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$$2\int_{0}^{\infty}{\cosh(x)-1\over x(e^{ax}-1)}\,\mathrm{d}x=\ln\left({\pi \over a\sin\left({\pi\over a }\right)}\right)$$

$$\cosh(x)=1+{x^2\over 2!}+{x^4\over 4!}+{x^6\over 6!}+\cdots$$

$${\cosh(x)-1\over x}={x\over 2!}+{x^3\over 4!}+{x^5\over 6!}+\cdots=\sum_{n=1}^{\infty}{x^{2n-1}\over (2n)!}$$

$$I=\sum_{n=1}^{\infty}{2\over (2n)!}\int_{0}^{\infty}{x^{2n-1}\over e^{ax-1}}\,\mathrm{d}x$$

Recall the Zeta function

$${a^k\over(k-1)!}\int_{0}^{\infty}{x^{k-1}\over e^{ax}-1}\,\mathrm{d}x=\zeta(k)$$

Now setting $k=2n$

$${a^{2k}\over(2k-1)!}\int_{0}^{\infty}{x^{2k-1}\over e^{ax}-1}\text{d}x=\zeta(2k)$$

Rearrange

$$I=\sum_{n=1}^{\infty}{2\over (2n)!}\cdot{(2n-1)!\zeta(2n)\over a^{2n}}=\sum_{n=1}^{\infty}{\zeta(2n)\over a^{2n}\cdot{n}}$$

So finally we have

$$\sum_{n=1}^{\infty}{\zeta(2n)\over a^{2n}\cdot{n}}=\ln\left({\pi \over a\sin\left({\pi\over a }\right)}\right)$$

This does not answer my question, it is only showing the transformation from integral into an infinite sum.

Anyone can help here to show that the infinite sum produced this closed form?

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    $\begingroup$ I suppose that, in the post, $\cos(x)$ is supposed to be $\cosh(x)$ $\endgroup$ – Claude Leibovici Jun 1 '16 at 6:13
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Hint. One may recall the ordinary generating function of the $\zeta(n)$ numbers expressed in terms of the digamma function, $$\psi(x):=\Gamma'(x)/\Gamma(x)=\left(\log \Gamma(x)\right)',$$ giving $$ \sum_{n=2}^{\infty}\zeta(n)\:x^n=-\gamma x -x\:\psi(1-x),\quad |x|<1, \tag1 $$ or $$ \sum_{n=2}^{\infty}(-1)^n\zeta(n)\:x^n=\gamma x +x\:\psi(1+x),\quad |x|<1, \tag2 $$ thus, by addition, $$ 2\sum_{n=1}^{\infty}\zeta(2n)\:x^{2n-1}=\psi(1+x)-\psi(1-x),\quad |x|<1, \tag3 $$ integrating $(3)$ gives, for $|x|<1$, $$ \sum_{n=1}^{\infty}\frac{\zeta(2n)}{n}\:x^{2n}=\log\left(\Gamma(1+x)\right)+\log\left(\Gamma(1-x)\right)=\log\left(\Gamma(1+x)\Gamma(1-x)\right) , \tag4 $$ and using the reflection relation $$ \Gamma(1+x)\Gamma(1-x)=\frac{\pi x}{\sin(\pi x)}, \quad |x|<1, \tag5 $$ we get

$$ \sum_{n=1}^{\infty}\frac{\zeta(2n)}{n}\:x^{2n}=\log\left(\frac{\pi x}{\sin(\pi x)}\right),\quad |x|<1 \tag6 $$

which is equivalent to the announced result.

