#  Unifying trigonometric and hyperbolic function derivatives via negative
integer order polylogarithms

Andrew Ducharme

(February 1, 2024)

###### Abstract

Special functions like the polygamma, Hurwitz zeta, and Lerch zeta functions
have sporadically been connected with the $n$th derivatives of trigonometric
functions. We show the polylogarithm $\operatorname{Li}_{s}(z)$, a function of
complex argument and order $z$ and $s$, encodes the $n$th derivatives of the
cotangent, tangent, cosecant and secant functions, and their hyperbolic
equivalents, at negative integer orders $s=-n$. We then show how at the same
orders, the polylogarithm represents the $n$th application of the operator
$x\frac{d}{dx}$ on the inverse trigonometric and hyperbolic functions.
Finally, we construct a sum relating two polylogarithms of order $-n$ to a
linear combination of polylogarithms of orders $s=0,-1,-2,...,-n$.

Department of Physics, University of Oregon, 1371 E 13th Avenue, Eugene,
Oregon 97403, USA<EMAIL_ADDRESS>
##  1 Introduction

The polylogarithm, or polylog, is defined as

$\operatorname{Li}_{s}(z)=\sum_{k=1}^{\infty}\frac{z^{k}}{k^{s}}.$ (1.1)

The order and argument $s$ and $z$ can be complex. The function converges
absolutely inside the unit disk $|z|\leq 1$ (except at $z=1$ for $\Re(s)\leq
1$) and can be extended to $|z|>1$ via analytic continuation. It is
principally valued on the complex plane with a branch cut along the positive
real axis $1\leq\Re(z)<\infty$.

Higher order polylogarithms, like the dilogarithm $\operatorname{Li}_{2}(z)$
and trilogarithm $\operatorname{Li}_{3}(z)$, which cannot be written in terms
of elementary functions, have been of interest since at least the mid-
eighteenth century [1]. Conversely, the negative integer order polylogarithms
are simply rational functions in $z$

$\operatorname{Li}_{-n}(z)=\left(z\frac{d}{dz}\right)^{n}\frac{z}{1-z}=\sum_{k=0}^{n}k!{n+1\brace
k+1}\left(\frac{z}{1-z}\right)^{k+1}$ (1.2)

where ${n\brace k}$ are Stirling numbers of the second kind. Its simplicity
likely explains the little discussion of these cases in the literature; Kummer
was the only scholar of the polylogarithm Truesdell notes who considered the
function at negative integer orders [2]. These cases also receive no mention
in Lewin’s authoritative text on the subject [3]. Negative order
polylogarithms were reintroduced by Wood [4], and found by Lee [5], who deemed
them “polypseudologs,” in his study of statistical mechanics. Cvijovic derived
multiple explicit formulas, one novel, of these functions [6].

Here, we demonstrate how the polylogarithm at negative integer orders is a
unifying structure behind the $n$th derivatives of the cotangent, tangent,
cosecant, and secant functions, and their hyperbolic cousins. The
polylogarithm also underlies the action of the operator
$\left(x\frac{d}{dz}\right)^{n}$ on all twelve of the inverse trigonometric
and hyperbolic functions. These results are easy to state: simply check
Section 2 for the elementary trigonometric derivatives barring $\sin x$ and
$\cos x$, Section 4 for the hyperbolic derivatives barring $\sinh x$ and
$\cosh x$, and Section 5 for all the inverse trigonometric and hyperbolic
functions. Eq. 1.2 simplifies the derivation as well. Simply find the proper
substitution, and voilá. What sets the polylogarithm apart is its flexibility
to represent these 20 functions more simply than other special functions like
the Lerch zeta or polygamma functions.

