4.3531757575591613140088085 - 0.53385166100905755261595669$\,i$
-1/1000 | LogGamma[z] |  -4.3322909095906129602545969 - 0.50110130347126170951651903$\,i$
| Bottom |  -4.3322909095906129602545969 - 0.50110130347126170951651903$\,i$
1/10000 | LogGamma[z] |  -4.3438006028809735966127763 - 0.51908338527968766540121412$\,i$
| $SD^{SS,U}_{1}(z)$ |  0.0000000065123040290213875 - 0.00000000003856476898298508$\,i$
| Top |  -4.3438006028809735966127763 - 0.51908338527968766540121412$\,i$
-1/10000 | LogGamma[z] |  -4.3417121085407199183370966 - 0.51580834470414165478538635$\,i$
| Bottom |  -4.3417121085407199183370966 - 0.51580834470414165478538635$\,i$
1/20000 | LogGamma[z] |  -4.3438006028809735966127763 - 0.51908338527968766540121412$\,i$
| $SD^{SS,U}_{1}(z)$ |  0.0000000065123851251757157 - 0.00000000001928245580002624$\,i$
| Top |  -4.3438006028809735966127763 - 0.51908338527968766540121412$\,i$
-1/20000 | LogGamma[z] |  -4.3422344065179726897501879 - 0.51662687288967352139359494$\,i$
| Bottom |  -4.3422344065179726897501879 - 0.51662687288967352139359494$\,i$
Table 1. Evaluation of $\ln\Gamma\\!\left(3\exp(i(1/2+\delta)\pi)\right)$ via
Eq. (70) for various values of $\delta$

In order to be consistent with the “Stokes smoothing” view, a relatively large
value of $|z|$ was chosen to execute the module, viz. $|z|\\!=\\!3$, which has
an optimal point of truncation, $N_{0}$, approximately equal to 10. This,
however, leads to another problem. At the present stage one does not know what
form the smoothed multiplier takes for small values of $|z|$. Hence, for small
values of $z$, one cannot possibly obtain exact values of $\ln\Gamma(z)$
according to the smoothing view. Presumably, small $|z|$ implies that there is
no optimal point of truncation, but there is no quantification from Paris on
this issue. Nevertheless, no such restriction applies to any of the Borel-
summed and MB-regularised asymptotic forms given here or in Ref. [2]. That is,
they are equally valid for $|z|>1$ and for $|z|\leq 1$. In addition, it should
be stressed that Table 1 only presents a small sample of the results from the
numerical investigation, in which numerous values of $\delta$, where
$\theta\\!=\\!(1/2\\!+\\!\delta)\pi$, were considered. The values of $\delta$
in the table were chosen because Stokes smoothing is expected to exhibit the
greatest deviation from the step-function postulated in the conventional view
of the Stokes phenomenon. Those very close to the Stokes line are given by
$|\delta|\leq 1/100$ in the table. In particular, those just below the Stokes
line correspond to ${\rm arg}\,z$ equal to $0.49\pi$, $0.499\pi$, $0.4999\pi$
and $0.49995\pi$, while for those just above it correspond to ${\rm arg}\,z$
equal to $0.51\pi$, $0.501\pi$, $0.5001\pi$ and $0.50005\pi$. Besides using
different values of $\delta$ in the study, the module was also run for
numerous values of the truncation parameter $N$.

For each positive value of $\delta$ in Table 1 there are three rows of values,
while for each negative value there are only two rows. This is because the
Stokes discontinuity term is zero for negative values of $\delta$ as indicated
by Eq. (98). The first row for each value of $\delta$ represents the value
obtained by using the LogGamma routine in Mathematica and is denoted by the
row with LogGamma[z] in the Method column. Depending upon whether $\delta$ is
positive or not, the second row presents the Stokes discontinuity term
according to the conventional view of the Stokes phenomenon. In general, this
term was found to possess real and imaginary parts of the order of $10^{-8}$
or a couple of orders lower. That is, the Stokes discontinuity term is very
small and would be neglected in standard Poincar$\acute{{\rm e}}$ asymptotics,
which is due to choosing a relatively large value of $|z|$. Even though the
Stokes discontinuity term is small, it is still necessary in order to give the
correct values of $\ln\Gamma(z)$ for the hyperasymptotic calculation to thirty
figures. The next value for each value of $\delta$ is labelled either Top or
Bottom in the Method column corresponding to whether the top or bottom
asymptotic form in Eq. (98) has been used to calculate $\ln\Gamma(z)$. It
should also be noted that the values of the truncated sum, the regularised
value of the remainder and the Stirling approximation were all evaluated in
the Mathematica module separately, but are not displayed here due to limited
space.

