# On the Typicality of Musical Sequences

###### Abstract

It has been shown in a recent publication that words in human-produced English
language tend to have an information content close to the conditional entropy.
In this paper, we show that the same is true for events in human-produced
monophonic musical sequences. We also show how “typical sampling” influences
the distribution of information around the entropy for single events and
sequences.

##  1 Introduction

It has recently been shown in [1] that human-produced natural English
sentences are composed of words which encode information close to the expected
information content (i.e., the entropy). Derived from this finding, the
authors of [1] also proposed a sampling strategy called “typical sampling”
that produced sentences that, compared to other sampling methods, were deemed
more natural by humans. In this paper, we evaluate if the trend towards the
expected information content can also be observed in music. To that end, we
reproduce some of the analyses performed in [1] and [2] for musical events and
sequences. More precisely, we show how the information of human-produced
musical material is distributed around the model entropy. We also evaluate how
typical sampling impacts these distributions and how it compares to
conventional, ancestral sampling.

##  2 Methods

In the following, we recap Information Theory concepts coined in the seminal
paper [3], which have been applied to music in [4] to explain expectation and
surprise. Let $p\left(x_{t}|x_{<t}\right)$ be the conditional probability of a
symbol $x_{t}$ given the past $x_{<t}$ observed symbols and $q$ a model fitted
to $p$, e.g., a neural network fitted with likelihood optimization. A symbol
could represent a word in natural language or a musical event in music. The
conditional information content (IC) is given by
$IC\left(x_{t}|x_{<t}\right)=-\log{q\left(x_{t}|x_{<t}\right)}$ and is a
measure of how surprising (i.e., high IC) or expected (i.e., low IC) the
symbol $x_{t}$ is according to the model $q$, given the past $x_{<t}$
observations. Furthermore, the IC of a whole sequence is defined as
$IC\left(x\right)=\sum_{t=0}^{\lvert x\rvert-1}IC\left(x_{t}|x_{<t}\right)$
and the information density $ID(x)=IC(x)/\lvert x\rvert$. We are also
interested in the expected information or entropy $H$ for the conditional
symbol distribution and the sequence distribution.

###  2.1 Typicality

In [1], the authors find that in natural language sentences, the conditional
word information is close to the expected conditional information, and in [2]
that the sequence information content is close to the (unconditional) entropy
or that

$\displaystyle\lvert\epsilon_{sym}\rvert$ $\displaystyle=\lvert
H\left(x_{t}|x_{<t}\right)-IC\left(x_{t}|x_{<t}\right)\rvert,\text{ and }$ (1)
$\displaystyle\lvert\epsilon_{seq}\rvert$ $\displaystyle=\lvert
H\left(x\right)-IC\left(x\right)\rvert,$ (2)

respectively, are small. We are interested in testing if the results transfer
from natural language sequences of words to human compositions represented as
sequences of musical events. As the lengths of musical sequences can vary a
lot, we hypothesize that $IC\left(x\right)$ (in eq. 2) has less relevance to
music than the (length-normalized) _information density_ , where the $ID(x)$
of individual songs should be close to the expected $ID(x)$, resulting in a
small

$\lvert\epsilon_{ID}\rvert=|\mathbb{E}\left[ID\left(x\right)\right]-ID\left(x\right)\rvert.$
(3)

###  2.2 Typical Sampling

In [1], a sampling strategy is proposed where the least typical
symbols111Symbols with highest value of $\epsilon_{sym}$ in eq. 1 of $q$ get
pruned. The obtained samples are reported to be more human-like than other
sampling strategies. In typical sampling, we initially identify the smallest
set of most typical symbols $V$ s.t. $q(V|x_{<t})\geq\tau$,

(a)

(b)

Figure 1: Distribution of typicality divergence for human-composed music
(Reference), conventional ancestral sampling (Conventional), and typical
sampling (Typical) for single events (a) and sequences (b). The distributions
are reported for datasets IF and GF.

where $\tau$ determines the amount of probability pruned. Secondly, the
probabilities of symbols $\notin V$ are zeroed. Finally, the resulting
function is normalized to yield a valid probability distribution. We wish to
investigate the effect of typical sampling on the typicality of generated
sequences (eqs. 1 and 3).

##  3 Experiments

Our experiments are performed on monophonic symbolic music. We use the dataset
of [5], consisting of $45849$ Irish folk music lead sheets (denoted IF). For
computational reasons, we discard the $5\%$ longest sequences, resulting in a
test set of $3618$ samples of length $\leq 691$. In addition, we use all
German folk songs from the Essen Folk Database [6, 7] (referred to as GF),
consisting of 5152 melodies. We partition both datasets in training,
validation, and test sets with proportions 10/12, 1/12, and 1/12, respectively
(all analyses are performed on the test set).

The compositions are encoded using a simplified variant of the tokenization
strategy described in [8]. We order the notes by absolute time and serialize a
note as a note-value (pitch) or rest token, followed by one or more duration
tokens indicating the note’s duration. Specifically, we use $128$ _change-
pitch tokens_ for pitches $C_{-1}-G_{9}$, $1$ _rest token_ , $100$ _duration
tokens_ quantized linearly from $10$ms to $1000$ms, and $1$ _end-of-sequence
token_. This results in a vocabulary of 230 tokens. For each dataset, we train
a Transformer decoder model [9] with relative attention [10, 11] in a self-
supervised prediction task. We sample sets of sequences from each model with a
maximum sequence length and size following their respective test set’s maximum
length and size. We perform conventional ancestral sampling from the
unmodified distribution and typical sampling with a low, medium, and high
$\tau$ value.

##  4 Results and Discussion

In fig. 1(a), we show the distributions of the information contents for single
events relative to the entropy (for typical sampling, the entropy and
information content of the events is always calculated based on the unpruned
distributions). In fig. 1(b), we show the distributions of Information
Densities (IDs) for sequences relative to the mean IDs (see eq. 3) of the
respective reference distribution. For both data sets, we show the
distributions of the human-generated melodies (Reference), conventional
ancestral sampling (Conventional), and typical sampling (Typical, with
different $\tau$ values).

Similar to [1], we find that the ICs of events are distributed densely around
the conditional entropy (see fig. 1(a)). When using typical sampling, this
trend gets more pronounced for decreasing $\tau$ values, showing that the
sampled musical events are indeed more typical.

However, also with conventional ancestral sampling, the highest density of the
event’s information is close to the entropy. In addition, the distributions of
Conventional follow those of Reference closer than Typical in both figures. In
fig. 1(b), typical sampling causes the ID distributions of generated sequences
to narrow and shift to the right with decreasing $\tau$ values. Note that in
fig. 1(b), $\epsilon$ is always calculated based on the entropy of the
reference distributions, meaning that _the generated sequences become more
probable when $\tau$ decreases_. This is reasonable, as typical sampling
prunes less likely events, as shown in fig. 1(a), where events to the left get
pruned. The Reference distribution of IF is not centred around $0$ in fig.
1(b) (i.e., it is shifted to the right). This suggests that a considerable
number of songs in this dataset have a high average ID (meaning that they are
rather unlikely given the dataset distribution, causing the overall entropy to
increase).

It is subject to future work (not shown in [1]) to investigate if the
typicality property of events is mainly linked to human-generated sequences.
That is not certain, considering that even events sampled from a uniform or
degenerate (i.e., deterministic) distribution are perfectly typical (with
$\epsilon=0$). If it turns out that the typicality property is trivial, the
reason why typical sampling yields sequences preferred by humans (as shown in
[1]) may also be explained by other factors, like a more stable random walk
(as the model is confronted with more probable context), or by the simple fact
that slightly higher probability sequences are more appealing.

##  5 Conclusion

We showed that musical events tend to be typical and that typical sampling
generally yields more likely sequences than conventional sampling. However, it
is unclear if the typicality is a property of human-generated data only and if
typicality of such sequences is a valid explanation for why typical sampling
results in sequences preferred by humans.††This work was conducted in a
collaboration between JKU and Sony Computer Science Laboratories Paris under a
research agreement.

## References

  * [1] C. Meister, T. Pimentel, G. Wiher, and R. Cotterell, “Typical decoding for natural language generation,” _arXiv preprint arXiv:2202.00666_ , 2022.
  * [2] C. Meister, G. Wiher, T. Pimentel, and R. Cotterell, “On the probability-quality paradox in language generation,” _CoRR_ , vol. abs/2203.17217, 2022. [Online]. Available: https://doi.org/10.48550/arXiv.2203.17217
  * [3] C. E. Shannon, “A mathematical theory of communication,” _Bell System Technical Journal_ , vol. 27, no. 3, pp. 379–423, 1948. [Online]. Available: https://onlinelibrary.wiley.com/doi/abs/10.1002/j.1538-7305.1948.tb01338.x
  * [4] L. B. Meyer, “Meaning in music and information theory,” _The Journal of Aesthetics and Art Criticism_ , vol. 15, no. 4, pp. 412–424, 1957. [Online]. Available: http://www.jstor.org/stable/427154
  * [5] B. L. Sturm, J. F. Santos, O. Ben-Tal, and I. Korshunova, “Music transcription modelling and composition using deep learning,” in _Proc. Conf. Computer Simulation of Musical Creativity_ , Huddersfield, UK, 2016.
  * [6] H. Schaffrath, “The esac electronic songbooks,” _Computing in Musicology_ , vol. 9, p. 78, 1994.
  * [7] ——, “The esac databases and mappet software,” _Computing in Musicology_ , vol. 8, no. 66, p. 1658, 1992.
  * [8] S. Oore, I. Simon, S. Dieleman, D. Eck, and K. Simonyan, “This time with feeling: Learning expressive musical performance,” _CoRR_ , vol. abs/1808.03715, 2018. [Online]. Available: http://arxiv.org/abs/1808.03715
  * [9] D. Bahdanau, K. Cho, and Y. Bengio, “Neural machine translation by jointly learning to align and translate,” in _3rd International Conference on Learning Representations, ICLR 2015, San Diego, CA, USA, May 7-9, 2015, Conference Track Proceedings_ , Y. Bengio and Y. LeCun, Eds., 2015. [Online]. Available: http://arxiv.org/abs/1409.0473
  * [10] P. Shaw, J. Uszkoreit, and A. Vaswani, “Self-attention with relative position representations,” in _Proceedings of the 2018 Conference of the North American Chapter of the Association for Computational Linguistics: Human Language Technologies, NAACL-HLT, New Orleans, Louisiana, USA, June 1-6, 2018, Volume 2 (Short Papers)_ , M. A. Walker, H. Ji, and A. Stent, Eds. Association for Computational Linguistics, 2018, pp. 464–468. [Online]. Available: https://doi.org/10.18653/v1/n18-2074
  * [11] C. A. Huang, A. Vaswani, J. Uszkoreit, N. Shazeer, C. Hawthorne, A. M. Dai, M. D. Hoffman, and D. Eck, “An improved relative self-attention mechanism for transformer with application to music generation,” _CoRR_ , vol. abs/1809.04281, 2018. [Online]. Available: http://arxiv.org/abs/1809.04281


# Mellin-Barnes integrals and the method of brackets

Ivan Gonzalez  Instituto de Física y Astronomía, Universidad de Valparaiso,
Gran Bretaña $1111$, Valparaiso, Chile<EMAIL_ADDRESS>,  Igor Kondrashuk
Grupo de Matemática Aplicada, Departamento de Ciencias Básicas, Universidad
del Bío-Bío, Campus Fernand May, Chillán, Chile<EMAIL_ADDRESS>,
Victor H. Moll  Department of Mathematics, Tulane University, New Orleans, LA
70118<EMAIL_ADDRESS>and  Luis M. Recabarren  Departamento de Física,
Universidad Santa Maria, Valparaiso, Chile<EMAIL_ADDRESS>
###### Abstract.

The method of brackets is a method for the evaluation of definite integrals
based on a small number of rules. This is employed here for the evaluation of
Mellin-Barnes integral. The fundamental idea is to transform these integral
representations into a bracket series to obtain their values. The expansion of
the gamma function in such a series constitute the main part of this new
application. The power and flexibility of this procedure is illustrated with a
variety of examples.

###### Key words and phrases:

Definite integrals, Mellin-Barnes representations, method of brackets

###### 1991 Mathematics Subject Classification:

Primary 33C05, Secondary 30E20, 33E20

##  1\. Introduction

A classical approaches to special functions developed in the $19^{th}$ century
was to classify them from the point of view of solutions to (second) order
differential equations with analytic coefficients

(1.1) $a(z)\frac{d^{2}u}{dz^{2}}+b(z)\frac{du}{dz}+c(z)u=0.$

Gauss understood that the singularities of this equation are central to this
classification. Moreover, the role of $\infty$ as a possible singularity of
(1.1) plays an important role. The notion of regular singularities and the
well-known Frobenius method to produce solutions arose from this point of
view. The reader will find information about these ideas in [16] and [23].

It is a classical result that any differential equation with $3$ regular
singular points (on $\mathbb{C}\cup\\{\infty\\}$) can be transformed to the
hypergeometric differential equation

(1.2) $z(1-z)\frac{d^{2}w}{dz^{2}}+[c-(a+b-1)z]\frac{dw}{dz}-abw=0,$

with singularities at $0,\,1,\,\infty$. The solution of (4.13), normalized by
$w(0)=1$ is the hypergeometric function, defined by the series expansion

(1.3)
${}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a\,\,\,b}{c}\bigg{|}{z}\right)=\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}\,n!}z^{n},$

where $(q)_{n}$ os the Pochhammer symbol defined by

(1.4) $(q)_{n}=\begin{cases}1&\textnormal{ if }n=0\\\
q(q+1)\cdots(q+n-1)&\textnormal{ if }n>0.\end{cases}$

Among the many representations of the hypergeometric function we single out
the so-called Mellin-Barnes integral

(1.5)
${}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a\,\,\,b}{c}\bigg{|}{z}\right)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\times\frac{1}{2\pi
i}\int_{\gamma}\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^{s}\,ds,$

where the contour of integration $\gamma$ joins $-i\infty$ to $+i\infty$ and
separates the poles of $\Gamma(-s)$ (namely $0,\,1,\,2,\ldots$) from those of
$\Gamma(a+s)\Gamma(b+s)$ (located at $-a,\,-a-1,\ldots,-b,-b-1,\ldots$). In
the general case, the poles of the gamma factors are located on horizontal
semi-axes. The contour $\gamma$ has to be chosen to separate those moving to
the right from those moving to the left. Examples may be found in [21, 22].

Many important functions are obtained from the hypergeometric function by
coalescing singularities. For instance, the so-called Whittaker function
$W_{\kappa,\mu}(z)$, defined from the equation

(1.6)
$\frac{d^{2}w}{dz^{2}}+\left(-\frac{1}{4}+\frac{k}{z}+\frac{1/4-m^{2}}{z^{2}}\right)w=0$

also has the integral representation

(1.7) $W_{\kappa,\mu}(z)=\frac{e^{-z/2}}{2\pi
i}\int_{\gamma}z^{-s}\frac{\Gamma\left(\tfrac{1}{2}+\mu+s\right)\Gamma\left(\tfrac{1}{2}-\mu+s\right)\Gamma\left(-\kappa-s\right)}{\Gamma\left(\tfrac{1}{2}+\mu-\kappa\right)\Gamma\left(\tfrac{1}{2}-\mu-\kappa\right)}\,ds$

with a similar condition on the contour of integration.

The main type of integrals considered in this work come from Mellin
transforms. This concept is recalled next.

######  Definition 1.1.

Given a function $f(x)$, defined on the positive real axis $\mathbb{R}^{+}$,
its Mellin transform is defined by

(1.8) $\varphi(s)=(\mathcal{M}f)(s)=\int_{0}^{\infty}x^{s-1}f(x)\,dx.$

This relation may be inverted by a line integral

(1.9) $f(x)=(\mathcal{M}^{-1}\varphi)(x)=\frac{1}{2\pi
i}\int_{\gamma}x^{-s}\varphi(s)\,ds,$

by an appropriate choice of the contour $\gamma$ of the type described above.