Consider Eq. 1.2 under the transformation $z\rightarrow e^{ix}$,
$x\in\mathbb{R}/\\{2\pi\ell\\}$, $\ell\in\mathbb{Z}$. Then
$\frac{d}{dz}\rightarrow(-i)^{n}\frac{d}{dx}$,
$\operatorname{Li}_{0}(z)=\frac{z}{1-z}\rightarrow\frac{1}{2}\left(i\cot\frac{x}{2}-1\right)$,
and for $n\geq 1$,

$\operatorname{Li}_{-n}(e^{ix})=\frac{i^{1-n}}{2}\left(\frac{d}{dx}\right)^{n}\cot\frac{x}{2}\\\
=\sum_{k=0}^{n}\frac{k!}{2^{k+1}}{n+1\brace
k+1}\left(i\cot\frac{x}{2}-1\right)^{k+1}$ (1.3)

This is just one of many connections of $n$th cotangent derivative with
special functions. Apostol wrote it in terms of the Lerch zeta function
$\ell_{1-n}(z)=\frac{i}{2(2\pi i)^{n-1}}\frac{d^{n-1}}{dz^{n-1}}\cot\pi z$ at
negative integer orders [7], a special case of the Lerch transcendent [8]. It
can be written as a polygamma reflection formula
$\psi_{n}(x)-(-1)^{n}\psi_{n}(1-x)=-\pi\left(\frac{d}{dx}\right)^{n}\cot\pi x$
[9, 10], or a Hurwitz zeta reflection formula
$\zeta(n,x)+(-1)^{n}\zeta(n,1-x)$
$=-\frac{\pi}{(n-1)!}\left(\frac{d}{dx}\right)^{n-1}\cot\pi x$ [11]. But note
how Eq. 1.3 more succinctly represents the cotangent derivative. Two instances
of the polygamma or Hurwitz zeta functions are needed where one polylogarithm
would suffice. Only one Lerch transcendent is required, but with more degrees
of freedom (three parameters) than the polylogarithm (two parameters).

The Lerch zeta function and the polylogarithm comparably represent the
cotangent derivatives because the polylogarithm substitution $z\rightarrow
e^{ix}$ nearly equates the two special functions:
$\ell_{s}(z)=\sum_{k=1}^{\infty}\frac{e^{2\pi
ikz}}{k^{s}}=\operatorname{Li}_{s}(e^{2\pi iz})$. But the Lerch zeta function
is too overconstrained to represent inverse trigonometric and hyperbolic
function derivatives. For example,
$\operatorname{Li}_{-n}(x)-\operatorname{Li}_{-n}(x)=2\left(x\frac{d}{dx}\right)^{n+1}\operatorname{arctanh}x$.
To write the left-hand side with Lerch zeta functions, we must undo the
complex exponential with a natural logarithm to get a gnarly
$\ell_{s}(\frac{1}{2\pi i}\log x)$. But the insertion of the logarithm
restricts the Lerch zeta function to exist only over the positive reals,
whereas the domain of $\operatorname{Li}_{-n}(x)$ is $x\in\mathbb{R}/\\{1\\}$.
With only one Lerch zeta function per side of the origin, it cannot represent
the $\operatorname{arctanh}x$ derivative. The polylogarithm prevails.

We note the $n$th trigonometric derivatives were connected to negative integer
order polylogarithms in [8] and negative integer order Lerch zeta and Legendre
chi functions in [12].

The rest of the paper works through the various trigonometric, hyperbolic,
inverse trigonometric, and inverse hyperbolic derivatives. As the difference
of mirrored polylogarithms becomes increasingly common in these relations, in
Section 3, we define the Legendre chi function $\chi_{s}(z)$ and the inverse
tangent integral $\operatorname{Ti}_{s}(z)$, special functions which
abbreviate these differences. In Section 6, we use trigonometric derivative
polynomials to construct a sum relating two polylogarithms of order $-n$ with
a linear combination of all polylogarithms of lesser magnitude negative
integer orders.

##  2 Trigonometric Derivative Polynomials

Trigonometric derivatives and their resulting derivative polynomials have long
been a problem of interest [13, 14, 15], and received a flurry of attention in
the late 2000s [11, 10, 16, 17, 18]

To get the $n$th tangent derivative, substitute $z\rightarrow-e^{ix}$ into Eq.
1.2 so for $n\geq 1$,

$-\operatorname{Li}_{-n}(-e^{ix})=\frac{i^{1-n}}{2}\left(\frac{d}{dx}\right)^{n}\tan\frac{x}{2}$
(2.1)

