Showing that $2\int_0^\infty {\cosh(x)-1\over x(e^{ax}-1)}\,\mathrm dx=\ln\left({\pi \over a\sin\left({\pi\over a }\right)}\right)$

Question

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

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

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

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

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?

Answer 1

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.

Answer 2

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!

Answer 3

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.

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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:

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\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}

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

Answer 5

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.

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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 $x$).