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    $\begingroup$ The generating function for zeta values was new for me! +1 $\endgroup$ – Paramanand Singh Jun 1 '16 at 16:26
  • $\begingroup$ @ParamanandSingh: if you differentiate both sides of $(6)$, you get that the values of $\zeta(2n)$ are given by the derivatives of the cotangent function: that should be a well-known fact, at least to you! ;) $\endgroup$ – Jack D'Aurizio Jun 2 '16 at 3:39
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    $\begingroup$ Thanks @JackD'Aurizio. You made the connection obvious. If we use the infinite product of $\sin(\pi x)$ and do logarithmic differentiation and collect powers of $x$ we can see that the identity $(6)$ is obvious. $\endgroup$ – Paramanand Singh Jun 2 '16 at 4:51
  • $\begingroup$ @JackD'Aurizio: I have given an answer here based on your comment. $\endgroup$ – Paramanand Singh Jun 2 '16 at 5:10
  • $\begingroup$ A proof of $(2)$ is available here math.stackexchange.com/a/1313088/72031 $\endgroup$ – Paramanand Singh Jun 2 '16 at 10:37
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Note that 2$\frac{\cosh(x)-1}{x}$ is of the the form $i(f(i x)-f(-i x))$ with $f(x)=\frac{\cos(x)-1}{x}$.

This allows us to apply, after a simple rescaling $x\rightarrow 2\pi x/a$, (a slightly generalized form of) the Abel-Plana formula:

$$ I(a)=\frac{ia}{2\pi}\int_{0}^{\infty}dx\frac{f(2 \pi i x/a)-f(-2 \pi i x/a)}{e^{2 \pi x}-1}=\lim_{R\rightarrow \infty}\left[\sum_{n=1}^{R}f(2 \pi n/a)-\frac{a}{2\pi}\int_0^R dx f(2\pi x/a)\right ] \quad (1) $$

where we have used the fact that $f(0)=0$ in two instances.

Splitting the integral at $x=1$ and using the fact $$\lim_{R\rightarrow \infty}\left[\int_1^{R}\frac{dx}{x}-\sum_{n=1}^{R}\frac{1}{n}\right]=\gamma$$($\gamma$ is the Marschoni-Euler constant )we can properly eliminate the divergent pieces at $R=\infty$ ending up with

$$ I(a)=\underbrace{\sum_{n=1}^{\infty}\frac{\cos(2\pi n/a)}{n}}_{S(a)}-\frac{a}{2\pi}\int_0^1f(2\pi x/a)dx-\int_1^{\infty}dx\frac{\cos(2\pi x/a)}{x}-\gamma $$

the sum is easy to handle using the taylor series of Log, yielding after some algebra

$$ S(a)=-\log(2\sin(\pi/a)) $$

In order to correctly calculate the remaining integrals we note that

$$ \frac{a}{2\pi}\int_0^1 f(2\pi x/a)dx-\int_1^{\infty}dx\frac{\cos(2\pi x/a)}{x}=-\gamma-\log(2\pi/a) $$ which follows from here (I can proof this if necessary). We therefore obtain

$$ I(a)=-\log(2\sin(\pi/a))-\gamma+\log(2 \pi/a)+\gamma=\\ \log\left( \frac{\pi/a}{\sin(\pi/a)}\right) $$

as announced!

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Another approach. By assuming $\text{Re}(a)>1$ we may simply compute: $$ I^+ = \int_{0}^{+\infty}\frac{e^x-1}{x(e^{ax}-1)}\,dx,\qquad I^-=\int_{0}^{+\infty}\frac{e^{-x}-1}{x(e^{ax}-1)}\,dx \tag{1}$$ by expanding the integrand function as a geometric series and applying Frullani's theorem. We get:

$$ I = I^{+}+ I^- = \log\prod_{n\geq 1}\left(\frac{na}{na+1}\cdot\frac{na}{na-1}\right)\tag{2}$$ but: $$ \prod_{n\geq 1}\left(1-\frac{1}{a^2 n^2}\right) = \frac{\sin\left(\frac{\pi}{a}\right)}{\frac{\pi}{a}}\tag{3}$$ due to the Weierstrass product for the sine function, and the claim follows.