The goal of the current work is to present a procedure to evaluate inverse
Mellin transforms based on the method of brackets. This is a method to
evaluate definite integrals and it is described in Section 2. The class of
functions considered here are of the form

(1.10)
$\varphi(s)=\frac{\prod\limits_{j=1}^{M}\Gamma\left(a_{j}+A_{j}s\right)}{\prod\limits_{j=1}^{N}\Gamma\left(b_{j}+B_{j}s\right)}\frac{\prod\limits_{j=1}^{P}\Gamma\left(c_{j}-C_{j}s\right)}{\prod\limits_{j=1}^{Q}\Gamma\left(d_{j}-D_{j}s\right)}.$

For simplicity the parameters $A_{j},\,B_{j},\,C_{j},\,D_{j}$ are taken real
and positive. Integrals of the form (1.9) with an integrand of the type (1.10)
will be referred as Mellin-Barnes integrals.

######  Remark 1.2.

In high energy physics Mellin-Barnes integrals appear frequently at the
intermediate steps in the process of calculations [2, 3, 10, 11, 12, 17, 14,
21]. These are contour integrals in which the integrands are quotients of
Euler Gamma functions, as in (1.10). This type of contour integrals represent
a typical representation of generalized hypergeometric functions. The Mellin-
Barnes representation of denominators in integrands corresponding to Feynman
diagrams in the momentum space representation is a standard procedure in
quantum field theory. This approach helps to calculate very difficult Feynman
diagrams [21] and this is the reason why Barnes integrals usually appear at
the penultimate step of calculation.

######  Remark 1.3.

In the previous work [13] it was mentioned that it would be interesting to
combine the Mellin-Barnes integrals and the method of brackets, due to the
similarity of these methods. This combination is presented here. In addition,
several known relations for generalized hypergeometric functions have been
reestablished by combining the method of brackets and Mellin-Barnes
transformations. Our investigation has been motivated by the work of Prausa
[19] which proposes an interesting symbolic procedure. Our work compares it
with the method of brackets.

##  2\. The method of brackets

This is a method that evaluates definite integrals over the half line
$[0,\,\infty)$. The application of the method consists of small number of
rules, deduced in heuristic form, some of which are placed on solid ground
[5].

For $a\in\mathbb{R}$, the symbol

(2.1) $\langle a\rangle=\int_{0}^{\infty}x^{a-1}\,dx$

is the bracket associated to the (divergent) integral on the right. The symbol

(2.2) $\phi_{n}=\frac{(-1)^{n}}{\Gamma(n+1)}$

is called the indicator associated to the index $n$. The notation
$\phi_{i_{1}i_{2}\cdots i_{r}}$, or simply $\phi_{12\cdots r}$, denotes the
product $\phi_{i_{1}}\phi_{i_{2}}\cdots\phi_{i_{r}}$.

Rules for the production of bracket series

The first part of the method is to associate to the integral

(2.3) $I(f)=\int_{0}^{\infty}f(x)\,dx$

a bracket series according to

${\mathbf{Rule\,\,P_{1}}}$. Assume $f$ has the expansion

(2.4) $f(x)=\sum_{n=0}^{\infty}\phi_{n}a_{n}x^{\alpha n+\beta-1}.$

Then $I(f)$ is assigned the bracket series

(2.5) $I(f)=\sum_{n\geq 0}a_{n}\langle\alpha n+\beta\rangle.$

${\mathbf{Rule\,\,P_{2}}}$. For $\alpha\in\mathbb{R}$, the multinomial power
$(a_{1}+a_{2}+\cdots+a_{r})^{\alpha}$ is assigned the $r$-dimension bracket
series

(2.6) $\sum_{n_{1}\geq 0}\sum_{n_{2}\geq 0}\cdots\sum_{n_{r}\geq
0}\phi_{n_{1}\,n_{2}\,\cdots n_{r}}a_{1}^{n_{1}}\cdots
a_{r}^{n_{r}}\frac{\langle-\alpha+n_{1}+\cdots+n_{r}\rangle}{\Gamma(-\alpha)}.$

${\mathbf{Rule\,\,P_{3}}}$. Each representation of an integral by a bracket
series has associated a complexity index of the representation via

(2.7) $\text{complexity index }=\text{number of sums }-\text{ number of
brackets}.$

It is important to observe that the complexity index is attached to a specific
representation of the integral and not just to integral itself. The experience
obtained by the authors using this method suggests that, among all
representations of an integral as a bracket series, the one with minimal
complexity index should be chosen. The level of difficulty in the analysis of
the resulting bracket series increases with the complexity index.

Rules for the evaluation of a bracket series

${\mathbf{Rule\,\,E_{1}}}$. Let $a,\,b\in\mathbb{R}$. The one-dimensional
bracket series is assigned the value

(2.8) $\sum_{n\geq 0}\phi_{n}f(n)\langle
an+b\rangle=\frac{1}{|a|}f(n^{*})\Gamma(-n^{*}),$

where $n^{*}$ is obtained from the vanishing of the bracket; that is, $n^{*}$
solves $an+b=0$. This is precisely the Ramanujan’s Master Theorem.

The next rule provides a value for multi-dimensional bracket series of index
$0$, that is, the number of sums is equal to the number of brackets.

${\mathbf{Rule\,\,E_{2}}}$. Let $a_{ij}\in\mathbb{R}$. Assuming the matrix
$A=(a_{ij})$ is non-singular, then the assignment is

$\sum_{n_{1}\geq 0}\cdots\sum_{n_{r}\geq 0}\phi_{n_{1}\cdots
n_{r}}f(n_{1},\cdots,n_{r})\langle
a_{11}n_{1}+\cdots+a_{1r}n_{r}+c_{1}\rangle\cdots\langle
a_{r1}n_{1}+\cdots+a_{rr}n_{r}+c_{r}\rangle$
$=\frac{1}{|\text{det}(A)|}f(n_{1}^{*},\cdots
n_{r}^{*})\Gamma(-n_{1}^{*})\cdots\Gamma(-n_{r}^{*})$

where $\\{n_{i}^{*}\\}$ is the (unique) solution of the linear system obtained
from the vanishing of the brackets. There is no assignment if $A$ is singular.

${\mathbf{Rule\,\,E_{3}}}$. The value of a multi-dimensional bracket series of
positive complexity index is obtained by computing all the contributions of
maximal rank by Rule $E_{2}$. These contributions to the integral appear as
series in the free indices. Series converging in a common region are added and
divergent/nulls series are discarded. There is no assignment to a bracket
series of negative complexity index. If all the resulting series are
discarded, then the method is not applicable.

######  Remark 2.1.

There is a small collection of formal operational rules for brackets. These
will be used in the calculations presented below.

######  Rule 2.1.

For any $\alpha\in\mathbb{R}$ the bracket satisfies
$\langle-\alpha\rangle=\langle\alpha\rangle$.

###### Proof.

This follows from the change of variables $x\mapsto 1/x$ in
$\displaystyle\langle-\alpha\rangle=\int_{0}^{\infty}x^{-\alpha-1}\,dx.$ ∎

A similar change of variables gives the next scaling rule:

######  Rule 2.2.

For any $\alpha,\,\beta,\,\gamma\in\mathbb{R}$ with $\alpha\neq 0$ the bracket
satisfies

(2.9)
$\langle\alpha\gamma+\beta\rangle=\frac{1}{|\alpha|}\left\langle\gamma+\frac{\beta}{\alpha}\right\rangle.$

This can be deduced from Rule $E_{1}$.

######  Rule 2.3.

For any $\alpha,\,\beta\in\mathbb{R}$ with $\alpha\neq 0$, for any
$n\in\mathbb{N}$ appearing as the index of a sum any allowable function $F$,
the identity

(2.10) $F(n)\langle\alpha
n+\beta\rangle=\frac{1}{|\alpha|}F\left(-\frac{\beta}{\alpha}\right)\left\langle
n+\frac{\beta}{\alpha}\right\rangle$

in the sense that any appearance of the left-hand side in a bracket series may
be replaced by the right-hand side.

###### Proof.

This follows directly from the rule $E_{1}$ to evaluate bracket series. ∎

##  3\. Some operational rules for integration

This section describes the relation between line integrals, like those
appearing in the inverse Mellin transform, and the method of brackets. These
complement those given for the discrete sums. The results presented here have
appeared in [19] as an extension of the method of brackets and used to produce
a minimal Mellin-Barnes representations of integrals appearing in connection
with Feynman diagrams.

###  3.1. The equivalence of brackets and the Dirac’s delta

Let $f$ be a function defined on $\mathbb{R}^{+}$ and consider its Mellin
transform

(3.1) $F(s)=\int_{0}^{\infty}x^{s-1}f(x)\,dx,$

with inversion rule

(3.2) $f(x)=\frac{1}{2\pi i}\int_{\gamma}x^{-s}F(s)\,ds,$

with the usual convention on the contour $\gamma$. Then, replacing (3.2) into
(3.1) produces

$\displaystyle F(s)$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}x^{s-1}\left[\frac{1}{2\pi
i}\int_{\gamma}x^{-s^{\prime}}F(s^{\prime})\,ds^{\prime}\right]\,dx$
$\displaystyle=$ $\displaystyle\frac{1}{2\pi
i}\int_{\gamma}F(s^{\prime})\left[\int_{0}^{\infty}x^{s-s^{\prime}-1}\,dx\right]\,ds^{\prime}$
$\displaystyle=$ $\displaystyle\frac{1}{2\pi
i}\int_{\gamma}F(s^{\prime})\langle s-s^{\prime}\rangle\,ds^{\prime}.$

This proves:

######  Rule 3.1.

The rule for integration respect to brackets is given by

(3.4) $\int_{\gamma}F(s)\langle s+\alpha\rangle\,ds=2\pi iF(-\alpha)$

where $\gamma$ is a contour of the usual type. The generalization

(3.5) $\int_{\gamma}F(s)\langle\beta s+\alpha\rangle\,ds=\frac{2\pi
i}{|\beta|}F\left(-\frac{\alpha}{\beta}\right),\quad\textnormal{with}\,\,\beta\in\mathbb{R}$

can be obtained from Rule 2.2.

######  Remark 3.1.

From an operational point of view, the result in Theorem 3.1 may be written as

(3.6) $\langle s+\alpha\rangle=2\pi i\delta(s+\alpha),$

where $\delta$ is Dirac’s delta function.

###  3.2. Mellin transform

This section contains a brief review of the Mellin transform. Recall that this
is defined by

(3.7) $\mathcal{M}(f)(z)=\int_{0}^{\infty}x^{z-1}f(x)~{}dx,$

where the arguments are the function $f$ to be transformed and the variable
$z$ appearing in the integral. The inverse Mellin transformation is

(3.8) $f(x)=\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}x^{-z}\mathcal{M}(f)(z)\,dz\quad\textnormal{ for
}x\in(0,\infty).$

The point $c$ associated to the contour of integration must be in the vertical
strip $c_{1}<c<c_{2}$, with boundaries determined by the condition that

(3.9) $\displaystyle\int_{0}^{1}x^{c_{1}-1}f(x)~{}dx$ $\displaystyle{\rm and}$
$\displaystyle\int_{1}^{\infty}x^{c_{2}-1}f(x)~{}dx$

must be finite. This is satisfied if the function $f$ satisfies the growth
conditions

$|f(x)|<1/x^{c_{1}}\quad{\rm when}\;x\to+0,\textnormal{ and
}\quad|f(x)|<1/x^{c_{2}}\quad\textnormal{ when}\;x\to+\infty.$

The conditions (3.9) imply that the Mellin transform $\mathcal{M}(f)(z)$ is
holomorphic in the vertical strip $c_{1}<{\rm Re}~{}z<c_{2}$. The asymptotic
behavior of the integrand is then used to determine the direction in which (a
finite segment) of the vertical line contour is closed in order to produce a
contour to apply Cauchy’s integral theorem. The singularities of the integrand
are then used to analyze the behavior of the integrals as the finite segment
becomes infinite.

One of the simplest examples of the Mellin transformation is

$\displaystyle\Gamma(z)=\int_{0}^{\infty}e^{-x}x^{z-1}~{}dx$
$\displaystyle{\rm and}$ $\displaystyle e^{-x}=\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}x^{-z}\Gamma(z)~{}dz.$

The contour in the complex plane is the vertical line with ${\rm Re}~{}z=c$ is
in the strip $0<c<A,$ where $A$ is a real positive number, the vertical line
contour must be closed to the left. It is convenient to include here a proof
of equations (3.7-3.8). This is classical and may be found in any textbook on
the theory of complex variable. It is reproduced here for pedagogical
arguments. First, use the fact that $\mathcal{M}(f)(z)$ is holomorphic in the
strip $c_{1}<{\rm Re}~{}z<c_{2}$, then taking $\delta>0$ to be infinitesimally
small,

$\displaystyle\mathcal{M}(f)(z)$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}x^{z-1}f(x)~{}dx$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi
i}\int_{0}^{\infty}x^{z-1}dx\int_{c-i\infty}^{c+i\infty}x^{-\omega}\mathcal{M}(f)(\omega)~{}d\omega$
$\displaystyle=$ $\displaystyle\frac{1}{2\pi
i}\int_{0}^{1}x^{z-1}dx\int_{c-i\infty}^{c+i\infty}x^{-\omega}\mathcal{M}(f)(\omega)~{}d\omega$
$\displaystyle\quad\quad+\frac{1}{2\pi
i}\int_{1}^{\infty}x^{z-1}dx\int_{c-i\infty}^{c+i\infty}x^{-\omega}\mathcal{M}(f)(\omega)~{}d\omega$
$\displaystyle=$ $\displaystyle\frac{1}{2\pi
i}\int_{0}^{1}x^{z-1}dx\int_{c_{1}-\delta-i\infty}^{c_{1}-\delta+i\infty}x^{-\omega}\mathcal{M}(f)(\omega)~{}d\omega$
$\displaystyle\quad\quad+\frac{1}{2\pi
i}\int_{1}^{\infty}x^{z-1}dx\int_{c_{2}+\delta-i\infty}^{c_{2}+\delta+i\infty}x^{-\omega}\mathcal{M}(f)(\omega)~{}d\omega$
$\displaystyle=\frac{1}{2\pi
i}\int_{c_{1}-\delta-i\infty}^{c_{1}-\delta+i\infty}\frac{\mathcal{M}(f)(\omega)}{z-\omega}~{}d\omega-\frac{1}{2\pi
i}\int_{c_{2}+\delta-i\infty}^{c_{2}+\delta+i\infty}\frac{\mathcal{M}(f)(\omega)}{z-\omega}~{}d\omega$
$\displaystyle=\frac{1}{2\pi
i}\int_{c_{1}-\delta+i\infty}^{c_{1}-\delta-i\infty}\frac{\mathcal{M}(f)(\omega)}{\omega-z}~{}d\omega+\frac{1}{2\pi
i}\int_{c_{2}+\delta-i\infty}^{c_{2}+\delta+i\infty}\frac{\mathcal{M}(f)(\omega)}{\omega-z}~{}d\omega$
$\displaystyle=\frac{1}{2\pi
i}\oint_{CR}\frac{\mathcal{M}(f)(\omega)}{\omega-z}~{}d\omega,$

where $CR$ is a rectangular contour constructed from the two vertical lines
from (LABEL:MT_direct) supplemented by two horizontal lines at the imaginary
complex infinities of the strip $c_{1}<{\rm Re}~{}z<c_{2}.$ The contour $CR$
is closed in the counterclockwise orientation.

