# Integral challenge #4

#### DreamWeaver

##### Well-known member
OK, OK, so I'll stop soon... lol This'll be the last one for a while. But hey, you all know what it's like; you just can't log on here and find too many interesting threads, so forgive me for getting carried away. I'm sorry... [liar] For $$\displaystyle 0 < a < \pi$$, and $$\displaystyle b \in \mathbb{R} > -1$$, show that

$$\displaystyle \int_0^{\infty}\frac{x^b}{\cosh x+\cos a}\,dx=\frac{2\Gamma(b+1)}{\sin a}\sum_{k=1}^{\infty}(-1)^{k+1}\frac{\sin ka}{k^{b+1}}$$

#### Random Variable

##### Well-known member
MHB Math Helper
I think it looks a little bit nicer if you express it as a Mellin transform.

$\displaystyle \int_{0}^{\infty} \frac{x^{b-1}}{\cosh x + \cos a} \ dx = \int_{0}^{\infty} \frac{x^{b-1}}{\frac{e^{x} +e^{-x}}{2} + \frac{e^{ia} + e^{-ia}}{2}} \ dx$

$\displaystyle = 2 \int_{0}^{\infty} \frac{x^{b-1}}{e^{-x} (e^{2x} +1 + e^{x+ia}+e^{x-ia})} \ dx = 2 \int_{0}^{\infty} \frac{e^{x} x^{b-1}}{(e^{x-ia} + 1)(e^{x+ia}+1)} \ dx$

$\displaystyle = 2 \int_{0}^{\infty} x^{b-1} \frac{1}{(e^{ia} - e^{-ia})} \Big( \frac{1}{e^{x-ia}+1} - \frac{1}{e^{x+ia}+1} \Big) dx$

$\displaystyle = \frac{1}{i \sin a} \Big( \int_{0}^{\infty} \frac{x^{b-1}}{e^{x-ia}+1} \ dx - \int_{0}^{\infty} \frac{x^{b-1}}{e^{x+ia}+1} \ dx \Big)$

$\displaystyle = \frac{1}{i \sin a} \Big( -\Gamma(b) \text{Li}_{b}(-e^{ia}) + \Gamma(b) \text{Li}_{b}(-e^{-ia}) \Big)$

$\displaystyle = \frac{\Gamma(b)}{i \sin(a)} \Big( - \sum_{k=1}^{\infty} (-1)^{k} \frac{e^{ika}}{k^{b}} + \sum_{k=1}^{\infty} (-1)^{k} \frac{e^{-ika}}{k^{b}} \Big)$

$\displaystyle = \frac{\Gamma(b)}{\sin a} \sum_{k=1}^{\infty} (-1)^{k} \frac{1}{k^{b}} \Big( \frac{-e^{ika} + e^{-ika}}{i} \Big) = \frac{2 \Gamma(b)}{\sin a} \sum_{k=1}^{\infty} (-1)^{k-1} \frac{\sin ka}{k^{b}}$

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#### DreamWeaver

##### Well-known member
I think it looks a little bit nicer if you express it as a Mellin transform.
^^ Agreed! Your Kung Fu is good, self-evidently...   