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Consider $(1)$

$$\int_{0}^{\infty}{e^x+e^{-x}-3\over (e^x+e^{-x})^2-1}\cdot \ln(x)\mathrm dx={\pi \ln(2)\over 3\sqrt{3}}\tag1$$

My try:

$x=-\ln(t)$ then $(1)$ becomes

$$\int_{0}^{1}{t^2-3t+1\over 2t+1}\cdot{\ln(-\ln t)}\mathrm dt\tag2$$

Split $(2):$

$$\color{blue}{\int_{0}^{1}{\ln(-\ln t)}\mathrm dt}+\int_{0}^{1}{t^2-5t\over 2t+1}\cdot{\ln(-\ln t)}\mathrm dt\tag3$$

$$\color{blue}{\gamma}+\int_{0}^{1}{t^2-5t\over 2t+1}\cdot{\ln(-\ln t)}\mathrm dt\tag4$$

Where $\gamma$ is Euler-Mascheroni Constant.

How may we prove $(1)?$

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  • $\begingroup$ This looks to me like something that could be solved with contour integration. $\endgroup$ – Tyberius May 7 '17 at 22:37
  • $\begingroup$ Only real analysis proof I can get so far uses partial fractions and is quite ugly and involved $\endgroup$ – Brevan Ellefsen May 7 '17 at 23:02
  • $\begingroup$ Hmm.. still can't find a better way than just splitting $$\int _1^{\infty}\left(\frac{2}{u^2+u+1}-\frac{1}{u^2-u+1}\right)\ln \left(\ln \left(u\right)\right)du$$... only problem is that things like the Gamma function come up in the "closed form" of each split integral. Not pretty... there must be symmetries here I am missing $\endgroup$ – Brevan Ellefsen May 7 '17 at 23:44
  • $\begingroup$ Another form is $$\int _1^\infty\frac{2u-3}{4u^2-1}\frac{\ln \left(\cosh ^{-1}u\right)}{\sqrt{u^2-1}}du$$ which might be attackable by trig substitution $\endgroup$ – Brevan Ellefsen May 7 '17 at 23:49
  • $\begingroup$ The integration can be performed as a contour integral along a $\mathsf{key\mbox{-}hole}$ which 'takes care' of the $\ln$-$\mathsf{Principal\ Branch}$. However, that is a $\mathsf{cumbersome\ task}$. I prefer the @mickep fine answer. $\endgroup$ – Felix Marin May 9 '17 at 1:13
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The factor $\ln x$ tells me that this is a Frullani integral, hidden with an integration by parts. Indeed (forget the constant) $$ \int\frac{e^x+e^{-x}-3}{(e^x+e^{-x})^2-1}\,dx =\frac{2}{\sqrt{3}}\biggl(\arctan\frac{1+2e^x}{\sqrt{3}}-\arctan\frac{1+2e^{2x}}{\sqrt{3}}\biggr). $$ This primitive will cancel the logaritmic singularities at $0$ and $+\infty$, and you are left with the Frullani integral $$ \begin{aligned} \int_0^{+\infty}\frac{1}{x}\frac{2}{\sqrt{3}}&\biggl(\arctan\frac{1+2e^{2x}}{\sqrt{3}}-\arctan\frac{1+2e^{x}}{\sqrt{3}}\biggr)\,dx\\ &=\frac{2}{\sqrt{3}}\biggl(\arctan\frac{1+2e^{+\infty}}{\sqrt{3}}-\arctan\frac{1+2e^0}{\sqrt{3}}\biggr)\ln\frac{2}{1}\\ &=\frac{2}{\sqrt{3}}\Bigl(\frac{\pi}{2}-\frac{\pi}{3}\Bigr)\ln 2 =\frac{\pi\ln 2}{3\sqrt{3}}. \end{aligned} $$

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    $\begingroup$ FANTASTIC answer. Thank you for posting this +1 $\endgroup$ – Brevan Ellefsen May 8 '17 at 16:25
  • $\begingroup$ Thank you @BrevanEllefsen, I'm glad it worked out so smoothly. $\endgroup$ – mickep May 8 '17 at 17:03
  • $\begingroup$ +1. Quite neat. Any contour integral is a cumbersome task. I started it but I left when I get bored !!!. $\endgroup$ – Felix Marin May 9 '17 at 1:13
  • $\begingroup$ Many thanks for the bounty! $\endgroup$ – mickep May 16 '17 at 19:01

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