$\left(\frac{d}{dx}\right)^{n}\tan
x=2^{n}i^{n-1}\sum_{k=0}^{n}\frac{(-1)^{k}k!}{2^{k}}{n+1\brace
k+1}\left(1+i\tan x\right)^{k+1}.$ (2.2)

The sum of Eqs. 1.3 and 2.1 is proportional to the $n$th derivative of
$\cot\frac{x}{2}+\tan\frac{x}{2}=2\csc x$, so

$\operatorname{Li}_{-n}(e^{ix})-\operatorname{Li}_{-n}(-e^{ix})=i^{1-n}\left(\frac{d}{dx}\right)^{n}\csc
x.$ (2.3)

Expanding both $(1+i\tan x)^{k+1}$ and $(i\cot x-1)^{k+1}$ by the Binomial
Theorem gives

$\left(\frac{d}{dx}\right)^{n}\csc
x=\frac{i^{n-1}}{2}\sum_{k=0}^{n}\frac{(-1)^{k}k!}{2^{k}}{n+1\brace k+1}\\\
\times\sum_{j=0}^{k+1}{k+1\choose
j}i^{j}\left[\tan^{j}\frac{x}{2}-(-1)^{j}\cot^{j}\frac{x}{2}\right].$ (2.4)

This is similar to the polylogarithm duplication formula
$\operatorname{Li}_{s}(z)+\operatorname{Li}_{s}(-z)=2^{1-s}\operatorname{Li}_{s}(z^{2})$
which under the same substitution $z\rightarrow e^{ix}$ proves a cotangent
double angle formula $2\cot 2x=\cot x-\tan x$.

Taking advantage of $\csc\left(x+\frac{\pi}{2}\right)=\sec x$, we get our
final trig derivative

$\operatorname{Li}_{-n}(ie^{ix})-\operatorname{Li}_{-n}(-ie^{ix})=i^{1-n}\left(\frac{d}{dx}\right)^{n}\sec
x.$ (2.5)

Because the identities for
$\tan\left(\theta+\frac{\pi}{4}\right)=\frac{\tan\theta+1}{1-\tan\theta}$ and
$\cot\left(\theta+\frac{\pi}{4}\right)=\frac{\cot\theta-1}{1+\cot\theta}$ are
not simply linear combinations of other trig functions, the $n$th secant
derivative can be written more succinctly as Eq. 2.4 with
$\frac{x}{2}\rightarrow\frac{x}{2}+\frac{\pi}{4}$. Alternatively, combining
the fractional forms of the quarter-period-shifted tangent and cotangent, then
binomially expanding $(\sec x\pm\tan x)^{j}$, produces the following three sum
equation

$2i^{1-n}\left(\frac{d}{dx}\right)^{n}\sec x=\\\
\sum_{k=0}^{n}\frac{(-1)^{k}k!}{2^{k}}{n+1\brace
k+1}\sum_{j=0}^{k+1}i^{j}{k+1\choose
j}\sum_{\ell=0}^{j}{j\choose\ell}\tan^{j-\ell}x\sec^{\ell}x(1-(-1)^{\ell}).$
(2.6)

Two polylogarithms is likely the fewest needed to build the cosecant and
secant derivatives because one, these even functions are constructed from the
odd cotangent and tangent, and two, there is no generic identity to reduce the
form $\operatorname{Li}_{s}(z)-\operatorname{Li}_{s}(-z)$, as it itself is a
special function, the Legendre chi function $\chi_{s}(x)$.

##  3 Legendre chi function and Inverse tangent integral

As infinite sums, the Legendre chi function

$\chi_{s}(z)=\sum_{k=0}^{\infty}\frac{z^{2k+1}}{(2k+1)^{s}}=\frac{1}{2}[\operatorname{Li}_{s}(z)-\operatorname{Li}_{s}(-z)]$
(3.1)

and inverse tangent integral

$\operatorname{Ti}_{s}(z)=\sum_{k=0}^{\infty}(-1)^{k}\frac{z^{2k+1}}{(2k+1)^{s}}=\frac{1}{2i}[\operatorname{Li}_{s}(iz)-\operatorname{Li}_{s}(-iz)]$
(3.2)

converge on the unit disk for all $s$, except at $z=\pm 1$ for $\chi_{s}(z)$,
and $z=\pm i$ for $\operatorname{Ti}_{s}(z)$ when $s\leq 1$. The functions are
the real and imaginary parts of the polylogarithm, and are related by
$\operatorname{Ti}_{s}(z)=-i\chi_{s}(iz)$. The inverse tangent integral is so
named because $\operatorname{Ti}_{2}(z)=\int_{0}^{x}\frac{\arctan t}{t}dt$.
Consequently, $\operatorname{Ti}_{1}(z)=\arctan z$ and
$\chi_{1}(z)=\operatorname{arctanh}z$. The first few functions of zeroth and
negative integer order are