All the calculations carried out in the study yielded the value of
$\ln\Gamma(z)$ to the hyperasymptotic accuracy as indicated in the table
except when $|\delta|$ was extremely small, e.g. for $\delta\leq 10^{-5}$.
Then the NIntegrate routine in Mathematica experiences convergence problems
because the numerical integration of the remainder $R_{N}^{SS}(z)$ is too
close to the singularities lying on the Stokes line. For example, when
$\delta=10^{-5}$, the module prints out a value of $\ln\Gamma(z)$ that agrees
with the actual value to 25 decimal places for the real part, but in the case
of the imaginary part the results only agree to 18 decimal places. The module,
however, does alert the user of the convergence problems that the NIntegrate
routine experiences. Although this calculation is not presented in the table,
it still represents a degree of success since the imaginary part of the Stokes
discontinuity term is of the order of $10^{-12}$. That is, the Stokes
discontinuity term had to be correct to the first six decimal places in order
to yield the value of the imaginary part of $\ln\Gamma(z)$ for this very small
value of $\delta$.

With the exception of the first value of $\delta$, which reflects the
situation as the error function begins to veer away from the step-function of
the conventional view, we expect for all other values of $\delta$ that the
real part of the smoothed Stokes multiplier given by Approximation (94) to be
close to 1/2 according to Figs. 1 in Refs. [1] and [2]. Note that Fig. 1 in
Ref. [1] gives the Stokes multiplier for $|z|=8$, which has an optimal point
of truncation that is approximately equal to 26. As a consequence, the
transition is far more rapid than in Fig. 1 of Ref. [2]. Whilst Paris presents
values of the Stokes multiplier for ${\rm arg}\,z$ ranging form 0.325 to 0.750
at intervals of 0.25, he does not give the values for the Stokes multiplier in
the vicinity of the Stokes line as in Table 1. Nevertheless, for ${\rm
arg}\,z=0.475$, he obtains a value of $0.2894310-0.0182669\,i$ for his
smoothed multiplier, while for ${\rm arg}\,z=0.525$, he obtains a value of
$0.7105689-0.0182669\,i$. Although these are not representative of the
situation in Table 1 and moreover, are nowhere near the accuracy needed to
conduct the hyperasymptotic investigation in Table 1, they indicate that the
Stokes discontinuity term should be at least 30 percent less than the figures
in the table for $\delta>0$ and that thirty percent of the Stokes
discontinuity needs to be added to the values in which $\delta<0$ according to
the smoothing view of the Stokes phenomenon. That is, the top asymptotic form
in Eq. (98) with about half the Stokes discontinuity term should be a far more
accurate approximation to the actual value of $\ln\Gamma(z)$. However, we see
the opposite. The first asymptotic form yields the exact value of
$\ln\Gamma(z)$ for all values of $\delta$ greater than zero despite the fact
that the Stokes discontinuity term has no effect on the first nine decimal
places. For $\delta<0$, according to the smoothing view the bottom asymptotic
form in Eq. (98) should not yield exact values of $\ln\Gamma(z)$ because it is
missing about half the Stokes discontinuity term. Once again, we observe the
opposite; the bottom asymptotic form yields exact values of $\ln\Gamma(z)$ for
all negative values of $\delta$ in the table. Thus, it is evident that there
is no smoothing of the Stokes phenomenon occurring in the vicinity of the
Stokes line at $\theta=\pi/2$ or else it would not been possible to give the
exact values of $\ln\Gamma(z)$ from Eq. (98).