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{a > 1}$: \begin{align} 2\int_{0}^{\infty}{\cosh\pars{x} - 1 \over x\pars{\expo{ax} - 1}}\,\dd x &= -a\int_{0}^{\infty}{1 + \pars{\expo{-ax}}^{2/a} - 2\pars{\expo{-ax}}^{1/a} \over \pars{-ax}\pars{1 - \expo{-ax}}}\,\pars{\expo{-ax}}^{1 - 1/a}\,\dd x \end{align}

With the sub$\ds{\ldots\ \expo{-ax} \equiv t\ \iff\ x = -\,{\ln\pars{t} \over a}}$, we'll have \begin{align} 2\int_{0}^{\infty}{\cosh\pars{x} - 1 \over x\pars{\expo{ax} - 1}}\,\dd x &= a\int_{0}^{1}{1 + t^{2/a} -2t^{1/a} \over \ln\pars{t}\pars{1 - t}} t^{1 - 1/a}\,\pars{-\,{\dd t \over at}} = \int_{0}^{1}{2 - t^{1/a} - t^{-1/a} \over 1 - t}{1 \over \ln\pars{t}}\,\dd t \\[3mm] & = \int_{0}^{1}{2 - t^{1/a} - t^{-1/a} \over 1 - t}\ \overbrace{\bracks{-\int_{0}^{\infty}t^{\mu}\,\dd\mu}} ^{\ds{{1 \over \ln\pars{t}}}}\,\ \dd t \\[3mm] & = \int_{0}^{\infty}\bracks{2\int_{0}^{1}{1 - t^{\mu} \over 1 - t}\,\,\dd t - \int_{0}^{1}{1 - t^{\mu + 1/a} \over 1 - t}\,\,\dd t - \int_{0}^{1}{1 - t^{\mu - 1/a} \over 1 - t}\,\,\dd t}\,\dd\mu \\[3mm] & = \int_{0}^{\infty}\bracks{2\Psi\pars{\mu + 1} - \Psi\pars{\mu + {1 \over a} + 1} - \Psi\pars{\mu - {1 \over a} + 1}}\,\dd\mu \end{align}

$\Psi$ is the Digamma function and we used the well known identity $\ds{\int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t = \Psi\pars{z} + \gamma}$. $\ds{\Re\pars{z} > 0}$. $\ds{\gamma}$ is the Euler-Mascheroni constant.

Since $\ds{\Psi\pars{z} \stackrel{\mbox{def.}}{=} \totald{\ln\pars{\Gamma\pars{z}}}{z}}$, the integration is a straightforward one:


\begin{align}\fbox{$\ds{\ % 2\int_{0}^{\infty}{\cosh\pars{x} - 1 \over x\pars{\expo{ax} - 1}}\,\dd x\ }$} & = \ln\pars{\Gamma\pars{{1 \over a} + 1}\Gamma\pars{-\,{1 \over a} + 1}} = \ln\pars{{1 \over a}\,\Gamma\pars{{1 \over a}}\Gamma\pars{1 -{1 \over a}}} \\[3mm] & = \fbox{$\ds{\ \ln\pars{{\pi \over a\sin\pars{\pi/a}}}\ }$} \end{align}


In the last steps we used the Gamma recurrence formula and the Euler reflection formula.

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Jack D'Aurizio comments on the answer given by Olivier Oloa and I expand his remark into an answer. Thanks to him for this simple and beautiful observation.


We know that $$\sin \pi x = \pi x\prod_{n = 1}^{\infty}\left(1 - \frac{x^{2}}{n^{2}}\right)\tag{1}$$ and hence \begin{align} \log\left(\frac{\pi x}{\sin \pi x}\right) &= -\sum_{n = 1}^{\infty}\log\left(1 - \frac{x^{2}}{n^{2}}\right)\notag\\ &= \sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty}\frac{x^{2k}}{kn^{2k}}\notag\\ &= \sum_{k = 1}^{\infty}\frac{x^{2k}}{k}\sum_{n = 1}^{\infty}\frac{1}{n^{2k}}\notag\\ &= \sum_{n = 1}^{\infty}\frac{\zeta(2n)}{n}x^{2n}\tag{2} \end{align} and this is what you need in your last equation (with $1/a$ in place of $x$).

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    $\begingroup$ @JackD'Aurizio: Thanks man! It appears that although I do know some facts about various series but I am not used to applying them in various contexts to derive many exciting identities in obvious manner, rather this requires a gentle push via your comments. I will try to improve my skills. BTW as far as this question is concerned your approach of Frullani's Theorem is damn simple!! $\endgroup$ – Paramanand Singh Jun 2 '16 at 5:22

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