$n$ $\chi_{n}(z)$ $\operatorname{Ti}_{n}(z)$ 0 $\frac{z}{1-z^{2}}$
$\frac{z}{1+z^{2}}$ -1 $\frac{z+z^{3}}{(1-z^{2})^{2}}$
$\frac{z-z^{3}}{(1+z^{2})^{2}}$ -2 $\frac{z+6z^{3}+z^{5}}{(1-z^{2})^{3}}$
$\frac{z-6z^{3}+z^{5}}{(1+z^{2})^{3}}$ -3
$\frac{z+23z^{3}+23z^{5}+z^{7}}{(1-z^{2})^{4}}$
$\frac{z-23z^{3}+23z^{5}-z^{7}}{(1+z^{2})^{4}}$ -4
$\frac{z+76z^{3}+230z^{5}+76z^{7}+z^{9}}{(1-z^{2})^{5}}$
$\frac{z-76z^{3}+230z^{5}-76z^{7}+z^{9}}{(1+z^{2})^{5}}$

A check of the OEIS finds that the coefficients of the numerator polynomial in
the Legendre chi function $S(n,k)$ is sequence A060187, describing the
Eulerian numbers of type B [19]. This pairs nicely with the similar
coefficients in the numerator of the polylogarithm being the Eulerian numbers
of type A (often just called the Eulerian numbers). A single sum form provided
by the OEIS is

$S(n,k)=\sum_{j=1}^{k}(-1)^{k-j}{n+1\choose k-j}(2j-1)^{n}.$ (3.3)

Closed forms of the Legendre chi function and inverse tangent integral over
non-positive integers $n\leq 0$ are

$\chi_{-n}(z)=\frac{1}{(1-z^{2})^{n+1}}\sum_{k=1}^{n+1}S(n,k)z^{2k-1}$ (3.4)

$\operatorname{Ti}_{-n}(z)=-\frac{1}{(1+z^{2})^{n+1}}\sum_{k=1}^{n+1}(-1)^{k}S(n,k)z^{2k-1}.$
(3.5)

Using these to clean up Eqs. 2.4 and 2.6, we get

$\left(\frac{d}{dx}\right)^{n}\csc
x=\frac{(-1)^{n}}{2^{n}}e^{-2ix}\csc^{n+1}x\sum_{k=1}^{n+1}S(n,k)e^{-i(n-2k)x}$
(3.6)

$\left(\frac{d}{dx}\right)^{n}\sec
x=-\frac{i^{n}}{2^{n}}e^{-2ix}\sec^{n+1}x\sum_{k=1}^{n+1}(-1)^{k}S(n,k)e^{-i(n-2k)x}$
(3.7)

##  4 Hyperbolic Function Derivatives

The hyperbolic functions have similar, but simpler, derivatives compared to
the trigonometric functions, lacking coefficients with $n$ dependence because
the substitution of the real exponential takes
$\left(z\frac{d}{dz}\right)^{n}$ to
$\left(e^{x}e^{-x}\frac{d}{dx}\right)^{n}=\frac{d^{n}}{dx^{n}}$, with no extra
factor of $-i^{n}$. These equations hold over $x\in\mathbb{R}$.