The inverse transformation proof is even simpler and may be used in order to
define Dirac $\delta$-function. Observe that

$\displaystyle f(x)$ $\displaystyle=$ $\displaystyle\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}x^{-z}\mathcal{M}(f)(z)~{}dz$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}x^{-z}~{}dz\int_{0}^{\infty}y^{z-1}f(y)~{}dy$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\delta(\ln{(y/x)})y^{-1}f(y)~{}dy=f(x),$

which is valid in view of the relation

$\displaystyle\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}e^{(x-y)z}dz$
$\displaystyle=$
$\displaystyle\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{(x-y)(c+i\tau)}d\tau$
$\displaystyle=$
$\displaystyle\frac{e^{(x-y)c}}{2\pi}\int_{-\infty}^{\infty}e^{i(x-y)\tau}d\tau$
$\displaystyle=$ $\displaystyle e^{(x-y)c}\delta(x-y)=\delta(x-y).$

This proof of the inverse Mellin transformation belongs to D. Hilbert and may
be found in any good textbook dedicated to complex analysis. In this paper we
show that the method of bracket when applied to Mellin integrals in some sense
is equivalent to this old proof of the inverse Mellin transformation. More
precisely, we may argue that by using the same trick like in this quite old
proof of the inverse Mellin transformation, namely to divide the integration
over the variable $x\in[0,\infty[$ in two parts, $x\in[0,1]$ and
$x\in[1,\infty[,$ and by creating in a such way a closed contour in the plane
of the complex variable we may map all the rules of the method of bracket to
the Cauchy integral formula.

In high energy particle physics, in order to solve integro-differential
equations representing evolution of important physical quantities, the
transformation to Mellin moment is frequently used [see, for example [4]]. The
inverse transformation of the Mellin moment has completely the same form like
the inverse Mellin transformation. The question may appear if the inverse
Mellin transformation returns us to some function but how we may know if we
came back to this function of a real variable from its Mellin moment or from
its Mellin transform in the complex plane? We may differ the Mellin moments
from the the Mellin transforms by studying the asymptotic behaviour at the
complex infinity of the given function in the complex plane.

######  Remark 3.2.

The question of complexity, as introduced in Rule $\mathbf{P_{3}}$ is now
extended. In the process of evaluating an integral by the method of brackets,
define $\sigma$ to be the number of sums plus the number of contour integrals
appearing and $\delta$ to be the number of brackets plus the number of
integrals on the half-line $[0,\,\infty)$ that appear. The (generalized) index
of complexity is $\iota=\sigma-\delta$. This index should be seen as a measure
of difficulty in the evaluation of the integral by the method of brackets. In
the case $\iota=0$, the answer is given by a single term. For $\iota>0$, the
gamma factors appearing in the numerators of the line integrals must be
expanded in bracket series. This guarantees that the method provides all
series representations of the solution. As usual series converging in a common
region must be added. It is the heuristic observation that the bracket/line
integral representations of a problem should aim to minimize the index
$\iota$.

###  3.3. Multiple integrals

The method discussed here can be extended to evaluate multiple integrals with
a bracket representation

(3.13) $J=\left(\frac{1}{2\pi
i}\right)^{N}\int_{\gamma_{1}}ds_{1}\cdots\int_{\gamma_{N}}ds_{N}\\\
F(s_{1},\cdots,s_{N})\langle a_{11}s_{1}+\cdots
a_{1N}s_{N}+c_{1}\rangle\cdots\langle a_{N1}s_{1}+\cdots
a_{NN}s_{N}+c_{N}\rangle,$

by using the one-dimensional rule in iterated form. The expression for $J$ has
the form

(3.14) $J=\frac{1}{|\det(A)|}F(s_{1}^{*},\cdots,s_{N}^{*})$

where $A=(a_{ij})$, with $a_{ij}\in\mathbb{R}$ and
$\\{s_{i}^{*}\\}\,\,(i=1,\ldots,N)$ is the solution of the system
$A\vec{s}=-\vec{c}$ produced by the vanishing of the arguments in the brackets
appearing in this process.

##  4\. Transforming Mellin-Barnes integrals to bracket series

This section evaluates Mellin-Barnes integrals by transforming them into a
bracket series. The rules of Section 2 are then used to produce an analytic
expression for the integral.

######  Lemma 4.1.

The gamma function has the bracket series representation

(4.1) $\Gamma(\alpha)=\sum_{n=0}^{\infty}\phi_{n}\langle\alpha+n\rangle.$

###### Proof.

This follows simply from expanding $e^{-t}$ in power series in the integral
representation of the gamma function to obtain

(4.2)
$\Gamma(\alpha)=\int_{0}^{\infty}t^{\alpha-1}e^{-t}\,dt=\int_{0}^{\infty}t^{\alpha-1}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}t^{n}\,dt=\sum_{n=0}^{\infty}\phi_{n}\langle\alpha+n\rangle.$

∎

In this section we present a systematic procedure to evaluate Mellin-Barnes
integrals of the form

(4.3) $I(x)=\frac{1}{2\pi
i}\int_{\gamma}x^{-s}\frac{\prod\limits_{j=1}^{M}\Gamma(a_{j}+A_{j}s)\prod\limits_{j=1}^{P}\Gamma(c_{j}-C_{j}s)}{\prod\limits_{j=1}^{N}\Gamma(b_{j}+B_{j}s)\prod\limits_{j=1}^{Q}\Gamma(d_{j}-D_{j}s)}\,ds$

where $\gamma$ is the usual vertical line contour. A similar argument has
appeared in [19]. The idea is to use the method of brackets to produce the
bracket series associated to (4.3). The parameters satisfy the rules
$a_{j},\,b_{j},\,c_{j},\,d_{j}\in\mathbb{C}$ and
$A_{j},\,B_{j},\,C_{j},\,D_{j}\in\mathbb{R}^{+}$ for the index $j$ in the
corresponding range; for instance, the $j$ associated to $a_{j},\,A_{j}$ vary
from $1$ to $M$.

The procedure to obtain the bracket series is systematic: the gamma factors in
the numerator are replaced using formula (4.1) (the gamma factors in the
denominator do not contribute):

$\displaystyle\prod_{j=1}^{M}\Gamma(a_{j}+A_{j}s)$ $\displaystyle=$
$\displaystyle\prod_{j=1}^{M}\left[\sum_{k_{j}}\phi_{k_{j}}\langle
a_{j}+A_{j}s+k_{j}\rangle\right]$ $\displaystyle=$
$\displaystyle\sum_{k_{1}}\cdots\sum_{k_{M}}\phi_{k_{1}\cdots
k_{M}}\prod_{j=1}^{M}\langle a_{j}+A_{j}s+k_{j}\rangle$

and similarly

$\displaystyle\prod_{j=1}^{P}\Gamma(c_{j}-C_{j}s)$ $\displaystyle=$
$\displaystyle\prod_{j=1}^{P}\left[\sum_{\ell_{j}}\phi_{\ell_{j}}\langle
c_{j}-C_{j}s+\ell_{j}\rangle\right]$ $\displaystyle=$
$\displaystyle\sum_{\ell_{1}}\cdots\sum_{\ell_{P}}\phi_{\ell_{1}\cdots\ell_{P}}\prod_{j=1}^{P}\langle
c_{j}-C_{j}s+\ell_{j}\rangle.$

The rules of the method of brackets described in Section 2 now yield a bracket
series associated with the integral (4.3). To illustrate these idea, introduce
the function

(4.6)
$G(s)=\frac{1}{\prod\limits_{j=1}^{N}\Gamma(b_{j}+B_{j}s)}\times\frac{1}{\prod\limits_{j=1}^{Q}\Gamma(d_{j}-D_{j}s)},$

The previous rules transform the integral $I(x)$ in (4.3) to

$\displaystyle I(x)$ $\displaystyle=$ $\displaystyle\frac{1}{2\pi
i}\sum_{k_{1}}\cdots\sum_{k_{M}}\sum_{\ell_{1}}\cdots\sum_{\ell_{P}}\phi_{k_{1}\cdots
k_{M}\ell_{1}\cdots\ell_{P}}$
$\displaystyle\times\int_{\gamma}x^{-s}G(s)\left[\prod_{j=1}^{M}\langle
a_{j}+A_{j}s+k_{j}\rangle\right]\,\left[\prod_{j=1}^{P}\langle
c_{j}-C_{j}s+\ell_{j}\rangle\right]\,ds.$

this will be written in the more compact form

(4.8) $I(x)=\frac{1}{2\pi
i}\sum_{\\{k\\}}\sum_{\\{\ell\\}}\phi_{\\{k\\},\,\\{\ell\\}}\int_{\gamma}x^{-s}G(s)\left[\prod_{j=1}^{M}\langle
a_{j}+A_{j}s+k_{j}\rangle\right]\,\left[\prod_{j=1}^{P}\langle
c_{j}-C_{j}s+\ell_{j}\rangle\right]\,ds.$

Now select the bracket $\langle a_{M}+A_{M}s+k_{M}\rangle$ to evaluate the
integral (4.8). Any other choice of bracket gives an equivalent value for
$I(x)$. Start with

(4.9)
$I(x)=\sum_{\\{k\\}}\sum_{\\{\ell\\}}\phi_{\\{k\\},\\{\ell\\}}\int_{\gamma}x^{-s}G(s)\\\
\left[\prod_{j=1}^{M-1}\langle
a_{j}+A_{j}s+k_{j}\rangle\right]\,\left[\prod_{j=1}^{P}\langle
c_{j}-C_{j}s+\ell_{j}\rangle\right]\,\frac{\langle
a_{M}+A_{M}s+k_{M}\rangle}{2\pi i}\,ds.$

The rules of the method of brackets given requires to solve the linear
equation coming from the vanishing of the last bracket and using Rule 3.1.
This produces

(4.10) $s^{*}=-\frac{a_{M}+k_{M}}{A_{M}}.$

Therefore

(4.11)
$I(x)=\frac{1}{|A_{M}|}\sum_{\\{k\\}}\sum_{\\{\ell\\}}\phi_{\\{k\\},\\{\ell\\}}x^{-s^{*}}G(s^{*})\prod_{j=1}^{M-1}\langle
a_{j}+A_{j}s^{*}+k_{j}\rangle\prod_{j=1}^{P}\langle
c_{j}-C_{j}s^{*}+\ell_{j}\rangle$

Therefore, the value of the integral $I(x)$ obtained from the selection of the
the bracket $\langle a_{M}+A_{M}s+k_{M}\rangle$ is given by

(4.12)
$I(x)=\frac{x^{a_{M}/A_{M}}}{|A_{M}|}\sum_{\\{k\\}}\sum_{\\{\ell\\}}\phi_{\\{k\\},\\{\ell\\}}x^{k_{M}/A_{M}}\\\
\frac{\prod\limits_{j=1}^{M-1}\left\langle
a_{j}-\frac{A_{j}a_{M}}{A_{M}}-\frac{A_{j}k_{M}}{A_{M}}+k_{j}\right\rangle\prod\limits_{j=1}^{P}\left\langle
c_{j}+\frac{C_{j}a_{M}}{A_{M}}+\frac{C_{j}k_{M}}{A_{M}}+\ell_{j}\right\rangle}{\prod\limits_{j=1}^{N}\left\langle
b_{j}-\frac{B_{j}a_{M}}{A_{M}}-\frac{B_{j}k_{M}}{A_{M}}\right\rangle\prod\limits_{j=1}^{Q}\left\langle
d_{j}+\frac{D_{j}a_{M}}{A_{M}}+\frac{D_{j}k_{M}}{A_{M}}\right\rangle}.$

Observe that one obtains a total of $M+P$ series representations for the
integral $I(x)$. There are $P$ of them in the argument $x^{-1/C_{j}}$ and the
remaining $M$ of them in the argument $x^{1/A_{j}}$. This procedure extends
without difficulty to multiple integrals.

These ideas are illustrated next.

######  Example 4.2.

The hypergeometric function ${{}_{2}}F_{1}$, defined by the series

(4.13)
${}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c}\bigg{|}{x}\right)=\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}\,n!}x^{n},$

for $|x|<1$, admits the Mellin-Barnes representation

(4.14)
${}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c}\bigg{|}{x}\right)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\frac{1}{2\pi
i}\int_{\gamma}\frac{\Gamma(-s)\Gamma(s+a)\Gamma(s+b)}{\Gamma(s+c)}(-x)^{s}\,ds$

as a contour integral. This appears as entry $9.113$ in [15].

The integral in (4.14) is now used to obtain the series representation (4.13).
As an added consequence, this method will also produce an analytic
continuation of the series (4.13) to the domain $|x|>1$.

The starting point is now the right-hand side of (4.14)

(4.15) $G(a,b,c;x)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\frac{1}{2\pi
i}\int_{\gamma}\frac{\Gamma(-s)\Gamma(s+a)\Gamma(s+b)}{\Gamma(s+c)}(-x)^{s}\,ds$

and using (4.1) in the three gamma factors yields

(4.16) $G(a,b,c;x)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\frac{1}{2\pi
i}\sum_{n_{1},n_{2},n_{3}}\phi_{123}\int_{\gamma}\frac{(-x)^{s}\langle-s+n_{1}\rangle\langle
s+a+n_{2}\rangle\langle s+b+n_{3}\rangle}{\Gamma(s+c)}ds.$

The gamma term in the denominator has no poles, so it is not expanded.

In order to evaluate the expression (4.16), select the bracket containing the
index $n_{1}$ and use Theorem 3.1 to obtain the bracket series:

(4.17)
$G(a,b,c;x)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\sum_{n_{1},n_{2},n_{3}}\phi_{123}\frac{(-x)^{n_{1}}}{\Gamma(n_{1}+c)}\langle
a+n_{1}+n_{2}\rangle\langle b+n_{1}+n_{3}\rangle.$

The evaluation of this series is done according to the rules given in Section
2.

Take $n_{1}$ as the free index. Then the indices $n_{2},\,n_{3}$ are
determined by the system

(4.18) $a+n_{1}+n_{2}=0\quad\text{ and }\quad b+n_{1}+n_{3}=0,$

which gives $n_{2}=-a-n_{1}$ and $n_{3}=-b-n_{1}$. Then (2.8) produces

(4.19)
$G_{1}(a,b,c;x)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\sum_{n_{1}=0}^{\infty}\phi_{1}\frac{(-x)^{n_{1}}}{\Gamma(n_{1}+c)}\Gamma(a+n_{1})\Gamma(b+n_{1}),$

where the index on $G_{1}$ is used to indicate that this sum comes from the
free index $n_{1}$. This reduces to (4.13), showing that

(4.20)
$G_{1}(a,b,c;x)=\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}\,n!}z^{n}.$

The series on the right converges for $|x|<1$. This recovers equation (4.13).

Take $n_{2}$ as the free index. Then the vanishing of the brackets give
$n_{1}=-a-n_{2}$ and $n_{3}=-b+a+n_{2}$. Then

(4.21)
$G_{2}(a,b,c;x)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\sum_{n_{2}=0}^{\infty}\phi_{2}\frac{(-x)^{-a-n_{2}}}{\Gamma(-a-n_{2}+c)}\Gamma(a+n_{2})\Gamma(b-a-
n_{2}).$

Using $\Gamma(u+m)=\Gamma(u)(u)_{m}$ converts (4.21) into

(4.22)
$G_{2}(a,b,c;x)=\frac{\Gamma(c)\Gamma(b-a)}{\Gamma(b)\Gamma(c-a)}\sum_{n_{2}=0}^{\infty}\phi_{2}\frac{(-x)^{-a-n_{2}}}{(c-a)_{-n_{2}}}(a)_{n_{2}}(b-a)_{-n_{2}}.$

The final step uses the transformation rule

(4.23) $(u)_{-n}=\frac{(-1)^{n}}{(1-u)_{n}}$

to eliminate the negative indices on the Pochhammer symbols and convert (4.22)
into

$\displaystyle\quad G_{2}(a,b,c;x)$ $\displaystyle=$
$\displaystyle\frac{\Gamma(c)\Gamma(b-a)}{\Gamma(b)\Gamma(c-a)}\sum_{n_{2}=0}^{\infty}\phi_{2}\frac{(-x)^{-a-n_{2}}}{(1-b+a)_{n_{2}}}(a)_{n_{2}}(1-c+a)_{n_{2}}$
$\displaystyle=$
$\displaystyle(-x)^{-a}\frac{\Gamma(c)\Gamma(b-a)}{\Gamma(b)\Gamma(c-a)}\sum_{n_{2}=0}^{\infty}\frac{(a)_{n_{2}}(1-c+a)_{n_{2}}}{(1-b+a)_{n_{2}}\,n_{2}!}x^{-n_{2}}.$

The series on the right is identified as a hypergeometric series and it yields

(4.25)
$G_{2}(a,b,c;x)=(-x)^{-a}\frac{\Gamma(c)\Gamma(b-a)}{\Gamma(b)\Gamma(c-a)}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a\,\,\,\,\,\,1-c+a}{1-b+a}\bigg{|}{\frac{1}{x}}\right)$

and this series converges for $|x|>1$.

