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Recently I went across an interesting integral $$\int\limits_0^{+\infty} \frac{(\coth x-1)(x\coth x-1)}{x} dx,$$ which numerically seems to equal $1+\gamma-\ln(2\pi)$. How can we prove it? I tried using the expansion $$\coth x=\frac1x+\sum_{n=1}^\infty\frac{2x}{x^2+\pi^2n^2},$$ but didn't succeed. Any other ways to approach this?

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  • $\begingroup$ Paul Enta, thanks, I fixed the typo ($x$ in the denominator instead $x^2$). $\endgroup$ – CuriousGuest Sep 26 '18 at 9:19
  • $\begingroup$ Try reducing $\cothx$ into exponentials and then integrate. $\endgroup$ – paulplusx Sep 26 '18 at 9:40
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    $\begingroup$ The result seems to be of the opposite sign. $\endgroup$ – Claude Leibovici Sep 26 '18 at 9:41
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I suppose that there is a simpler way, but here's one possible route:

We use your pole expansion $$ \coth (x) = \frac{1}{x}+\sum \limits_{n=1}^\infty \frac{2x}{x^2+\pi^2 n^2} $$ and the geometric series $$ \coth (x) = 1 + 2 \sum \limits_{k=1}^\infty \mathrm{e}^{-2 k x} $$ in the integral to obtain (monotone convergence theorem) $$ I \equiv \int \limits_0^\infty \left[\coth(x)-1\right] \left[\coth(x) - \frac{1}{x}\right] \, \mathrm{d} x = 4 \sum \limits_{k,n=1}^\infty \int \limits_0^\infty \frac{x}{x^2 + \pi^2 n^2} \mathrm{e}^{-2 k x} \, \mathrm{d} x \, .$$ Letting $x = \pi n t$ and employing the Laplace transform identity $$ \int \limits_0^\infty \frac{y}{1+y^2} \mathrm{e}^{-p y} \, \mathrm{d} y = \sin (p) \left[\frac{\pi}{2} - \operatorname{Si} (p)\right] - \cos(p) \operatorname{Ci}(p) \, , \, \operatorname{Re}(p) > 0 \, , $$ we find $$ I = - 4 \sum \limits_{k,n=1}^\infty \operatorname{Ci}(2 \pi k n) \, .$$ The definition of the cosine integral yields $$ I = 4 \sum \limits_{k,n=1}^\infty ~ \int \limits_{2 \pi k n}^\infty \frac{\cos(x)}{x} \, \mathrm{d} x = 4 \sum \limits_{k,n=1}^\infty ~ \int \limits_{2 \pi k n}^\infty \frac{\sin(x)}{x^2} \, \mathrm{d} x = \sum \limits_{k,n=1}^\infty \frac{2}{\pi k n} \int \limits_1^\infty \frac{\sin(2 \pi k n t)}{t^2} \, \mathrm{d} t \, .$$ Now we can make use of the Fourier series $$1 - 2 \{y\} = \sum \limits_{k=1}^\infty \frac{2}{\pi k} \sin(2 \pi k y) \, , y \in \mathbb{R} \, , $$ where $\{y\} = y - \lfloor y \rfloor$ is the fractional part of $y$ . Then the integral can be written as $$ I = \sum \limits_{n=1}^\infty \frac{1}{n} \int \limits_1^\infty \frac{1 - 2 \{n t\}}{t^2} \, \mathrm{d} t \, . $$ The integral $$\int \limits_1^\infty \frac{\{n t\}}{t^2} \, \mathrm{d} t = n \left[H_n - \ln(n) - \gamma\right]$$ can be computed from the definition of the fractional part and we are left with $$ I = - 2 \sum \limits_{n=1}^\infty \left[H_n - \ln(n) - \gamma -\frac{1}{2n}\right] = \ln(2\pi) - 1 -\gamma \, . $$ The last series has been evaluated in this question.

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  • $\begingroup$ Neat, thanks. It's indeed of the opposite sign. $\endgroup$ – CuriousGuest Sep 26 '18 at 18:32
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Another approach using the limits of Gamma and Zeta functions: $$ \coth{x}=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}=\frac{e^{2x}+1}{e^{2x}-1}=1+\frac{2}{e^{2x}-1} $$ Hence: $$ \begin{align} I&=\int_{0}^{\infty}\left(\coth{x}-1\right)\left(\coth{x}-1/x\right)\,dx \\[2mm] &=\int_{0}^{\infty}\coth^2{x}-\coth{x}-\coth{x}/x+1/x\,\,\,dx \\[2mm] &=\int_{0}^{\infty}\frac{4}{(e^{2x}-1)^2}+\frac{2-2/x}{e^{2x}-1}\,\,\,dx =\int_{0}^{\infty}\frac{2}{(e^x-1)^2}+\frac{1-2/x}{e^x-1}\,\,\,dx \\[2mm] &=\int_{0}^{\infty}\frac{2}{(e^x-1)^2}+\frac{1-2/x}{e^x-1}\,+\frac{2e^x-2e^x}{(e^x-1)^2}+\frac{2-2}{x^2}+\frac{1-1}{xe^x}\,\,\,dx \\[2mm] &=\color{red}{\int_{0}^{\infty}\left(\frac{2e^x}{(e^x-1)^2}-\frac{2}{x^2}\right)\,dx} -\color{blue}{\int_{0}^{\infty}\left(\frac{1}{e^x-1}-\frac{1}{xe^x}\right)\,dx} -\color{magenta}{\int_{0}^{\infty}\frac1x\left(\frac{2}{e^x-1}-\frac{2}{x}+\frac{1}{e^x}\right)\,dx} \\[2mm] &=\color{red}{\lim_{x\to0}\left[2\Gamma(x+1)\zeta(x)\right]} -\color{blue}{\lim_{x\to0}\left[\Gamma(x+1)\zeta(x+1)-\Gamma(x)\right]} -\color{magenta}{\lim_{x\to0}\left[2\Gamma(x)\zeta(x)+\Gamma(x)\right]} \\[2mm] &=\color{red}{-1}\color{blue}{-\gamma}\color{magenta}{+\log{2\pi}} \end{align} $$


Where $\,\displaystyle{\Gamma(s)\zeta(s)=\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^x-1}-\frac{1}{x}\right)dx\space\colon\space\Re{s}\in(0,1)}\,$
& IBP $\,\displaystyle{\left\{u=\frac{1}{e^x-1}-\frac{1}{x},\,dv=x^{s-1}\right\}}\,$ gives the first (red) term.

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