$\operatorname{Li}_{-n}(e^{x})=-\frac{1}{2}\left(\frac{d}{dx}\right)^{n}\coth\frac{x}{2}$
(4.1)

$\operatorname{Li}_{-n}(-e^{x})=-\frac{1}{2}\left(\frac{d}{dx}\right)^{n}\tanh\frac{x}{2}$
(4.2)

$\operatorname{Li}_{-n}(e^{x})-\operatorname{Li}_{-n}(-e^{x})=2\chi_{-n}(e^{x})=-\left(\frac{d}{dx}\right)^{n}\operatorname{csch}x,$
(4.3)

$\operatorname{Li}_{-n}(ie^{x})-\operatorname{Li}_{-n}(-ie^{x})=2i\operatorname{Ti}_{-n}(e^{x})=i\left(\frac{d}{dx}\right)^{n}\operatorname{sech}x,$
(4.4)

$\left(\frac{d}{dx}\right)^{n}\coth
x=2^{n}\sum_{k=0}^{n}\frac{(-1)^{k}k!}{2^{k}}{n+1\brace k+1}\left(1+\coth
x\right)^{k+1}$ (4.5)

$\left(\frac{d}{dx}\right)^{n}\tanh
x=2^{n}\sum_{k=0}^{n}\frac{(-1)^{k}k!}{2^{k}}{n+1\brace k+1}\left(1+\tanh
x\right)^{k+1}$ (4.6)

$\left(\frac{d}{dx}\right)^{n}\operatorname{csch}x=\frac{(-1)^{n}}{2^{n}}e^{2x}\operatorname{csch}^{n+1}x\sum_{k=1}^{n+1}S(n,k)e^{(n-2k)x}$
(4.7)

$\left(\frac{d}{dx}\right)^{n}\operatorname{sech}x=-\frac{(-1)^{n}}{2^{n}}e^{2x}\operatorname{sech}^{n+1}x\sum_{k=1}^{n+1}(-1)^{k}S(n,k)e^{(n-2k)x}$
(4.8)

##  5 Inverse Trig and Hyperbolic Functions

The substitution of exponentials in the previous sections are useful because
it reduces the operator $z\frac{d}{dz}$ to a derivative, but are not the sole
use cases of Eq. 1.2. We can build the $n$th application of $x\frac{d}{dx}$ to
inverse trigonometric and hyperbolic functions from
$\operatorname{Ti}_{0}(x)=x\frac{d}{dx}\arctan x$ and
$\chi_{0}(x)=x\frac{d}{dx}\operatorname{arctanh}x$ for $n\geq 0$ and, unless
otherwise noted, $x\in\mathbb{R}$.

$\chi_{-n}(x)=\left(x\frac{d}{dx}\right)^{n+1}\operatorname{arctanh}x$ (5.1)

$\chi_{-n}\left(\frac{1}{x}\right)=\left(-x\frac{d}{dx}\right)^{n+1}\operatorname{arccoth}x$
(5.2)

$\chi_{-n}\left(\frac{x}{\sqrt{1+x^{2}}}\right)=\left((x+x^{3})\frac{d}{dx}\right)^{n+1}\operatorname{arcsinh}x$
(5.3)

$\chi_{-n}\left(\frac{1}{x\sqrt{1+x^{-2}}}\right)=\left(-\left(x+\frac{1}{x}\right)\frac{d}{dx}\right)^{n+1}\operatorname{arccsch}x$
(5.4)

Note for $x<0$, $x\sqrt{1+x^{-2}}=-\sqrt{1+x^{2}}$.

$\chi_{-n}\left(\frac{\sqrt{x^{2}-1}}{x}\right)=\left(x\left(x^{2}-1\right)\frac{d}{dx}\right)^{n+1}\operatorname{arccosh}x,\;x\geq
1$ (5.5)

$\chi_{-n}\left(\sqrt{1-x^{2}}\right)=\left(\left(x-\frac{1}{x}\right)\frac{d}{dx}\right)^{n+1}\operatorname{arccosh}x,\;0<x\leq
1.$ (5.6)

$\operatorname{Ti}_{-n}(x)=\left(x\frac{d}{dx}\right)^{n+1}\arctan x$ (5.7)

$\operatorname{Ti}_{-n}(x)=-\left(x\frac{d}{dx}\right)^{n+1}\operatorname{arccot}x$
(5.8)

$\operatorname{Ti}_{-n}\left(\frac{x}{\sqrt{1-x^{2}}}\right)=\left((x-x^{3})\frac{d}{dx}\right)^{n+1}\arcsin
x,\;|x|\leq 1$ (5.9)

