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While I was playing with Wolfram Alpha online calculator I wondered about a possible closed-form of the first few cases for integers $n\geq 1$ in

$$\int_0^\pi\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx.\tag{1}$$ My problem (as aficionado) here that I think that it could be in the literature, and I am curious (today and sometimes in past ocassions about series and integrals) to know how to find some of my creations.

Question 1. Do you know how to find if the integral $$\int_0^\pi\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx$$ is in the literature? I am asking about the tools for an aficionado, where I should to search, with what key words in Internet? What sites? Optionally if you know this integral feel free to add the reference answering this question as a reference request, and I search and read such closed-form from the literature. Many thanks.

I'm almost sure that I've seen this kind of integrals, or that should be easy to get, but I don't remember where. I know (without a justification) cases as $n=1,2$ and $n=3$.

Question 2. Calculate a good approximation of $$\int_0^\pi\frac{\cos^{3}(x)}{e^{\cos(x)}-1}dx.\tag{2}$$ Provide your justification. Many thanks.

There were a typo, see the first comment.

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  • $\begingroup$ I'm sorry there was a typo @YuriyS $\endgroup$ – user243301 Apr 6 '18 at 11:44
  • $\begingroup$ What does 'aficionado' mean in this context? I'm not a native English speaker, and I'm not sure I understand the meaning $\endgroup$ – Yuriy S Apr 6 '18 at 12:00
  • $\begingroup$ I am out of the university-world. Each person has different tools to search information, I imagine that an editor or a professor has different tools or access to tables where he/she can search by means of key words what integrals are known. Many thanks @YuriyS $\endgroup$ – user243301 Apr 6 '18 at 12:12
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    $\begingroup$ These integrals resemble the the generating function of the Bernouli polynomials. It may be possible to perform the integration term wise on the series expansion and perhaps close up the result after. en.wikipedia.org/wiki/Bernoulli_polynomials $\endgroup$ – Graham Hesketh Apr 6 '18 at 13:56
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    $\begingroup$ I do not think that an expansion in terms of Bernoulli numbers is a good idea. Except the first ones, Bernoulli numbers tend to grow pretty fast, while $\int_{0}^{\pi}\cos^{2n}(x)\,dx$ just behaves like $\frac{1}{\sqrt{n}}$, so termwise integration might lead to a non-convergent representation. $\endgroup$ – Jack D'Aurizio Apr 6 '18 at 14:01
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Q1. This integral has closed form and for integer $n$, it reduces to a simple Beta function $\left(B\right)$

$$\int_0^\pi\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx=\int_0^{\pi/2}\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx+\int_{\pi/2}^\pi\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx \tag{1}$$

$$\int_0^\pi\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx=\int_0^{\pi/2}\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx+\int_{0}^{\pi/2}\frac{(-\cos(x))^{2n}}{e^{-\cos(x)}-1}dx \tag{2}$$

$$\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}+\frac{(-\cos(x))^{2n}}{e^{-\cos(x)}-1}=-\cos(x)^{2n} \quad n\in \mathcal{Z} \tag{3}$$

$$\int_0^\pi\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx=-\int_0^{\pi/2}\cos^{2n}(x)dx=-\frac{1}{2}B\!\left(\frac{1}{2},n+\frac{1}{2}\right)=-\frac{\pi\left(2n\right)!}{n!^{2}\,2^{2n+1}} \tag{4}$$

Q2. More generally, term wise integration of the generating function of the Bernoulli polynomials, with $b_{n}$ the $n^{th}$ Bernoulli number, gives:

$$ \int_{0}^{\pi}\!{\frac { \cos^v \left( x \right)}{{ {\rm e}^{\cos \left( x \right) }}-1}}\,{\rm d}x=-\frac{\left( 1+ \left( -1 \right) ^{v} \right)}{4}\,B\! \left( \frac{1}{2} ,\frac{v+1}{2} \right) +\frac{\left( 1- \left( -1 \right) ^{v} \right)}{2}\, \sum _{n=0}^{\infty }{\frac {b_{2n} \,B\! \left( \frac{1}{2},n+\frac{v}{2} \right) }{ \left( 2\,n \right) !}} $$ $$\tag{5}$$ hence: $$\int_{0}^{\pi}\!{\frac { \cos^3 \left( x \right) }{{ {\rm e}^{\cos \left( x \right) }}-1}}\,{\rm d}x=\pi\sum _{n=0}^{\infty}{\frac { \left( 2\,n+1 \right) \left( n+1 \right) b_{2n} }{{2}^{2\,n+1} \left( \left( n+1 \right) ! \right) ^{2}} }\approx {\frac {58705}{110592}}\pi \quad\quad\text{(4 terms)} \tag{6} $$

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  • $\begingroup$ I was waiting the reference request or advices to find this kind of integrals, any case many thanks for your attention an help $\endgroup$ – user243301 Apr 7 '18 at 13:25

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