Integral Challenge #3: Proving Li$_{2m+1}$ w/ Clausen Function

[DreamWeaver]

Prove the following integral representation of Polylogarithm, in terms of the Clausen function:\(\displaystyle \text{Li}_{2m+1}(e^{-\theta})=\frac{2}{\pi}\int_0^{\pi /2}\text{Cl}_{2m+1}(\theta \tan x)\, dx\)
NB. You're unlikely to find this is any books... I worked it out a while back, and haven't seen it anywhere else. That said, it's actually a lot easier than it looks... (Heidy)

 

[DreamWeaver]

I don't want to spoil this one, or prematurely give the answer - as I know many folks don't exactly log on every day - but for those of you who'd like a bit of a hint, here are two...

Spoiler #1:

Spoiler

Consider the classic integral

\(\displaystyle \int_0^{\infty}\frac{\cos ax}{1+x^2}\,dx=\frac{\pi}{2}e^{-a} \, \quad 0 < a < \infty \in \mathbb{R}\)

Spoiler #2:

Spoiler

Let \(\displaystyle a\) be an integer \(\displaystyle \ge 1\). Now sum over \(\displaystyle a \in \mathbb{Z}^{+}\), and consider the infinite sum of these integrals in \(\displaystyle a\) in terms of the series definition of the Clausen function...

Broader implications (Spoiler #3):

Spoiler

Integrals of a similar type to that in Spoiler #1 can be used to derive more complex results, such as:

\(\displaystyle (1) \int_0^{\infty}\frac{x \text{Cl}_{2m}(x \theta)}{(b^2+x^2)^2}\,dx=\frac{\pi \theta}{4b}\text{Li}_{2m-1}(e^{-b\theta})\)

\(\displaystyle (1) \int_0^{\infty}\frac{x \text{Sl}_{2m+1}(x \theta)}{(b^2+x^2)^2}\,dx=\frac{\pi \theta}{4b}\text{Li}_{2m}(e^{-b\theta})\)