$\operatorname{Ti}_{-n}\left(\frac{\sqrt{1-x^{2}}}{x}\right)=\left((x^{3}-x)\frac{d}{dx}\right)^{n+1}\arccos
x,\;|x|\leq 1$ (5.10)

$\operatorname{Ti}_{-n}\left(\frac{1}{x\sqrt{1-x^{-2}}}\right)=\left(\left(\frac{1}{x}-x\right)\frac{d}{dx}\right)^{n+1}\operatorname{arccsc}x,\;|x|\geq
1$ (5.11)

$\operatorname{Ti}_{-n}\left(x\sqrt{1-x^{-2}}\right)=\left(\left(x-\frac{1}{x}\right)\frac{d}{dx}\right)^{n+1}\operatorname{arcsec}x,\;|x|\geq
1$ (5.12)

Alternatively, one could simplify the operand of
$\left(z\frac{d}{dz}\right)^{n}$ at the expense of complicating the operator.
To transform the operand into a generic $f(x)$, choose
$z\rightarrow\frac{f(x)}{1+f(x)}$ so

$\operatorname{Li}_{-n}\left(\frac{f(x)}{1+f(x)}\right)=\left(\frac{f(x)}{f^{\prime}(x)}(1+f(x))\frac{d}{dx}\right)^{n}f(x)\\\
=\sum_{k=0}^{n}k!{n+1\brace k+1}f(x)^{k+1}.$ (5.13)

$\operatorname{Li}_{n}(z)$ diverges for $z=1$, but $\frac{f(x)}{1+f(x)}$ can
never be unitary. $\operatorname{Li}_{-n}\left(\frac{f(x)}{1+f(x)}\right)$
does diverge when $f(x)=-1$.

As an example, we will choose operands $f(x)=\sin x$ and $\cos x$.

$\operatorname{Li}_{-n}\left(\frac{\sin x}{1+\sin x}\right)=\left(\tan
x(1+\sin x)\frac{d}{dx}\right)^{n}\sin x\\\ =\sum_{k=0}^{n}k!{n+1\brace
k+1}\sin^{k+1}x$ (5.14)

$\operatorname{Li}_{-n}\left(\frac{\cos x}{1+\cos x}\right)=\left(-\cot
x(1+\cos x)\frac{d}{dx}\right)^{n}\cos x\\\ =\sum_{k=0}^{n}k!{n+1\brace
k+1}\cos^{k+1}x.$ (5.15)

##  6 A New Ladder-like Sum

Polylogarithmic ladders are extraordinary numerical connections between linear
combinations of polylogarithms of fixed order $n$ evaluated at different
powers of some $x$ with the natural logarithm
$\log(y)=-\operatorname{Li}_{1}(1-y)$. [20] provides the example

$-\pi^{2}\log(2)=\frac{35}{2}\operatorname{Li}_{3}(1)+36\operatorname{Li}_{3}\left(\frac{1}{2}\right)\\\
-18\operatorname{Li}_{3}\left(\frac{1}{4}\right)-4\operatorname{Li}_{3}\left(\frac{1}{8}\right)+\operatorname{Li}_{3}\left(\frac{1}{64}\right).$

We will instead bridge two (fixed) negative integer order polylogarithms $-n$
to a linear combination of individual polylogarithms of each integer order
from $s=0$ to $s=-n$. An illustrating example is

$\frac{z}{2}\left(\operatorname{Li}_{-6}(z)-\operatorname{Li}_{-6}(-z)\right)=\\\
\operatorname{Li}_{0}(z^{2})-12\operatorname{Li}_{-1}(z^{2})+60\operatorname{Li}_{-2}(z^{2})-160\operatorname{Li}_{-3}(z^{2})\\\
+240\operatorname{Li}_{-4}(z^{2})-192\operatorname{Li}_{-5}(z^{2})+64\operatorname{Li}_{-6}(z^{2}).$
(6.1)

We derive the general form by combining Eq. 2.3 with an alternate way Risomar
Sousa used get the $n$th derivative of $\csc x$ [8]: apply the General Leibniz
rule to $\csc x=e^{-ix}(i+\cot x)$ so