Finally take $n_{3}$ as the free index. This case is similar to the previous
one and it produces

(4.26)
$G_{3}(a,b,c;x)=(-x)^{-b}\frac{\Gamma(c)\Gamma(a-b)}{\Gamma(a)\Gamma(c-b)}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{b\,\,\,\,\,\,1-c+b}{1-a+b}\bigg{|}{\frac{1}{x}}\right)$

and this series also converges for $|x|>1$.

The rules in Section 2 state that if in the evaluation of an integral one
obtains a collection of series, coming from choices of free indices, those
converging in a common region must be added. Thus, the integral $G(a,b,c;x)$
in (4.17) has the representations

(4.27)
$G(a,b,c;x)={}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a\,\,b}{c}\bigg{|}{x}\right)\quad\text{for}\,\,|x|<1$

and

(4.28)
$G(a,b,c;x)=(-x)^{-a}\frac{\Gamma(c)\Gamma(b-a)}{\Gamma(b)\Gamma(c-a)}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a\,\,\,\,\,\,1-c+a}{1-b+a}\bigg{|}{\frac{1}{x}}\right)\\\
+(-x)^{-b}\frac{\Gamma(c)\Gamma(a-b)}{\Gamma(a)\Gamma(c-b)}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{b\,\,\,\,\,\,1-c+b}{1-a+b}\bigg{|}{\frac{1}{x}}\right),\quad\text{for}\,\,|x|>1.$

Therefore we have obtained an analytic continuation of the hypergeometric
function ${}_{2}F_{1}(a,b,c;x)$ from $|x|<1$ to the exterior of the unit
circle. The identity (4.28) appears as entry $9.132.2$ in [15].

##  5\. Inverse Mellin transforms

The method of brackets is now used to evaluate integrals of the form (1.9)

(5.1) $f(x)=\frac{1}{2\pi i}\int_{\gamma}x^{-s}\varphi(s)\,ds,$

where $\gamma$ is a vertical line on $\mathbb{C}$, adjusted to each problem.
Given $\varphi(s)$, the function $f(x)$ has Mellin transform $\varphi$.

######  Example 5.1.

Consider the function $\varphi(s)=\Gamma(s-a).$ Its inverse Mellin transform
is given by

(5.2) $f(x)=\frac{1}{2\pi i}\int_{\gamma}x^{-s}\Gamma(s-a)\,ds.$

Now use (4.1) to write

(5.3) $\Gamma(s-a)=\sum_{n}\phi_{n}\langle s-a+n\rangle$

and (5.2) yields

(5.4) $f(x)=\sum_{n}\phi_{n}\frac{1}{2\pi i}\int_{\gamma}x^{-s}\langle
s-a+n\rangle\,ds$

Theorem 3.1 now gives

(5.5) $f(x)=\sum_{n}\phi_{n}x^{n-a}=x^{-a}e^{-x}.$

This is written as

(5.6) $\frac{1}{2\pi i}\int_{\gamma}x^{-s}\Gamma(s-a)\,ds=x^{-a}e^{-x},$

or equivalently

(5.7) $\int_{0}^{\infty}x^{s-a-1}e^{-x}\,dx=\Gamma(s-a).$

Replacing $s-a$ by $s$, this is the integral definition of the gamma function.

######  Example 5.2.

The inverse Mellin transform of $\varphi(s)=\Gamma(a-s)$ is obtained as in
Example 5.1. The result is

(5.8) $f(x)=x^{-a}e^{-1/x},$

also written as

(5.9) $\frac{1}{2\pi i}\int_{\gamma}x^{-s}\Gamma(a-s)\,ds=x^{-a}e^{-1/x},$

or equivalently

(5.10) $\int_{0}^{\infty}x^{s-1}x^{-a}e^{-1/x}\,dx=\Gamma(a-s).$

The change of variables $u=x^{-1}$ gives the integral representation of the
gamma function.

######  Example 5.3.

The inversion of $\varphi(s)=\Gamma(s-a)\Gamma(s-b)$ amounts to the evaluation
of the line integral

(5.11) $f(x)=\mathcal{M}^{-1}(\Gamma(s-a)\Gamma(s-b))(x)=\frac{1}{2\pi
i}\int_{\gamma}x^{-s}\Gamma(s-a)\Gamma(s-b)\,ds.$

Now use (4.1) to write

(5.12) $\Gamma(s-a)=\sum_{n_{1}}\phi_{n_{1}}\langle
s-a+n_{1}\rangle\quad\text{and}\quad\Gamma(s-b)=\sum_{n_{2}}\phi_{n_{2}}\langle
s-b+n_{2}\rangle$

and produce

(5.13) $f(x)=\frac{1}{2\pi
i}\int_{\gamma}x^{-s}\sum_{n_{1},n_{2}}\phi_{12}\langle
s-a+n_{1}\rangle\langle s-b+n_{2}\rangle.$

Now select the bracket containing the index $n_{1}$ and write (5.13) using
Theorem 3.1 as

$\displaystyle f(x)$ $\displaystyle=$
$\displaystyle\sum_{n_{1},n_{2}}\phi_{12}\frac{1}{2\pi
i}\int_{\gamma}\left(x^{-s}\langle s-b+n_{2}\rangle\right)\langle
s-a+n_{1}\rangle$ $\displaystyle=$
$\displaystyle\sum_{n_{1},n_{2}}\phi_{12}\,x^{-a+n_{1}}\langle
a-n_{1}-b+n_{2}\rangle.$

This is a two-dimensional bracket series and its evaluation is achieved using
the rules in Section 2:

$n_{1}$ is a free index. Then $n_{2}=n_{1}-a+b$ and this produces the value

$\displaystyle f_{1}(x)$ $\displaystyle=$
$\displaystyle\sum_{n_{1}=0}^{\infty}\phi_{1}x^{-a+n_{1}}\Gamma(-n_{1}+a-b)$
$\displaystyle=$ $\displaystyle
x^{-a}\Gamma(a-b)\sum_{n_{1}=0}^{\infty}\phi_{1}(a-b)_{-n_{1}}$
$\displaystyle=$ $\displaystyle
x^{-a}\Gamma(a-b)\sum_{n_{1}=0}^{\infty}\frac{x^{n_{1}}}{n_{1}!\,(1-a+b)_{n_{1}}}$
$\displaystyle=$ $\displaystyle
x^{-a}\Gamma(a-b)\,{}_{0}F_{1}\left(\genfrac{}{}{0.0pt}{}{-}{1-a+b}\bigg{|}{x}\right).$

$n_{2}$ is a free index. A similar argument gives

(5.16)
$f_{2}(x)=x^{-b}\Gamma(b-a)\,{}_{0}F_{1}\left(\genfrac{}{}{0.0pt}{}{-}{1-b+a}\bigg{|}{x}\right).$

Since both representations always converge, one obtains

(5.17)
$f(x)=x^{-a}\Gamma(a-b)\,{}_{0}F_{1}\left(\genfrac{}{}{0.0pt}{}{-}{1-a+b}\bigg{|}{x}\right)+x^{-b}\Gamma(b-a)\,{}_{0}F_{1}\left(\genfrac{}{}{0.0pt}{}{-}{1-b+a}\bigg{|}{x}\right).$

The function ${}_{0}F_{1}$ is now expressed in terms of the modified Bessel
function $I_{\nu}(z)$. This is defined in [18, 10.25.2] by the power series

(5.18)
$I_{\nu}(z)=\left(\frac{z}{2}\right)^{\nu}\sum_{k=0}^{\infty}\frac{1}{k!\,\Gamma(\nu+k+1)}\left(\frac{z^{2}}{4}\right)^{k}.$

######  Lemma 5.4.

For $\alpha\in\mathbb{R}$, the identity

(5.19)
${}_{0}F_{1}\left(\genfrac{}{}{0.0pt}{}{-}{\alpha}\bigg{|}{x}\right)=\Gamma(\alpha)x^{(1-\alpha)/2}I_{\alpha-1}(2\sqrt{x})$

holds.

###### Proof.

This follows directly from (5.18). ∎

Replacing the expression in Lemma 5.4 in (5.17) gives

(5.20)
$f(x)=\frac{\pi}{\sin(\pi(a-b))}x^{-(a+b)/2}\left(I_{b-a}(2\sqrt{x})-I_{a-b}(2\sqrt{x})\right).$

The relation [18, 10.27.4]

(5.21) $K_{\nu}(z)=\frac{\pi}{2}\frac{I_{-\nu}(z)-I_{\nu}(z)}{\sin(\pi\nu))}$

(which is here as the definition of $K_{\nu}(z)$) now implies

(5.22) $f(x)=2x^{-\nu/2-b}K_{\nu}(2\sqrt{x});$

with $\nu=a-b$. This is

(5.23) $\frac{1}{2\pi
i}\int_{\gamma}x^{-s}\Gamma(s-a)\Gamma(s-b)\,ds=2x^{-(a+b)/2}K_{a-b}(2\sqrt{x}).$

After some elementary changes, this is written as

(5.24) $K_{\nu}(x)=\frac{1}{4\pi
i}\left(\frac{x}{2}\right)^{\nu}\int_{\gamma}\Gamma(s)\Gamma(s-\nu)\left(\frac{x}{2}\right)^{-2s}\,ds,$

the form appearing in [18, 10.32.13]. The expression (5.24) is now written in
the equivalent form

(5.25)
$\int_{0}^{\infty}x^{s-1}K_{\nu}(x)\,dx=2^{s-2}\Gamma\left(\frac{s+\nu}{2}\right)\Gamma\left(\frac{s-\nu}{2}\right).$

######  Example 5.5.

Consider the inversion of $\varphi(s)=\Gamma(s-a)\Gamma(b-s)$. Observe that
there is a change in the order of the argument of the second gamma factor with
respect to Example 5.3. To evaluate this example, expand the term
$\Gamma(s-a)$ in a bracket series using (4.1) to obtain

(5.26) $f(x)=\sum_{n}\phi_{n}\frac{1}{2\pi i}\int_{\gamma}x^{-s}\langle
s-a+n\rangle\Gamma(b-s)\,ds.$

Theorem 3.1 yields

(5.27) $f(x)=x^{-a}\sum_{n=0}^{\infty}\phi_{n}\Gamma(b-a+n)x^{n}.$

To simplify this answer write $\Gamma(b-a+n)=\Gamma(b-a)(b-a)_{n}$, use

(5.28)
${}_{1}F_{0}\left(\genfrac{}{}{0.0pt}{}{\alpha}{-}\bigg{|}{x}\right)=(1-x)^{-\alpha}$

and conclude that

(5.29) $f(x)=\frac{\Gamma(b-a)}{x^{a}(1+x)^{b-a}}.$

This is equivalent to the evaluation

(5.30) $\frac{1}{2\pi
i}\int_{\gamma}x^{-s}\Gamma(s-a)\Gamma(b-s)\,ds=\frac{\Gamma(b-a)}{x^{a}(1+x)^{b-a}}.$

Expanding the other gamma factors produces the same analytic expression for
the integral.

######  Example 5.6.

This example considers the simplest case of an integrand where a quotient of
gamma factors appears. This is the inversion of

(5.31) $\varphi(s)=\frac{\Gamma(s-a)}{\Gamma(s-b)}.$

The usual formulation now gives

(5.32) $f(x)=\sum_{n}\phi_{n}\frac{1}{2\pi
i}\int_{\gamma}\left(\frac{x^{-s}}{\Gamma(s-b)}\right)\langle
s-a+n\rangle\,ds.$

Theorem 3.1 now yields

(5.33) $f(x)=\sum_{n=0}^{\infty}\phi_{n}\frac{x^{n-a}}{\Gamma(a-b-n)}.$

This expression is simplified using

(5.34)
$\Gamma(a-b-n)=\Gamma(a-b)(a-b)_{-n}=(-1)^{n}\frac{\Gamma(a-b)}{(1-a+b)_{n}}$

to obtain

$\displaystyle f(x)$ $\displaystyle=$
$\displaystyle\frac{x^{-a}}{\Gamma(a-b)}\sum_{n=0}^{\infty}\frac{(1-a+b)_{n}}{n!}x^{n}$
$\displaystyle=$ $\displaystyle\frac{1}{\Gamma(a-b)}x^{-a}(1-x)^{-1+a-b}.$

This can be written as

(5.36) $\frac{1}{2\pi
i}\int_{\gamma}x^{-s}\frac{\Gamma(s-a)}{\Gamma(s-b)}\,ds=\frac{1}{\Gamma(a-b)}x^{-a}(1-x)^{-1+a-b}.$

######  Example 5.7.

The inverse Mellin transform $f(x)$ of

(5.37) $\varphi(s)=\frac{\Gamma(s-a)}{\Gamma(b-s)}$

is computed from the line integral

(5.38) $f(x)=\frac{1}{2\pi
i}\int_{\gamma}x^{-s}\frac{\Gamma(s-a)}{\Gamma(b-s)}\,ds.$

The usual procedure now yields

(5.39)
$f(x)=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{n-a}}{n!\,\Gamma(b-a+n)}=\frac{x^{-a}}{\Gamma(b-a)}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!\,(b-a)_{n}}x^{n}.$

The series is identified as an ${{}_{0}F_{1}}$ and using the identity

(5.40)
$J_{\nu}(z)=\frac{1}{\Gamma(\nu+1)}\left(\frac{z}{2}\right)^{\nu}{}_{0}F_{1}\left(\genfrac{}{}{0.0pt}{}{-}{\nu+1}\bigg{|}{-\frac{z^{2}}{4}}\right)$

produces

(5.41) $f(x)=x^{(1-a-b)/2}J_{-1-a+b}(2\sqrt{x})$

and gives the evaluation

(5.42) $\frac{1}{2\pi
i}\int_{\gamma}x^{-s}\frac{\Gamma(s-a)}{\Gamma(b-s)}\,ds=x^{(1-a-b)/2}J_{-1-a+b}(2\sqrt{x}).$

######  Example 5.8.

The inverse Mellin transform $f(x)$ of

(5.43) $\varphi(s)=\frac{\Gamma(a-s)}{\Gamma(s-b)}$

is computed as in Example 5.7. The result is

(5.44) $\frac{1}{2\pi
i}\int_{\gamma}x^{-s}\frac{\Gamma(a-s)}{\Gamma(s-b)}\,ds=x^{-(a+b+1)/2}J_{a-b-1}\left(\frac{2}{\sqrt{x}}\right)$

and thus

(5.45)
$\int_{0}^{\infty}x^{s-1}x^{-(a+b+1)/2}J_{a-b-1}\left(\frac{2}{\sqrt{x}}\right)\,dx=\frac{\Gamma(a-s)}{\Gamma(s-b)}.$