From the results in the table we see that by using the conventional view of
the Stokes multiplier we are able to obtain thirty figure accuracy for
$\ln\Gamma(z)$ in the vicinity of the Stokes line. Moreover, we could have
considered more decimal places, if this was really necessary. To achieve
higher levels of accuracy all one has to do is alter the working precision
plus the precision and accuracy goals in the Mathematica module. This will
come at the expense of the computing time. Despite this, has Paris done the
same in Ref. [1]? If the smoothing point of view for the Stokes phenomenon is
indeed correct, then it should not only be vastly superior to the conventional
view by yielding more accurate values of $\ln\Gamma(z)$, it should also expose
where the errors or deficiencies occur in the conventional view. After all,
there is simply no point in offering an alternative view to the mathematical
community if it is unable to provide an improvement on the existing
view/approach. Despite all the aggrandisement coming from Paris, we do not see
any numerical evidence demonstrating how smoothing is able to match or even
provide more accurate values of $\ln\Gamma(z)$ in the vicinity of a Stokes
line. All we see in Paris’s comments is a table giving the Stokes multiplier
(both real and imaginary parts) to six decimal figures for a much larger value
of $|z|$ than in Table 1, presumably because he was unable to obtain this
level of accuracy for the Stokes multiplier for $|z|\\!=\\!3$. In addition,
the analysis in Ref. [2] was conducted for numerous values of the truncation
parameter $N$ far away from the optimal point whereas Paris’s computation
require an optimal point of truncation. This means that his approach is
useless for $|z|<1$, a limitation that should not apply.

The issues mentioned in the preceding paragraph are the ones that Paris needs
to address before attacking the results of $\ln\Gamma(z)$. Until his smoothing
approach is able to provide hyperasymptotic values of a special function
irrespective of whether the variable is large or small and without the need
for optimal truncation, there is no point adopting his smoothing approach to
asymptotics, which is not only vague and limited, but also very unwieldy.
Therefore, the challenge to Paris is that he should at least match the results
of Table 1 before he can claim that the smoothing concept is the correct
alternative to the conventional view of the Stokes phenomenon. The
particularly annoying aspect of Ref. [1] is that he chose a value of $|z|=8$
when he knew full well that $|z|\\!=\\!3$ was chosen in Ref. [2] because it
had a significantly lower optimal point of truncation. Nevertheless, even for
$|z|\\!=\\!8$, he does not present a numerical study as his Table 1 where he
has used Eq. (97) to determine hyperasymptotic values of $\ln\Gamma(z)$.

At the end of Sec. 2 in Ref. [1] Paris criticises the numerical computations
with the Borel-summed results or Eq. (98) carried out in Sec. 3 of Ref. [2].
There he states that his computations using Eq. (7) only took a fraction of a
second to compute compared with the hours taken when the Borel-summed
remainder or the second term on the rhs of Equivalence (87) is computed.
However, the first computation of $\ln\Gamma(z)$ in Ref. [2] is performed with
Eq. (7) under the same conditions as those for the Borel-summed remainder
given by Eq. (101). There it is stated that the calculations using the
incomplete gamma function in Mathematica took just as long as those for the
Borel-summed remainder. It was also stated that the calculations could be sped
up dramatically for the computations involving the incomplete gamma function
by wrapping the terms being calculated in the module by using the Mathematica
construct, N[expr,50]. Then the calculations take only a minute to perform,
but it can affect the accuracy of the final result for $\ln\Gamma(z)$,
particularly when the remainder begins to diverge, which is far more rapid and
earlier for small values of $|z|$. The primary consideration in the
calculations was to maintain the same accuracy whether $|z|$ was large or
small without the need to change the program or module in the process. As for
Paris’s computations, these are not the same as those reported in Ref. [2],
which are concerned with obtaining $\ln\Gamma(z)$ to 50 decimal places for
$|z|\\!=\\!3$ and $|z|\\!=\\!1/10$. For these calculations it was necessary to
consider much higher values of $k$ than those given by Paris in Table 1 of
Ref. [1]. Furthermore, it is stated later in the paper that the fastest method
of computing $\ln\Gamma(z)$ to hyperasymptotic accuracy is by using the MB-
regularised asymptotic forms, which are presented in Sec. 4 of Ref. [2] and
investigated numerically in Sec. 5. The MB-regularised results were obtained
in only a few seconds.