$\left(\frac{d}{dx}\right)^{n}\csc x=2i^{n-1}e^{-ix}\sum_{k=0}^{n}{n\choose
k}(-1)^{n-k}2^{k}\operatorname{Li}_{-n}(e^{2ix}).$ (6.2)

The combination is

$\operatorname{Li}_{-n}(e^{ix})-\operatorname{Li}_{-n}(-e^{ix})=2e^{-ix}\sum_{k=0}^{n}{n\choose
k}(-1)^{n-k}2^{k}\operatorname{Li}_{-k}(e^{2ix}),$ (6.3)

which holds for $n\geq 0$. $e^{ix}\rightarrow z$ gives the final relation

$\operatorname{Li}_{-n}(z)-\operatorname{Li}_{-n}(-z)=\frac{2}{z}\sum_{k=0}^{n}{n\choose
k}(-1)^{n-k}2^{k}\operatorname{Li}_{-k}(z^{2}).$ (6.4)

This could be interpreted as the negative integer version of

$\operatorname{Li}_{n}(z)=\frac{(-1)^{n-1}}{(n-2)!}\int_{1}^{z}\frac{\log(t^{n-2})\log(1-t)}{t}dt+\operatorname{Li}_{n}(1)\\\
+\sum_{k=1}^{n-2}\frac{(-1)^{k-1}}{k!}\operatorname{Li}_{n-k}(z)\log^{k}(z)$
(6.5)

from [21].

The first few sums are

$\operatorname{Li}_{0}(z)-\operatorname{Li}_{0}(-z)=\frac{2}{z}\operatorname{Li}_{0}(z^{2})$
(6.6)

$\operatorname{Li}_{-1}(z)-\operatorname{Li}_{-1}(-z)=\frac{2}{z}\left[-\operatorname{Li}_{0}(z^{2})+2\operatorname{Li}_{-1}(z^{2})\right]$
(6.7)

$\operatorname{Li}_{-2}(z)-\operatorname{Li}_{-2}(-z)=\frac{2}{z}\left[\operatorname{Li}_{0}(z^{2})-4\operatorname{Li}_{-1}(z^{2})+4\operatorname{Li}_{-2}(z^{2})\right]$
(6.8)

$\operatorname{Li}_{-3}(z)-\operatorname{Li}_{-3}(-z)=\\\
\frac{2}{z}\left[-\operatorname{Li}_{0}(z^{2})+6\operatorname{Li}_{-1}(z^{2})-12\operatorname{Li}_{-2}(z^{2})+8\operatorname{Li}_{-3}(z^{2})\right]$
(6.9)

$\operatorname{Li}_{-4}(z)-\operatorname{Li}_{-4}(-z)=\\\
\frac{2}{z}\left[\operatorname{Li}_{0}(z^{2})-8\operatorname{Li}_{-1}(z^{2})+24\operatorname{Li}_{-2}(z^{2})-32\operatorname{Li}_{-3}(z^{2})+16\operatorname{Li}_{-4}(z^{2})\right]$
(6.10)

The same approach replicated on $\sec x=e^{-ix}(1+i\tan x)$ gives a near-
identical result

$\operatorname{Li}_{-n}(ie^{ix})-\operatorname{Li}_{-n}(-ie^{ix})=-2ie^{-ix}\sum_{k=0}^{n}{n\choose
k}(-1)^{n-k}2^{k}\operatorname{Li}_{-k}(-e^{2ix}),$ (6.11)

which is Eq. 6.3 under $e^{ix}\rightarrow ie^{ix}$ ($z\rightarrow iz$).
Writing Eq. 6.4 as a Legendre chi function

$z\chi_{-n}(z)=\sum_{k=0}^{n}{n\choose
k}(-1)^{n-k}2^{k}\operatorname{Li}_{-k}(z^{2})$ (6.12)

demonstrates we have overcomplicated the same identity derivation for the
inverse tangent integral

$z\operatorname{Ti}_{-n}(z)=-\sum_{k=0}^{n}{n\choose
k}(-1)^{n-k}2^{k}\operatorname{Li}_{-k}(-z^{2}).$ (6.13)

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