The identity (5.44), with $a=0$ and $b=-\nu-1$, can be written as

(5.46) $J_{\nu}(x)=\frac{1}{2\pi
i}\int_{\gamma}\frac{\Gamma(-s)}{\Gamma(s+\nu+1)}\left(\frac{x}{2}\right)^{2s+\nu}\,ds.$

######  Example 5.9.

The Mellin inversion of the function

(5.47) $\varphi(s)=\frac{\Gamma(s)\Gamma(1-s)}{\Gamma(\beta-\alpha s)}$

is given by

(5.48) $f(x)=\frac{1}{2\pi
i}\int_{\gamma}x^{-s}\frac{\Gamma(s)\Gamma(1-s)}{\Gamma(\beta-\alpha s)}\,ds.$

The standard procedure gives

(5.49) $f(x)=\sum_{n_{1},n_{2}}\phi_{12}\,\frac{1}{2\pi
i}\int_{\gamma}\left(\frac{\langle 1-s+n_{2}\rangle
x^{-s}}{\Gamma(\beta-\alpha s)}\right)\,\langle s+n_{1}\rangle\,ds.$

Theorem 3.1 then produces

(5.50) $f(x)=\sum_{n_{1},n_{2}}\phi_{12}\frac{x^{n_{1}}}{\Gamma(\beta+\alpha
n_{1})}\langle 1+n_{1}+n_{2}\rangle.$

To evaluate this two-dimensional bracket series proceed as in Example 5.3.
This gives

(5.51) $f(x)=\sum_{n=0}^{\infty}\frac{(-x)^{n}}{\Gamma(\beta+\alpha
n)}\quad\text{when}\,\,|x|<1$

and

(5.52) $f(x)=\frac{1}{x}\sum_{n=0}^{\infty}\frac{1}{\Gamma(\beta-\alpha-\alpha
n)}\frac{(-1)^{n}}{x^{n}}\quad\text{when}\,\,|x|>1.$

The function appearing in (5.51) is the Mittag-Leffler function, defined in
[18] by

(5.53) $E_{\alpha,\beta}(z)=\sum_{n=0}^{\infty}\frac{z^{n}}{\Gamma(\alpha
n+\beta)}.$

produces the final expression

(5.54) $f(x)=\begin{cases}E_{\alpha,\beta}(-x)&\quad\text{if}\,\,|x|<1\\\
x^{-1}E_{-\alpha,\beta-\alpha}(-1/x)&\quad\text{if}\,\,|x|>1.\end{cases}$

##  6\. Direct computations of Mellin transforms

This section describes how to use the method of brackets to produce the
evaluation of the Mellin transform

(6.1) $\mathcal{M}(f(x))(s)=\int_{0}^{\infty}x^{s-1}f(x)\,dx.$

######  Example 6.1.

Example 5.3 has produced the evaluation of

(6.2)
$\int_{0}^{\infty}x^{\alpha-1}K_{\nu}(x)\,dx=2^{\alpha-1}\Gamma\left(\frac{\alpha+\nu}{2}\right)\Gamma\left(\frac{\alpha-\nu}{2}\right),$

from the Mellin inversion of $\Gamma(s-a)\Gamma(s-b)$.

######  Example 6.2.

The next example evaluates

(6.3)
$I(\alpha,\mu,\nu)=\int_{0}^{\infty}x^{\alpha-1}K_{\mu}(x)K_{\nu}(x)\,dx.$

by the methods developed here.

Entry $10.32.19$ in [18] contains the representation

(6.4) $K_{\mu}(x)K_{\nu}(x)=\\\ \frac{1}{8\pi
i}\int_{\gamma}\left(\frac{x}{2}\right)^{-2s}\frac{1}{\Gamma(2s)}\Gamma\left(s+\frac{\mu+\nu}{2}\right)\Gamma\left(s+\frac{\mu-\nu}{2}\right)\Gamma\left(s-\frac{\mu-\nu}{2}\right)\Gamma\left(s-\frac{\mu+\nu}{2}\right)\,ds$

Replacing this in (6.3) and identifying the $x$-integral as a bracket, yields

(6.5) $I(\alpha,\mu,\nu)=\frac{1}{8\pi
i}\int_{\gamma}\frac{2^{2s}}{\Gamma(2s)}\Gamma\left(s+\frac{\mu+\nu}{2}\right)\Gamma\left(s+\frac{\mu-\nu}{2}\right)\\\
\Gamma\left(s-\frac{\mu-\nu}{2}\right)\Gamma\left(s-\frac{\mu+\nu}{2}\right)\langle\alpha-2s\rangle\,ds$

Since this problem contains one bracket and one contour integral, there is no
need to expand the gamma terms in bracket series and the result is obtained
directly from Theorem 3.1. The result is

(6.6) $\int_{0}^{\infty}x^{\alpha-1}K_{\mu}(x)K_{\nu}(x)\,dx\\\
=\frac{2^{\alpha-3}}{\Gamma(\alpha)}\Gamma\left(\frac{\alpha+\mu+\nu}{2}\right)\Gamma\left(\frac{\alpha+\mu-\nu}{2}\right)\Gamma\left(\frac{\alpha-\mu+\nu}{2}\right)\Gamma\left(\frac{\alpha-\mu-\nu}{2}\right).$

######  Example 6.3.

The evaluation of

(6.7)
$\int_{0}^{\infty}x^{2a-1}K_{\nu}^{2}(x)\,dx=\frac{\sqrt{\pi}\,\Gamma(a+\nu)\Gamma(a-\nu)\Gamma(a)}{4\Gamma\left(a+\tfrac{1}{2}\right)}$

is the special case of Example 6.2 with $\mu=\nu$. Note that the parameter $a$
has been replaced by $2a$, in order to write the answer in a more compact
form. In particular, with $\nu=0$, this becomes

(6.8)
$\int_{0}^{\infty}x^{2a-1}K_{0}^{2}(x)\,dx=\frac{\sqrt{\pi}\Gamma^{3}(a)}{4\,\Gamma\left(a+\tfrac{1}{2}\right)}.$

The final special case mentioned here has $a=\tfrac{1}{2}$:

(6.9) $\int_{0}^{\infty}K_{0}^{2}(x)\,dx=\frac{\pi^{2}}{4}.$

These examples have been evaluated in [9] by a different procedure.

######  Example 6.4.

Now consider the integral

(6.10) $\varphi_{3}(a)=\int_{0}^{\infty}K_{0}^{3}(ax)\,dx$

with an auxiliary parameter $a$ that naturally can be scaled out.

The evaluation begins with a more general problem

(6.11)
$I=I(a,b;\mu,\nu,\alpha)=\int_{0}^{\infty}K_{\mu}(ax)K_{\nu}(ax)K_{\alpha}(bx)\,dx$

and then

(6.12)
$\varphi_{3}(a)=\lim\limits_{\stackrel{{\scriptstyle\alpha=\mu=\nu\rightarrow
0}}{{b\rightarrow a}}}I(a,b;\mu,\nu,\alpha).$

The Mellin-Barnes representations of the factors in the integrand

(6.13) $K_{\mu}(ax)K_{\nu}(ax)=\\\ \frac{1}{8\pi
i}\int_{\gamma}\Gamma\left(t+\frac{\mu+\nu}{2}\right)\Gamma\left(t+\frac{\mu-\nu}{2}\right)\Gamma\left(t-\frac{\mu+\nu}{2}\right)\Gamma\left(t-\frac{\mu-\nu}{2}\right)\left(\frac{ax}{2}\right)^{-2t}\frac{dt}{\Gamma(2t)},$

and

(6.14) $K_{\alpha}(bx)=\frac{1}{4\pi
i}\left(\frac{bx}{2}\right)^{\alpha}\int_{\gamma}\Gamma(s)\Gamma(s-\alpha)\left(\frac{bx}{2}\right)^{-2s}\,ds.$

Replacing in (6.11) gives

(6.15) $I=\frac{1}{8(2\pi
i)^{2}}\int_{\gamma_{1}}\int_{\gamma_{2}}\frac{\Gamma\left(t+\frac{\mu+\nu}{2}\right)\Gamma\left(t+\frac{\mu-\nu}{2}\right)\Gamma\left(t-\frac{\mu+\nu}{2}\right)\Gamma\left(t-\frac{\mu-\nu}{2}\right)\Gamma(s)\Gamma(s-\alpha)}{a^{2t}b^{2s-\alpha}2^{-2t-2s+\alpha}\Gamma(2t)}\\\
\times\langle-2s-2t+\alpha+1\rangle\,dt\,ds.$

Now replace the gamma factors in the denominator by their corresponding
bracket series to obtain

$\displaystyle I$ $\displaystyle=$
$\displaystyle\frac{1}{8}\sum_{\\{n\\}}\phi_{n_{1}\cdots n_{6}}\frac{1}{(2\pi
i)^{2}}\int_{\gamma_{1}}\int_{\gamma_{2}}\frac{1}{a^{2t}b^{2s-\alpha}2^{-2t-2s+\alpha}\Gamma(2t)}$
$\displaystyle\times$ $\displaystyle\langle
t+\frac{\mu+\nu}{2}+n_{1}\rangle\langle
t+\frac{\mu-\nu}{2}+n_{2}\rangle\langle t-\frac{\mu+\nu}{2}+n_{3}\rangle$
$\displaystyle\times$ $\displaystyle\langle
t-\frac{\mu-\nu}{2}+n_{4}\rangle\langle s+n_{5}\rangle\langle
s-\alpha+n_{6}\rangle\langle-2s-2t+\alpha+1\rangle\,dt\,ds.$

To evaluate this expression choose to eliminate the sums with indices $n_{1}$
and $n_{5}$. The resulting sums now depend on four indices
$n_{2},\,n_{3},\,n_{4}$ and $n_{6}$ and the variables of integration $t$ and
$s$ take the values

(6.17) $t^{*}=-\frac{\mu+\nu}{2}-n_{1}\quad\textnormal{and}\,\,s^{*}=-n_{5}.$

This yields

(6.18) $I=\frac{1}{8}\sum_{\\{n\\}}\phi_{n_{2}n_{3}n_{4}n_{6}}\times\\\
\frac{\langle t^{*}+\frac{\mu-\nu}{2}+n_{2}\rangle\langle
t^{*}-\frac{\mu+\nu}{2}+n_{3}\rangle\langle
t^{*}-\frac{\mu-\nu}{2}+n_{4}\rangle\langle
s^{*}-\alpha+n_{6}\rangle\langle-2s^{*}-2t^{*}+\alpha+1\rangle}{a^{2t^{*}}b^{2s^{*}-\alpha}2^{-2t^{*}-2s^{*}+\alpha}\Gamma(2t^{*})}.$

Under the assumption $|4a^{2}|<|b^{2}|$ the integral in (6.11) is expressed as

(6.19) $I=I(a,b;\mu,\nu,\alpha)=T_{1}+T_{2}+T_{3}+T_{4}$

with

$\displaystyle\quad T_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{8}\frac{a^{\mu+\nu}}{b^{1+\mu+\nu}}\Gamma\left(\frac{1-\alpha+\mu+\nu}{2}\right)\Gamma\left(\frac{1+\alpha+\mu+\nu}{2}\right)\Gamma(-\mu)\Gamma(-\nu)$
$\displaystyle\quad\quad\times{}_{4}F_{3}\left(\genfrac{}{}{0.0pt}{}{1+\tfrac{\mu+\nu}{2}\,\,\tfrac{1+\mu+\nu}{2}\,\,\frac{1+\alpha+\mu+\nu}{2}\,\,\frac{1-\alpha+\mu+\nu}{2}}{1+\mu\,\,1+\nu\,\,1+\mu+\nu}\bigg{|}{\frac{4a^{2}}{b^{2}}}\right)$
$\displaystyle\quad T_{2}$ $\displaystyle=$
$\displaystyle\frac{1}{8}\frac{a^{\mu-\nu}}{b^{1+\mu-\nu}}\Gamma\left(\frac{1-\alpha+\mu-\nu}{2}\right)\Gamma\left(\frac{1+\alpha+\mu-\nu}{2}\right)\Gamma(-\mu)\Gamma(\nu)$
$\displaystyle\quad\quad\times{}_{4}F_{3}\left(\genfrac{}{}{0.0pt}{}{1+\tfrac{\mu-\nu}{2}\,\,\tfrac{1+\mu-\nu}{2}\,\,\frac{1-\alpha+\mu-\nu}{2}\,\,\frac{1+\alpha+\mu-\nu}{2}}{1+\mu\,\,1-\nu\,\,1+\mu-\nu}\bigg{|}{\frac{4a^{2}}{b^{2}}}\right)$
$\displaystyle\quad T_{3}$ $\displaystyle=$
$\displaystyle\frac{1}{8}\frac{b^{\mu+\nu-1}}{a^{\mu+\nu}}\Gamma\left(\frac{1-\alpha-\mu-\nu}{2}\right)\Gamma\left(\frac{1+\alpha-\mu-\nu}{2}\right)\Gamma(\mu)\Gamma(\nu)$
$\displaystyle\quad\quad\times{}_{4}F_{3}\left(\genfrac{}{}{0.0pt}{}{1-\tfrac{\mu+\nu}{2}\,\,\tfrac{1-\mu-\nu}{2}\,\,\frac{1-\alpha-\mu-\nu}{2}\,\,\frac{1+\alpha-\mu-\nu}{2}}{1-\mu\,\,1-\nu\,\,1-\mu-\nu}\bigg{|}{\frac{4a^{2}}{b^{2}}}\right)$
$\displaystyle\quad T_{4}$ $\displaystyle=$
$\displaystyle\frac{1}{8}\frac{b^{\mu-\nu-1}}{a^{\mu-\nu}}\Gamma\left(\frac{1-\alpha-\mu+\nu}{2}\right)\Gamma\left(\frac{1+\alpha-\mu+\nu}{2}\right)\Gamma(\mu)\Gamma(-\nu)$
$\displaystyle\quad\quad\times{}_{4}F_{3}\left(\genfrac{}{}{0.0pt}{}{1-\tfrac{\mu-\nu}{2}\,\,\tfrac{1-\mu+\nu}{2}\,\,\frac{1-\alpha-\mu+\nu}{2}\,\,\frac{1+\alpha-\mu+\nu}{2}}{1-\mu\,\,1+\nu\,\,1-\mu+\nu}\bigg{|}{\frac{4a^{2}}{b^{2}}}\right).$

Now passing to the limit as $\alpha,\,\nu,\,\mu\rightarrow 0$ yields

$\displaystyle\int_{0}^{\infty}K_{0}^{2}(ax)K_{0}(bx)\,dx$ $\displaystyle=$
$\displaystyle\lim\limits_{\mu\rightarrow
0}\frac{1}{8b}\left[\left(\frac{a^{2}}{b^{2}}\right)^{\mu}\Gamma\left(\frac{1+2\mu}{2}\right)^{2}\Gamma(-\mu)^{2}{}_{3}F_{2}\left(\genfrac{}{}{0.0pt}{}{\tfrac{1+2\mu}{2}\,\,\tfrac{1+2\mu}{2}\,\,\tfrac{1+2\mu}{2}}{1+\mu\,\,1+2\mu}\bigg{|}{\frac{4a^{2}}{b^{2}}}\right)\right.$
$\displaystyle\quad\quad+2\pi\Gamma(-\mu)\Gamma(\mu){}_{3}F_{2}\left(\genfrac{}{}{0.0pt}{}{\tfrac{1}{2}\,\,\,\tfrac{1}{2}\,\,\,\tfrac{1}{2}}{1+\mu\,\,1-\mu}\bigg{|}{\frac{4a^{2}}{b^{2}}}\right)$
$\displaystyle\left.\quad\quad+\left(\frac{a^{2}}{b^{2}}\right)^{-\mu}\Gamma^{2}\left(\tfrac{1-2\mu}{2}\right)\Gamma^{2}(\mu){}_{3}F_{2}\left(\genfrac{}{}{0.0pt}{}{\tfrac{1-2\mu}{2}\,\,\tfrac{1-2\mu}{2}\,\,\tfrac{1-2\mu}{2}}{1-\mu\,\,\,1-2\mu}\bigg{|}{\frac{4a^{2}}{b^{2}}}\right)\right]$