Paris has also overlooked the fact that the Borel-summed asymptotic forms are
required in order to provide a crucial comparison or check on the MB-
regularised results, especially when the argument of variable $z$ lies outside
the principal branch. This is because the LogGamma routine in Mathematica
cannot be used to check the values of the special function for such values of
${\rm arg}\,z$. The issue of considering the asymptotic behaviour outside the
principal branch is conveniently passed over by Paris as is the issue of
evaluating $\ln\Gamma(z)$ from his asymptotic forms on a Stokes line.

In the introduction to Ref. [2] it is stated that computational expediency was
the reason why MB regularisation was first devised in Ref. [11], although it
should be pointed out this is not always the case as discussed in Ref. [9]. In
relation to $\ln\Gamma(z)$ the superiority of the MB-regularised asymptotic
forms is primarily due to the fact that the Borel-summed asymptotic forms
possess an infinite sum over all positive integers with an intractable
denominator, e.g., $H(l,z)=\sum_{n=1}^{\infty}1/n^{l}(n^{2}+z^{2})$. The
corresponding sum in the MB-regularised asymptotic forms is replaced by the
Riemann zeta function, while in the case of Eq. (7), the integration has been
replaced by the incomplete gamma function. So, it is really a tribute to the
programming team at Wolfram Research for developing powerful routines such as
Zeta[s] and Gamma[N,z], which are capable of calculating these special
functions to incredible accuracy extremely quickly. If it were not for the
Zeta routine, then the zeta function in the MB-regularised asymptotic forms
for $\ln\Gamma(z)$ would have to be replaced by its Dirichlet series form and
we would be on the same level as the Borel-summed asymptotic forms since we
would be required to evaluate a similar number of integrals in an infinite sum
given by Eq. (98) to achieve the hyperasymptotic accuracy of Table 1.
Moreover, if there were no Gamma routine in Mathematica, then it would become
a formidable exercise to evaluate Eq. (7) extremely quickly. Alternatively, if
someone were to devise a special routine that could calculate $H(l,z)$ for all
values of $z$, then the calculations via Eq. (101) would be greatly
accelerated. Nevertheless, it should be emphasised that exactification of the
asymptotic forms for $\ln\Gamma(z)$ was the main aim of Ref. [2], while
computational expediency represented only a secondary issue. After all, the
work was concerned with developing the discipline of asymptotics, not the
field of numerical mathematics.

##  4\. Conclusion

In this paper we have refuted the two mains claims made by Paris [1]
concerning Ref. [2]. These are that there is no need for the concept of
regularisation to be employed in the asymptotics of $\ln\Gamma(z)$ and that
the “Stokes smoothing” view originated by Berry [15], and subsequently
supported by Olver [25], represents the correct view for interpreting
asymptotic behaviour near a Stokes line. In regard to the first claim, we have
seen that unbeknownst to him, Paris has actually employed the concept of
regularisation to arrive at the forms that he derives for the remainder
$R_{N}(z)$ given by Eqs. (4)- (7) here. With regard to the second claim, we
have seen that the Borel-summed regularised values of $\ln\Gamma(z)$ in Ref.
[2] are based entirely on the conventional view of the Stokes phenomenon,
which holds that the Stokes multiplier of the subdominant part of a complete
asymptotic expansion behaves as a step-function at a Stokes line. Moreover,
Paris has not provided a hyperasymptotic study to demonstrate that the
smoothing view is indeed superior to the conventional view. Therefore, the
challenge to Paris is that he should demonstrate that his asymptotic results
in Ref. [1] at least reproduce or even better still improve upon the results
in Table 1, which represent the results from a hyperasymptotic analysis of
$\ln\Gamma(z)$ in Ref. [2] based on the conventional view of the Stokes
phenomenon. Only then can he claim that smoothing represents a viable concept
in asymptotics. After all, if the smoothing approach is indeed correct and the
conventional view false, then this should become evident when one wishes to
obtain hyperasymptotic values of a function from its asymptotic forms not only
near, but also far away from a Stokes line, regardless of whether optimal
truncation has been invoked or not. In addition, it should also be able to
handle all values of $|z|$, not just $|z|\gg 1$ as in Ref. [1].

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