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 $(1)?$

  • $\begingroup$ 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

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 $$ \begin{aligned} \int_0^{+\infty}\frac{1}{x}\frac{2}{\sqrt{3}}&\biggl(\arctan\frac{1+2e^{2x}}{\sqrt{3}}-\arctan\frac{1+2e^{x}}{\sqrt{3}}\biggr)\,dx\\ &=\frac{2}{\sqrt{3}}\biggl(\arctan\frac{1+2e^{+\infty}}{\sqrt{3}}-\arctan\frac{1+2e^0}{\sqrt{3}}\biggr)\ln\frac{2}{1}\\ &=\frac{2}{\sqrt{3}}\Bigl(\frac{\pi}{2}-\frac{\pi}{3}\Bigr)\ln 2 =\frac{\pi\ln 2}{3\sqrt{3}}. \end{aligned} $$

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.