In a similar form, in the case $|4a^{2}|>|b^{2}|$ one obtains

(6.21) $I=I(a,b;\mu,\nu,\alpha)=T_{5}+T_{6}$

with

$\displaystyle\quad T_{5}$ $\displaystyle=$
$\displaystyle\frac{1}{8}\frac{b^{\alpha}}{a^{\alpha+1}}$
$\displaystyle\frac{\Gamma(-\alpha)}{\Gamma(\alpha+1)}\Gamma\left(\frac{1+\alpha-\mu+\nu}{2}\right)\Gamma\left(\frac{1+\alpha-\mu-\nu}{2}\right)\Gamma\left(\frac{1+\alpha+\mu-\nu}{2}\right)\Gamma\left(\frac{1+\alpha+\mu+\nu}{2}\right)$
$\displaystyle\quad\quad\times{}_{4}F_{3}\left(\genfrac{}{}{0.0pt}{}{\tfrac{1+\alpha+\mu+\nu}{2}\,\,\tfrac{1+\alpha-\mu+\nu}{2}\,\,\frac{1+\alpha+\mu-\nu}{2}\,\,\frac{1+\alpha-\mu-\nu}{2}}{1+\alpha\,\,1+\tfrac{\alpha}{2}\,\,\tfrac{1+\alpha}{2}}\bigg{|}{\frac{b^{2}}{4a^{2}}}\right)$
$\displaystyle\quad T_{6}$ $\displaystyle=$
$\displaystyle\frac{1}{8}\frac{a^{\alpha-1}}{b^{\alpha}}$
$\displaystyle\frac{\Gamma(\alpha)}{\Gamma(1-\alpha)}\Gamma\left(\frac{1-\alpha-\mu+\nu}{2}\right)\Gamma\left(\frac{1-\alpha-\mu-\nu}{2}\right)\Gamma\left(\frac{1-\alpha+\mu-\nu}{2}\right)\Gamma\left(\frac{1-\alpha+\mu+\nu}{2}\right)$
$\displaystyle\quad\quad\times{}_{4}F_{3}\left(\genfrac{}{}{0.0pt}{}{\tfrac{1-\alpha+\mu+\nu}{2}\,\,\tfrac{1-\alpha-\mu+\nu}{2}\,\,\frac{1-\alpha+\mu-\nu}{2}\,\,\frac{1-\alpha-\mu-\nu}{2}}{1-\alpha\,\,1-\tfrac{\alpha}{2}\,\,\tfrac{1-\alpha}{2}}\bigg{|}{\frac{b^{2}}{4a^{2}}}\right)$

and then letting $\mu=\nu=0,\,b=a$ and $\alpha\rightarrow 0$ yields, after
scaling the parameter $a$

(6.22) $\int_{0}^{\infty}K_{0}^{3}(x)=\lim\limits_{\alpha\rightarrow
0}\frac{1}{8}\left[\frac{\Gamma(-\alpha)\Gamma^{4}\left(\frac{1+\alpha}{2}\right)}{\Gamma(1+\alpha)}{}_{3}F_{2}\left(\genfrac{}{}{0.0pt}{}{\tfrac{1+\alpha}{2}\,\,\tfrac{1+\alpha}{2}\,\,\tfrac{1+\alpha}{2}}{1+\alpha\,\,1+\tfrac{\alpha}{2}}\bigg{|}{\frac{1}{4}}\right)\right.\\\
\left.+\frac{\Gamma(\alpha)\Gamma^{4}\left(\tfrac{1-\alpha}{2}\right)}{\Gamma(1-\alpha)}{}_{3}F_{2}\left(\genfrac{}{}{0.0pt}{}{\tfrac{1-\alpha}{2}\,\,\tfrac{1-\alpha}{2}\,\,\tfrac{1-\alpha}{2}}{1-\alpha\,\,1-\tfrac{\alpha}{2}}\bigg{|}{\frac{1}{4}}\right)\right].$

The authors have been unable to produce a simpler analytic expression for this
limiting value.

######  Example 6.5.

A similar argument as the one presented in the previous example yields

(6.23)
$\int_{0}^{\infty}K_{0}^{4}(x)\,dx=\frac{\sqrt{\pi}}{16}\lim\limits_{\alpha\rightarrow
0}\left[\frac{\Gamma^{2}(-\alpha)\Gamma^{2}\left(\alpha+\tfrac{1}{2}\right)\Gamma\left(2\alpha+\tfrac{1}{2}\right)}{\Gamma(2\alpha+1)}{}_{4}F_{3}\left(\genfrac{}{}{0.0pt}{}{\tfrac{1}{2}\,\,\alpha+\tfrac{1}{2}\,\,\alpha+\tfrac{1}{2}\,\,2\alpha+\tfrac{1}{2}}{1+\alpha\,\,1+\alpha\,\,1+2\alpha}\bigg{|}{1}\right)\right.\\\
\left.+2\sqrt{\pi}\Gamma(-\alpha)\Gamma(\alpha)\Gamma(\tfrac{1}{2}+\alpha)\Gamma(\tfrac{1}{2}-\alpha){}_{4}F_{3}\left(\genfrac{}{}{0.0pt}{}{\tfrac{1}{2}\,\,\,\,\tfrac{1}{2}\,\,\,\,\tfrac{1}{2}+\alpha\,\,\,\,\tfrac{1}{2}-\alpha}{1\,\,\,\,1+\alpha\,\,\,\,1-\alpha}\bigg{|}{1}\right)\right.\\\
\left.+\frac{\Gamma^{2}(\alpha)\Gamma^{2}(\tfrac{1}{2}-\alpha)\Gamma(\tfrac{1}{2}-2\alpha)}{\Gamma(1-2\alpha)}{}_{4}F_{3}\left(\genfrac{}{}{0.0pt}{}{\tfrac{1}{2}\,\,\,\,\tfrac{1}{2}-\alpha\,\,\,\,\tfrac{1}{2}-\alpha\,\,\,\,\tfrac{1}{2}-2\alpha}{1-\alpha\,\,\,\,1-\alpha\,\,\,\,1-2\alpha}\bigg{|}{1}\right)\right]$

Details are omitted.

##  7\. Mellin transforms of products

This section presents a method to evaluate the Mellin transform

(7.1) $I(s,b)=\int_{0}^{\infty}x^{s-1}f(x)g(bx)\,dx.$

knowing a series for $f$ of the form

(7.2) $f(x)=\sum_{n}\phi_{n}F(n)x^{\beta n}$

and the inverse Mellin transform

(7.3) $g(x)=\frac{1}{2\pi i}\int_{\gamma}x^{-s}\varphi(s)\,ds.$

For $\kappa>0$, the change of variables $s=\kappa s^{\prime}$ gives

(7.4) $g(x)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}x^{-\kappa
s}\widetilde{\varphi}(s)\,ds,$

where $\widetilde{\varphi}(s)=\kappa\varphi(\kappa s)$. The formula (7.4) is
now written as

(7.5) $g(x)=\frac{1}{2\pi i}\int_{\gamma}x^{-\kappa s}\varphi(s)\,ds,$

that is, the tilde notation is dropped and the parameter $\kappa$ is kept.

Then

$\displaystyle I$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}x^{\alpha-1}f(x)g(bx)\,dx$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi i}\int_{\gamma}\varphi(s)b^{-\kappa
s}\left[\sum_{n}\phi_{n}F(n)\left[\int_{0}^{\infty}x^{\alpha+\beta n-\kappa
s-1}\,dx\right]\right]\,ds$ $\displaystyle=$ $\displaystyle\frac{1}{2\pi
i}\int_{\gamma}\varphi(s)b^{-\kappa
s}\left[\sum_{n}\phi_{n}F(n)\langle\alpha+\beta n-\kappa s\rangle\right]\,ds$
$\displaystyle=$ $\displaystyle\frac{1}{2\pi
i}\int_{\gamma}\frac{\varphi(s)b^{-\kappa
s}}{|\beta|}\left[\sum_{n}\phi_{n}F(n)\left\langle\frac{\alpha}{\beta}-\frac{\kappa
s}{\beta}+n\right\rangle\right]\,ds$ $\displaystyle=$
$\displaystyle\frac{1}{|\beta|}\frac{1}{2\pi
i}\int_{\gamma}\varphi(s)b^{-\kappa s}\Gamma\left(\frac{\alpha-\kappa
s}{\beta}\right)F\left(\frac{\kappa s-\alpha}{\beta}\right)\,ds.$

This is stated as a theorem.

######  Theorem 7.1.

Assume the function $f(x)$ has an expansion given by

(7.7) $f(x)=\sum_{n}\phi_{n}F(n)x^{\beta n}$

and the function $g(x)$ is given by rescaled version of the inverse Mellin
transform

(7.8) $g(x)=\frac{1}{2\pi i}\int_{\gamma}\varphi(s)x^{-\kappa s}\,ds.$

Then

$\int_{0}^{\infty}x^{\alpha-1}f(x)g(bx)\,dx=\frac{1}{|\beta|}\frac{1}{2\pi
i}\int_{\gamma}\varphi(s)\Gamma\left(\frac{\alpha-\kappa
s}{\beta}\right)F\left(\frac{\kappa s-\alpha}{\beta}\right)b^{-\kappa s}\,ds.$

######  Example 7.2.

Entry $6.532.4$ in [15] states that

(7.9) $\int_{0}^{\infty}\frac{x\,J_{0}(Ax)\,dx}{x^{2}+k^{2}}=K_{0}(Ak).$

Theorem 7.1 is now used to establish this evaluation.

Start with the expansion

(7.10)
$f(x)=\frac{1}{x^{2}+k^{2}}=\sum_{n}\phi_{n}\left[\Gamma(n+1)k^{-2n-2}\right]x^{2n}.$

This gives (7.7) with $F(n)=\Gamma(n+1)k^{-2n-2}$ and $\beta=2$.

Now use [18, 10.9.23]

(7.11) $J_{\nu}(z)=\frac{1}{2\pi
i}\int_{\gamma}\frac{\Gamma(t)}{\Gamma(\nu-t+1)}\left(\frac{z}{2}\right)^{\nu-2t}\,dt,$

and replace $z$ by $Ax$ to produce the desired representation of the Bessel
function:

(7.12) $J_{0}(Ax)=\frac{1}{2\pi
i}\int_{\gamma}\frac{(Ax)^{2s}\Gamma(s)}{2^{2s}\Gamma(1-s)}\,ds.$

In the notation of (7.8), the parameters are $\kappa=2,\,\beta=2,\,\alpha=2$
and $b=A$ and $\displaystyle\varphi(s)=\frac{\Gamma(s)}{2^{2s}\,\Gamma(1-s)}$.
Theorem 7.1 now gives

(7.13) $\int_{0}^{\infty}\frac{xJ_{0}(Ax)}{x^{2}+k^{2}}\,dx=\frac{1}{4\pi
i}\int_{\gamma}\Gamma^{2}(s)\left(\frac{Ak}{2}\right)^{-2s}\,ds.$

Now the formula [18, 10.32.13])

(7.14) $K_{\nu}(z)=\frac{1}{4\pi
i}\left(\frac{z}{2}\right)^{\nu}\int_{\gamma}\Gamma(t)\Gamma(t-\nu)\left(\frac{z}{2}\right)^{-2t}\,dt,$

In particular for $\nu=0$ this becomes

(7.15) $K_{0}(z)=\frac{1}{4\pi
i}\int_{\gamma}\Gamma^{2}(t)\left(\frac{z}{2}\right)^{-2t}\,dt.$

Formula (7.9) now follows from (7.13) and (7.15).

######  Example 7.3.

The next evaluation is

(7.16)
$I=\int_{0}^{\infty}xK_{\nu}(ax)J_{\nu}(bx)\,dx=\frac{b^{\nu}}{a^{\nu}(a^{2}+b^{2})}.$

This is entry $6.521.2$ in [15].

Formulas (5.24) and (5.46) give, respectively, the representations

(7.17) $K_{\nu}(z)=\frac{1}{4\pi
i}\int_{\gamma}\Gamma(s)\Gamma(s-\nu)\,\left(\frac{z}{2}\right)^{-2s+\nu}\,ds$

and

(7.18) $J_{\nu}(z)=\frac{1}{2\pi
i}\int_{\gamma}\frac{\Gamma(-s)}{\Gamma(s+\nu+1)}\left(\frac{z}{2}\right)^{2s+\nu}\,ds.$

Replacing in (7.16) and recognizing the $x$-integral into a bracket yields

(7.19) $I=\frac{1}{2(2\pi
i)^{2}}\left(\frac{ab}{4}\right)^{\nu}\int_{\gamma_{1}}\int_{\gamma_{2}}\frac{\Gamma(-t)\Gamma(s)\Gamma(s-\nu)}{\Gamma(t+\nu+1)}\left(\frac{a}{2}\right)^{-2s}\left(\frac{b}{2}\right)^{2t}\\\
\times\frac{1}{2}\langle 1+\nu+t-s\rangle\,ds\,dt.$

The bracket is now used to eliminate the $s$-integral, the result is

(7.20) $I=\frac{1}{a^{2}}\left(\frac{b}{a}\right)^{\nu}\frac{1}{2\pi
i}\int_{\gamma}\Gamma(-t)\Gamma(1+t)\left(\frac{b^{2}}{a^{2}}\right)^{t}\,dt.$

The gamma terms are expanded as bracket series using (4.1) and then eliminate
the line integral with one of the brackets to obtain

$\displaystyle I$ $\displaystyle=$
$\displaystyle\frac{b^{\nu}}{a^{\nu+2}}\frac{1}{2\pi
i}\int_{\gamma}\sum_{n_{1},n_{2}}\left(\frac{b^{2}}{a^{2}}\right)^{t}\phi_{n_{1}n_{2}}\langle
n_{1}-t\rangle\langle n_{2}+1+t\rangle\,dt$ $\displaystyle=$
$\displaystyle\frac{b^{\nu}}{a^{\nu+2}}\sum_{n_{1},n_{2}}\phi_{n_{1}n_{2}}\left(\frac{b^{2}}{a^{2}}\right)^{n_{1}}\langle
n_{2}+n_{1}+1\rangle.$

The method of brackets now gives two different expressions obtained by using
$n_{1}$ or $n_{2}$ as the free index of summation. These options give series
converging in disjoint regions $|b|<|a|$ and $|b|>|a|$. The rules of the
method of brackets shows that each sum gives the value of the integral in the
corresponding region. In this case, both cases give the same expression

(7.22) $I=\frac{b^{\nu}}{a^{\nu}(a^{2}+b^{2})},$

for the value of the integral. This confirms (7.16).

##  8\. An integral involving exponentials and Bessel modified functions

The method developed in this work is now applied to the computation of some
definite integrals. The idea is relatively simple: given a function $f(x)$
with a Mellin transform $\varphi(s)$ containing gamma factors, the integral

(8.1) $F=\int_{0}^{\infty}e^{-x}f(x)\,dx$

can be evaluated by writing the exponential as

(8.2) $e^{-x}=\sum_{n}\phi_{n}x^{n}$

and then using the method of brackets. The examples below illustrates this
process.

Integrals involving the Bessel function $K_{\nu}(x)$ have been presented in
[9]. The computation there is based on the concept of totally null/divergent
representations. The first type includes

(8.3)
$K_{0}(x)=\frac{1}{x}\sum_{n=0}^{\infty}\frac{\Gamma^{2}(n+\tfrac{1}{2})}{n!\,\Gamma(-n)}\left(-\frac{4}{x^{2}}\right)^{n}$

in which every term vanishes. There is a similar expression for a series in
which every term diverges. In spite of the lack of rigor, these series have
shown to be useful in the evaluation of definite integrals. See [9] for
details. A second technique used before is the integral representation of
$K_{\nu}(x)$ such as

(8.4)
$K_{\nu}(x)=\frac{2^{\nu}\Gamma\left(\nu+\tfrac{1}{2}\right)}{\Gamma(\tfrac{1}{2})}x^{\nu}\int_{0}^{\infty}\frac{\cos
t\,dt}{(x^{2}+t^{2})^{\nu+1/2}},$

and then apply the methods of brackets. The method present next is an
improvement over those employed in [9].

######  Example 8.1.

Consider the integral

(8.5) $F(b,\nu)=\int_{0}^{\infty}e^{-x}K_{\nu}(bx)\,dx.$

Naturally the parameter $b$ can be scaled out, but it is instructive to leave
it as is.

Write the exponential function as in (8.2) and the Bessel function from (5.24)
as

(8.6) $K_{\nu}(bx)=\frac{1}{4\pi
i}\left(\frac{bx}{2}\right)^{\nu}\int_{-i\infty}^{i\infty}\Gamma(s)\Gamma(s-\nu)\left(\frac{bx}{2}\right)^{-2s}\,ds.$

Then

$\displaystyle F(b,\nu)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{n_{1}}\phi_{n_{1}}\frac{1}{2\pi
i}\int_{\gamma}\left(\frac{b}{2}\right)^{\nu-2s}\Gamma(s)\Gamma(s-\nu)\left(\int_{0}^{\infty}x^{n_{1}-2s+\nu}\,dx\right)\,ds$
$\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{n_{1}}\phi_{n_{1}}\frac{1}{2\pi
i}\int_{\gamma}\left(\frac{b}{2}\right)^{\nu-2s}\Gamma(s)\Gamma(s-\nu)\langle
n_{1}-2s+\nu+1\rangle\,ds$

Now replace the gamma factors by their bracket expansions as in (4.1) to
produce

$F(b,\nu)=\frac{1}{2}\sum_{n_{1}}\phi_{123}\frac{1}{2\pi
i}\int_{\gamma}\left(\frac{b}{2}\right)^{\nu-2s}\langle
s+n_{2}\rangle\,\langle s-\nu+n_{3}\rangle\,\langle n_{1}-2s+\nu+1\rangle\,ds$

Using the bracket $\langle s+n_{2}\rangle$ to apply Theorem 3.1 gives a
bracket series for the integral $F(b,\nu)$:

(8.7)
$F(b,\nu)=\frac{1}{2}\sum_{\vec{n}}\phi_{123}\left(\frac{b}{2}\right)^{\nu+2n_{2}}\langle-
n_{2}-\nu+n_{3}\rangle\langle n_{1}+2n_{2}+\nu+1\rangle,$

where $\vec{n}=\\{n_{1},n_{2},n_{3}\\}$.

The method of brackets is now used to produce three sums for (8.7):

(8.8) $\displaystyle T_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\Gamma\left(\frac{1-\nu+n}{2}\right)\Gamma\left(\frac{1+\nu+n}{2}\right)\left(\frac{b}{2}\right)^{-n-1}$
(8.9) $\displaystyle T_{2}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\Gamma(-n-\nu)\Gamma(1+2n+\nu)\left(\frac{b}{2}\right)^{\nu+2n}$
(8.10) $\displaystyle T_{3}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\Gamma(\nu-n)\Gamma(1+2n-\nu)\left(\frac{b}{2}\right)^{-\nu+2n}$

Observe that $T_{1}$ diverges since it is of type ${{}_{2}}F_{0}$ (this sum
plays a role in the asymptotic study of the integral when
$b\rightarrow\infty$, not described here) and the series $T_{2},\,T_{3}$
converge when $|b|<1$. The method of brackets now states that
$F(b,\nu)=T_{2}+T_{3}$, when $|b|<1$. Under this assumption and since
$T_{3}(b,\nu)=T_{2}(b,-\nu)$, it suffices to obtain an expression for
$T_{2}(b,\nu)$. This is the next result.

######  Proposition 8.2.

The function $T_{2}(b,\nu)$ is given by

(8.11)
$T_{2}(b,\nu)=-\frac{\pi}{2\sin(\pi\nu)}\frac{b^{\nu}(1+\sqrt{1-b^{2}})^{-\nu}}{\sqrt{1-b^{2}}}.$

###### Proof.

The proof is divided in a sequence of steps.

Step 1. The function $T_{2}(b,\nu)$ is given by

(8.12) $T_{2}(b,\nu)=-\frac{\pi
b^{\nu}}{2^{\nu+1}\sin(\pi\nu)}\sum_{n=0}^{\infty}\frac{\left(\frac{1+\nu}{2}\right)_{n}\,\left(1+\frac{\nu}{2}\right)_{n}}{(1+\nu)_{n}}\,\frac{b^{2n}}{n!}.$

Proof. In the definition of $T_{2}(b,\nu)$ use
$(w)_{2n}=2^{2n}\left(\frac{w}{2}\right)_{n}\,\left(\frac{w+1}{2}\right)_{n},$

$\Gamma(-n-\nu)=\frac{(-1)^{n}\Gamma(-\nu)}{(1+\nu)_{n}},\,\,\Gamma(1+\nu+2n)=\Gamma(1+\nu)2^{2n}\left(\frac{1+\nu}{2}\right)_{n}\left(1+\frac{\nu}{2}\right)_{n}$

and $\Gamma(-\nu)\Gamma(1+\nu)=-\pi/\sin(\pi\nu)$, to obtain the result.

This identity shows that the statement of the Proposition is equivalent to
proving

(8.13)
$\frac{1}{2^{\nu}}\sum_{n=0}^{\infty}\frac{\left(\frac{1+\nu}{2}\right)_{n}\,\left(1+\frac{\nu}{2}\right)_{n}}{(1+\nu)_{n}}\frac{b^{2n}}{n!}=\frac{(1+\sqrt{1-b^{2}}\,\,)^{-\nu}}{\sqrt{1-b^{2}}}.$

Step 2. Formula (2.5.16) in H. Wilf [24]

(8.14)
$\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{k}=\sum_{n=0}^{\infty}\frac{k\,(2n+k-1)!}{n!\,(n+k)!}x^{n}$

and the identity $(1+\sqrt{1-b^{2}})(1-\sqrt{1-b^{2}})=b^{2}$ shows, after
some elementary simplifications, that the statement of the Proposition is
equivalent to

(8.15)
$\sum_{n=0}^{\infty}\frac{\left(\frac{1+\nu}{2}\right)_{n}\,\left(1+\frac{\nu}{2}\right)_{n}}{(1+\nu)_{n}}\frac{c^{n}}{n!}=\frac{1}{\sqrt{1-c}}\sum_{n=0}^{\infty}\frac{\left(\frac{\nu}{2}\right)_{n}\,\left(1+\frac{\nu}{2}\right)_{n}}{(1+\nu)_{n}}\frac{c^{n}}{n!},$

where $c=b^{2}$.

Step 3. The identity at the end of Step 2 is equivalent to

(8.16)
${}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1+\nu}{2}\,\,1+\frac{\nu}{2}}{1+\nu}\bigg{|}{u}\right)={}_{1}F_{0}\left(\genfrac{}{}{0.0pt}{}{\tfrac{1}{2}}{-}\bigg{|}{u}\right)\times{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{1+\nu}{2}\,\,\frac{\nu}{2}}{1+\nu}\bigg{|}{u}\right).$

Proof. Simply use the binomial theorem

(8.17) $(1-t)^{-\alpha}=\sum_{n=0}^{\infty}\frac{(\alpha)_{n}}{n!}t^{n}.$

with $\alpha=1/2$.

Step 4. The conclusion of the Proposition is equivalent to the identity

(8.18)
$\sum_{j=0}^{n}\frac{\binom{n}{j}}{(1+\nu)_{n-j}}\left(\frac{1}{2}\right)_{j}\left(\frac{1+\nu}{2}\right)_{n-j}\left(\frac{\nu}{2}\right)_{n-j}=\frac{\left(\frac{\nu}{2}\right)_{n}\,\left(1+\frac{\nu}{2}\right)_{n}}{(1+\nu)_{n}},$

for every $n\in\mathbb{N}$.

∎

###### Proof.

The umbral method [8, 20] shows that the identity is equivalent to the one
formed by replacing Pochhammer symbols $(a)_{k}$ by $a^{k}$. In this case,
(8.18) becomes

(8.19) $\sum_{j=0}^{n}\binom{n}{j}2^{j}\nu^{n-j}=(\nu+2)^{n},$

and this follows from the binomial theorem. ∎

###### Proof.

An alternative proof of (8.18) is presented next. Expressing the Pochhammer
symbols in terms of binomial coefficients, it is routine to check that the
desired identity is equivalent to

(8.20)
$\sum_{j=0}^{n}\binom{2j}{j}\binom{\nu+2(n-j)}{n-j}\frac{\nu}{\nu+2(n-j)}=\binom{\nu+2n}{n}.$

This identity is interpreted as the coefficient of $x^{n}$ in the product of
the two series

(8.21) $A(x)=\sum_{j=0}^{\infty}\binom{2j}{j}x^{j}\quad\text{and}\quad
B(x)=\sum_{k=0}^{\infty}\nu\binom{\nu+2k}{k}\frac{x^{k}}{\nu+2k}.$

The first sum is given by the binomial theorem as

(8.22) $A(x)=\frac{1}{\sqrt{1-4x}}.$

To obtain an analytic expression for $B(x)$ start with entry $2.5.15$ in [24]

(8.23)
$\frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{\nu}=\sum_{n=0}^{\infty}\binom{\nu+2k}{k}x^{k}$

where the term in brackets is the generating function of the Catalan numbers.
Some elementary manipulations give

(8.24) $B(x)=\nu
x^{-\nu/2}\int_{0}^{\sqrt{x}}\frac{t^{\nu-1}}{\sqrt{1-4t^{2}}}\left(\frac{1-\sqrt{1-4t^{2}}}{2t^{2}}\right)^{\nu}\,dt.$

Then (8.20) is equivalent to the identity

(8.25)
$\int_{0}^{\sqrt{x}}\frac{t^{\nu-1}}{\sqrt{1-4t^{2}}}\left(\frac{1-\sqrt{1-4t^{2}}}{2t^{2}}\right)^{\nu}\,dt=\frac{1}{\nu}x^{\nu/2}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{\nu}.$

This can now be verified by observing that both sides vanish at $x=0$ and a
direct computation shows that the derivatives match. ∎

The integral is stated next. It appears as entry $6.611.3$ in [15].

######  Corollary 8.1.

For $\nu,\,b\in\mathbb{R}$

(8.26) $\int_{0}^{\infty}e^{-x}K_{\nu}(bx)\,dx=\\\
\frac{\pi}{2b^{\nu}\sin(\pi\nu)}\left[\frac{(1+\sqrt{1-b^{2}})^{\nu}-(1-\sqrt{1-b^{2}})^{\nu}}{\sqrt{1-b^{2}}}\right].$

In particular, as $b\rightarrow 1$,

(8.27) $\int_{0}^{\infty}e^{-x}K_{\nu}(x)\,dx=\\\
\frac{\pi\nu}{\sin(\pi\nu)},$

and as $\nu\rightarrow 0$,

(8.28)
$\int_{0}^{\infty}e^{-x}K_{0}(bx)\,dx=\frac{\log(1+\sqrt{1-b^{2}})-\log(1-\sqrt{1-b^{2}})}{2\sqrt{1-b^{2}}}.$

Finally, letting $\nu\rightarrow 0$ and $b\rightarrow 1$ gives

(8.29) $\int_{0}^{\infty}e^{-x}K_{0}(x)\,dx=1.$

##  9\. Flexibility of the method of brackets

This final section illustrates the flexibility of the method of brackets. To
achieve this goal, we present four different evaluations of the integral

(9.1) $I(a,b)=\int_{0}^{\infty}\frac{e^{-a^{2}x^{2}}\,dx}{x^{2}+b^{2}}.$

The parameters $a,\,b$ are assumed to be real and positive. This formula
appears as entry $3.466.1$ in [15] with value

(9.2) $I(a,b)=\frac{\pi}{2b}e^{a^{2}b^{2}}(1-\textnormal{Erf}(ab))$

and the reader will find in [1] an elementary proof of it. This section
presents $4$ different ways to use the method of brackets to evaluate this
integral.

Method 1. Start with the bracket series representations

(9.3) $\displaystyle\exp(-a^{2}x^{2})$ $\displaystyle=$
$\displaystyle\sum_{n_{1}}\phi_{n_{1}}a^{2n_{1}}x^{2n_{1}}$
$\displaystyle\frac{1}{x^{2}+b^{2}}$ $\displaystyle=$
$\displaystyle\sum_{n_{2}}\sum_{n_{3}}\phi_{n_{2}n_{3}}b^{2n_{2}}x^{2n_{3}}\langle
1+n_{2}+n_{3}\rangle$

and produce the bracket series

(9.4)
$I(a,b)=\sum_{n_{1}}\sum_{n_{2}}\sum_{n_{3}}\phi_{n_{1}n_{2}+n_{3}}a^{2n_{1}}b^{2n_{2}}\langle
1+n_{2}+n_{3}\rangle\,\langle 2n_{1}+2n_{3}+1\rangle.$

Method 2. The second form begins with the Mellin-Barnes representations

(9.5) $\displaystyle\exp(-a^{2}x^{2})$ $\displaystyle=$
$\displaystyle\frac{1}{4\pi
i}\int_{\gamma}x^{-t}a^{-t}\Gamma\left(\frac{t}{2}\right)\,dt$
$\displaystyle\frac{1}{x^{2}+b^{2}}$ $\displaystyle=$
$\displaystyle\frac{1}{4\pi
b^{2}i}\int_{\gamma}x^{-s}b^{s}\Gamma\left(\frac{s}{2}\right)\Gamma\left(1-\frac{s}{2}\right)\,ds$

Replacing in (9.1) a bracket appears and one obtains the representation

(9.6) $I(a,b)=\frac{1}{4b^{2}(2\pi
i)^{2}}\int_{\gamma_{1}}\int_{\gamma_{2}}a^{-t}b^{s}\Gamma\left(\tfrac{t}{2}\right)\Gamma\left(\tfrac{s}{2}\right)\Gamma\left(1-\tfrac{s}{2}\right)\langle
1-t-s\rangle\,ds\,dt.$

The bracket is used to evaluate the $s$-integral to produce $s^{*}=1-t$ and
the expression

(9.7) $I(a,b)=\frac{1}{4b^{2}(2\pi
i)}\int_{\gamma}a^{-t}b^{1-t}\Gamma\left(\tfrac{t}{2}\right)\Gamma\left(\tfrac{1-t}{2}\right)\Gamma\left(\tfrac{1+t}{2}\right)\,dt.$

The next step is to express the gamma factors in the numerator as bracket
series to obtain

$I(a,b)=\frac{1}{4b^{2}(2\pi
i)}\sum_{n_{1}}\sum_{n_{2}}\sum_{n_{3}}\phi_{n_{1}n_{2}n_{3}}\int_{\gamma}a^{-t}b^{1-t}\langle\tfrac{t}{2}+n_{1}\rangle\langle\tfrac{1}{2}-\tfrac{t}{2}+n_{2}\rangle\langle\tfrac{1}{2}+\tfrac{t}{2}+n_{3}\rangle\,dt$

and for instance selecting the bracket in $n_{1}$ to eliminate the integral
and using

(9.8) $\left\langle\tfrac{t}{2}+n_{1}\right\rangle=2\langle t+2n_{1}\rangle$

gives

(9.9)
$I(a,b)=\frac{1}{2b}\sum_{n_{1}}\sum_{n_{2}}\sum_{n_{3}}\phi_{n_{1}n_{2}n_{3}}a^{2n_{1}}b^{2n_{1}}\langle\tfrac{1}{2}+n_{1}+n_{2}\rangle\langle\tfrac{1}{2}-n_{1}+n_{3}\rangle.$

Method 3. The next way to evaluate the integral (9.1) is a mixture of the
previous two. Using

(9.10) $\displaystyle\frac{1}{x^{2}+b^{2}}$ $\displaystyle=$
$\displaystyle\sum_{n_{1}}\sum_{n_{2}}\phi_{n_{1}n_{2}}b^{2n_{1}}x^{2n_{2}}\langle
1+n_{1}+n_{2}\rangle$ $\displaystyle\exp(-a^{2}x^{2})$ $\displaystyle=$
$\displaystyle\frac{1}{4\pi
i}\int_{\gamma}x^{-t}a^{-t}\Gamma\left(\tfrac{t}{2}\right)\,dt$

and the usual procedures gives

$I(a,b)=\frac{1}{4\pi i}\sum_{n_{1}}\sum_{n_{2}}b^{2n_{1}}\langle
1+n_{1}+n_{2}\rangle\int_{\gamma}a^{-t}\Gamma\left(\tfrac{t}{2}\right)\langle
2n_{2}-t+1\rangle\,dt$

Now use the bracket in $n_{2}$ to eliminate the integral and produce

(9.11)
$I(a,b)=\frac{1}{2}\sum_{n_{1}}\sum_{n_{2}}\phi_{n_{1}n_{2}}a^{-2n_{2}-1}b^{2n_{1}}\langle
1+n_{1}+n_{2}\rangle\Gamma\left(n_{2}+\tfrac{1}{2}\right)$

and replacing the gamma factor by its bracket series yields

(9.12)
$I(a,b)=\frac{1}{2}\sum_{n_{1}}\sum_{n_{2}}\sum_{n_{3}}\phi_{n_{1}n_{2}n_{3}}b^{2n_{1}}a^{-2n_{2}-1}\langle
1+n_{1}+n_{2}\rangle\langle\tfrac{1}{2}+n_{2}+n_{3}\rangle.$

Method 4. The final form uses the representations

(9.13) $\displaystyle\frac{1}{x^{2}+b^{2}}$ $\displaystyle=$
$\displaystyle\frac{1}{4\pi
b^{2}i}\int_{\gamma}x^{-s}b^{s}\Gamma\left(\frac{s}{2}\right)\Gamma\left(1-\frac{s}{2}\right)\,ds$
$\displaystyle\exp(-a^{2}x^{2})$ $\displaystyle=$
$\displaystyle\sum_{n_{1}}\phi_{n_{1}}a^{2n_{1}}x^{2n_{1}}$

to write

(9.14) $I(a,b)=\frac{1}{4b^{2}\pi
i}\sum_{n_{1}}\phi_{n_{1}}a^{2n_{1}}\int_{\gamma}b^{s}\Gamma\left(\tfrac{s}{2}\right)\Gamma\left(1-\tfrac{s}{2}\right)\langle
2n_{1}-s+1\rangle\,ds,$

where the last bracket comes the original integral. Now write the gamma
factors as bracket series to produce

$I(a,b)=\frac{1}{2b^{2}(2\pi
i)}\sum_{n_{1}}\sum_{n_{2}}\sum_{n_{3}}\phi_{n_{1}n_{2}n_{3}}a^{2n_{1}}\left[\int_{\gamma}b^{s}\langle\tfrac{s}{2}+n_{2}\rangle\langle
1-\tfrac{s}{2}+n_{3}\rangle\langle 2n_{1}-s+1\rangle\,ds\right],$

and using the $n_{2}$-bracket to eliminate the integral yields

(9.15)
$I(a,b)=\frac{1}{b^{2}}\sum_{n_{1}}\sum_{n_{2}}\sum_{n_{3}}\phi_{n_{1}n_{2}n_{3}}a^{2n_{1}}b^{-2n_{2}}\langle
1+n_{2}+n_{3}\rangle\,\langle 2n_{1}+2n_{2}+1\rangle.$

These four bracket series lead to the terms

$\displaystyle T_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{2b}\sum_{k=0}^{\infty}\frac{1}{k!}\Gamma(\tfrac{1}{2}-k)\Gamma(\tfrac{1}{2}+k)(-a^{2}b^{2})^{k}$
$\displaystyle=$ $\displaystyle\frac{\pi}{2b}\exp(a^{2}b^{2})$ $\displaystyle
T_{2}$ $\displaystyle=$
$\displaystyle\frac{a}{2}\sum_{k=0}^{\infty}\frac{1}{k!}\Gamma(1+k)\Gamma(-\tfrac{1}{2}-k)(-a^{2}b^{2})^{k}$
$\displaystyle=$
$\displaystyle-\frac{\pi}{2b}\exp(a^{2}b^{2})\textnormal{erf}(ab)$
$\displaystyle T_{3}$ $\displaystyle=$
$\displaystyle\frac{1}{2ab^{2}}\sum_{k=0}^{\infty}\frac{1}{k!}\Gamma(1+k)\Gamma(\tfrac{1}{2}+k)\left(-\frac{1}{a^{2}b^{2}}\right)^{k}$
$\displaystyle=$
$\displaystyle\frac{\sqrt{\pi}}{2ab^{2}}{}_{2}F_{0}\left(\genfrac{}{}{0.0pt}{}{1\,\,\,\,\tfrac{1}{2}}{-}\bigg{|}{-\frac{1}{a^{2}b^{2}}}\right).$

The values $T_{1}$ and $T_{2}$ and Rule $E_{3}$ in Section 2 are combined to
give the value

(9.19)
$I(a,b)=\frac{\pi}{2b}\exp(a^{2}b^{2})\left[1-\textnormal{erf}(ab)\right],$

conforming (9.2). The value of $T_{3}$ is useful in the asymptotic study of
$I(a,b)$ as $a^{2}b^{2}\rightarrow\infty$.

The final example illustrates a different combination ideas.

######  Example 9.1.

Consider the two-dimensional integral

(9.20)
$J(a,b)=\int_{0}^{\infty}\int_{0}^{\infty}x^{a-1}y^{b-1}\textnormal{Ei}(-x^{2}y)K_{0}\left(\frac{x}{y}\right)\,dx\,dy$

where Ei is the exponential integral defined by

(9.21)
$\textnormal{Ei}(-x)=-\int_{1}^{\infty}\frac{\exp(-xt)}{t}\,dt\quad\textnormal{for
}x>0$

with divergent representation

(9.22) $\textnormal{Ei}(-x^{2}y)=\sum_{\ell\geq
0}\phi_{\ell}\frac{x^{2\ell}y^{\ell}}{\ell}$

(see [9] for the concept of divergent expansion in the context of the method
of brackets). Using this series and the Mellin-Barnes representation of
$K_{0}$

(9.23) $K_{0}\left(\frac{x}{y}\right)=\frac{1}{4\pi
i}\int_{\gamma}\Gamma^{2}(t)\left(\frac{x}{2y}\right)^{-2t}\,dt$

yields

(9.24) $J(a,b)=\frac{1}{4\pi i}\sum_{\ell\geq
0}\phi_{\ell}\frac{1}{\ell}\int_{\gamma}\Gamma^{2}(t)2^{2t}\langle
a+2\ell-2t\rangle\langle b+\ell+2t\rangle\,dt,$

where the two brackets come from the integrals on the half-line. To evaluate
this integral use the bracket $\langle a+2\ell-2t\rangle$ to produce

$\displaystyle J(a,b)$ $\displaystyle=$ $\displaystyle\frac{1}{8\pi
i}\sum_{\ell\geq
0}\frac{\phi_{\ell}}{\ell}\left[\int_{\gamma}\Gamma^{2}(t)2^{2t}\langle\tfrac{a}{2}+\ell-t\rangle\langle
b+\ell+2t\rangle\,dt\right]$ $\displaystyle=$ $\displaystyle\frac{1}{8\pi
i}\sum_{\ell\geq 0}\frac{\phi_{\ell}}{\ell}\left[2\pi
i\Gamma^{2}(t)2^{2t}\langle b+\ell+2t\rangle\right]_{t=\tfrac{a}{2}+\ell}$
$\displaystyle=$ $\displaystyle\frac{1}{4}\sum_{\ell\geq
0}\frac{\phi_{\ell}}{\ell}\Gamma^{2}\left(\tfrac{a}{2}+\ell\right)2^{a+2\ell}\langle
a+b+3\ell\rangle,$

and eliminating the series with the bracket yields the value

(9.26)
$J(a,b)=-\frac{4^{-1-b/3+a/6}}{a+b}\Gamma^{2}\left(\frac{a-2b}{6}\right)\Gamma\left(\frac{a+b}{3}\right).$

##  10\. Feynman diagrams

A variety of interesting integrals appear in the evaluation of Feynman
diagrams. See [21, 22] for details. The example corresponds to a loop diagram
with a propagator in the scalar theory with internal propagators of equal
masses [7]:

Figure 1. The loop diagram

The integral attached to this diagram is

(10.1) $J(\alpha,\beta,m,m)=\int
d^{D}Q\frac{1}{\left[Q^{2}-m^{2}\right]^{\alpha}\left[(p+Q)^{2}-m^{2}\right]^{\beta}}$

with Mellin-Barnes representation given by

(10.2)
$J(\alpha,\beta,m,m)=\pi^{D/2}i^{1-D}\frac{(-m^{2})^{D/2-\alpha-\beta}}{\Gamma(\alpha)\Gamma(\beta)}\\\
\times\frac{1}{2\pi
i}\int_{\gamma}(p^{2})^{{}_{u}}(-m^{2})^{u}\frac{\Gamma(-u)\Gamma(\alpha+u)\Gamma(\beta+u)\Gamma(\alpha+\beta-D/2+u)}{\Gamma(\alpha+\beta+2u)}\,du$

Replacing the Gamma factors in the numerator by its corresponding bracket
series gives

(10.3)
$J(\alpha,\beta,m,m)\leftarrow\pi^{D/2}i^{1-D}\frac{(-m^{2})^{D/2-\alpha-\beta}}{\Gamma(\alpha)\Gamma(\beta)}\sum_{\\{n\\}}\phi_{n_{1},\ldots,n_{d}}\\\
\frac{1}{2\pi
i}\int_{\gamma}(p^{2})^{u}(-m^{2})^{u}\frac{\langle-u+n_{1}\rangle\langle\alpha+u+n_{2}\rangle\langle\beta+u+n_{3}\rangle\langle\alpha+\beta-D/2+u+n_{4}\rangle}{\Gamma(\alpha+\beta+2u)}\,du.$

To evaluate the integral, the vanishing of the bracket
$\langle-u+n_{1}\rangle$ gives $u^{*}=n_{1}$ and $J$ is expressed as a bracket
series

(10.4)
$J(\alpha,\beta,m,m)=\pi^{D/2}i^{1-D}\frac{(-m^{2})^{D/2-\alpha-\beta}}{\Gamma(\alpha)\Gamma(\beta)}\\\
\times\sum_{\\{n\\}}\phi_{n_{1},\ldots,n_{4}}(p^{2})^{u^{*}}(-m^{2})^{u^{*}}\frac{\langle\alpha+u^{*}+n_{2}\rangle\langle\beta+u^{*}+n_{3}\rangle\langle\alpha+\beta-D/2+u^{*}+n_{4}\rangle}{\Gamma(\alpha+\beta+2u^{*})}.$

The terms obtained from the bracket series above are given as hypergeometric
values

$\displaystyle T_{1}$ $\displaystyle=$
$\displaystyle\pi^{D/2}i^{1-D}(-m^{2})^{D/2-\alpha-\beta}\frac{\Gamma(\alpha+\beta-D/2)}{\Gamma(\alpha+\beta)}{}_{3}F_{2}\left(\genfrac{}{}{0.0pt}{}{\alpha\,\,\,\,\,\,\beta\,\,\,\,\,\,\alpha+\beta-\tfrac{D}{2}}{\tfrac{\alpha+\beta}{2}\,\,\,\,\,\,\tfrac{\alpha+\beta+1}{2}}\bigg{|}{\frac{p^{2}}{4m^{2}}}\right)$
$\displaystyle T_{2}$ $\displaystyle=$
$\displaystyle\pi^{D/2}i^{1-D}(-m^{2})^{D/2-\beta}(p^{2})^{-\alpha}\,\frac{\Gamma(\beta-D/2)}{\Gamma(\beta)}{}_{3}F_{2}\left(\genfrac{}{}{0.0pt}{}{\alpha\,\,\,\,\,\,1+\frac{\alpha-\beta}{2}\,\,\,\,\,\,\tfrac{1+\alpha-\beta}{2}}{1+\alpha-\beta\,\,\,\,\,\,1-\beta+\tfrac{D}{2}}\bigg{|}{\frac{4m^{2}}{p^{2}}}\right)$
$\displaystyle T_{3}$ $\displaystyle=$
$\displaystyle\pi^{D/2}i^{1-D}(-m^{2})^{D/2-\alpha}(p^{2})^{-\beta}\,\frac{\Gamma(\alpha-D/2)}{\Gamma(\alpha)}{}_{3}F_{2}\left(\genfrac{}{}{0.0pt}{}{\beta\,\,\,\,\,\,1+\frac{\beta-\alpha}{2}\,\,\,\,\,\,\tfrac{1+\beta-\alpha}{2}}{1+\beta-\alpha\,\,\,\,\,\,1-\alpha+\tfrac{D}{2}}\bigg{|}{\frac{4m^{2}}{p^{2}}}\right)$
$\displaystyle T_{4}$ $\displaystyle=$
$\displaystyle\pi^{D/2}i^{1-D}(p^{2})^{D/2-\alpha-\beta}\,\frac{\Gamma(D/2-\alpha)\Gamma(D/2-\beta)\Gamma(\alpha+\beta-D/2)}{\Gamma(\alpha)\Gamma(\beta)\Gamma(D-\alpha-\beta)}$
$\displaystyle\quad\quad\quad\quad\quad\quad\times{}_{3}F_{2}\left(\genfrac{}{}{0.0pt}{}{\alpha+\beta-\frac{D}{2}\,\,\,\,\,\,1+\tfrac{\alpha+\beta-D}{2}\,\,\,\,\,\tfrac{1+\alpha+\beta-D}{2}}{1+\beta-\tfrac{D}{2}\,\,\,\,\,1+\alpha-\tfrac{D}{2}}\bigg{|}{\frac{4m^{2}}{p^{2}}}\right).$

The conclusion is the evaluation of the Feynman diagram shown in Figure 1 is
given by

(10.5)
$J(\alpha,\beta,m,m)=\begin{cases}T_{1}\quad&\textnormal{for}\,\,\left|\frac{p^{2}}{4m^{2}}\right|<1\\\
T_{2}+T_{3}+T_{4}\quad&\textnormal{for}\,\,\left|\frac{4m^{2}}{p^{2}}\right|<1.\end{cases}$

.

##  11\. Conclusions

The work presented here gives a new procedure to evaluate Mellin-Barnes
integral representations. The main idea is to replace the Gamma factors
appearing in such representations by its corresponding bracket series. The
method of brackets is then used to evaluate these integrals. This method has
been shown to be simple and efficient, given its algorithmic nature. A
collection of representative examples has been provided.

##  12\. Acknowledgments

I.K. was supported in part by Fondecyt (Chile) Grants Nos. 1040368, 1050512
and 1121030, by DIUBB (Chile) Grant Nos. 102609, GI 153209/C and GI 152606/VC.
I. G. wishes to thank the hospitality of the Department of Mathematics at
Tulane University, where part of this work was conducted.

The Feynman diagram has been drawn using JaxoDraw [6].

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