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When I met the integral, $$\int _0^{\infty }\frac{x}{1+2\cosh x}dx,$$ I just wondered whether there is a simpler method to evaluate it and started with an usual substitution: $y=e^{-x}$ and obtained $$ \begin{aligned} \int _0^{\infty }\frac{x}{1+2\cosh x}dx& =\int_0^1 \frac{-\ln y}{1+y^{-1}+y} \frac{d y}{y} \\ & =-\int_0^1 \frac{\ln y}{y^2+y+1} d y\\ & =-\int_0^1 \frac{(1-y) \ln y}{1-y^ 3} d y \\ & =-\int_0^1(1-y) \ln y \sum_{n=0}^{\infty} y^{3n} d y \quad \textrm{ (via expansion)} \\ & =-\sum_{n=0}^{\infty} \int_0^1\left(y^{3 n}-y^{3 n+1}\right) \ln y d y \\ & =\sum_{n=0}^{\infty}\left[\frac{1}{(3 n+1)^2}-\frac{1}{(3 n+2)^2}\right] \quad (\textrm{ via IBP twice}) \\ & =\frac{1}{9} \sum_{n=0}^{\infty}\left[\frac{1}{\left(n+\frac{1}{3}\right)^2}-\frac{1}{\left(n+\frac{2}{3}\right)^2}\right] \\ & =\frac{1}{9}\left[\psi^{\prime}\left(\frac{1}{3}\right)-\psi^{\prime}\left(\frac{2}{3}\right)\right] \end{aligned} $$

My Question:

Whether there is a simpler answer or method to evaluate the integral?

Your comments and alternative methods are highly appreciated.

Lai
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  • "Long" time no see. Simpler, I do not think (even if the antiderivative does exist. – Claude Leibovici Oct 01 '25 at 08:21
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    $\begin{array}{l} \int\limits_0^\infty {\frac{x}{{1 + 2\cosh \left( x \right)}}dx} = \frac{2}{{\sqrt 3 }}\int\limits_0^\infty {x\left( {\sum\limits_{n = 1}^\infty {\sin \left( {\frac{{2n\pi }}{3}} \right){e^{ - nx}}} } \right)dx} = \frac{2}{{\sqrt 3 }}\sum\limits_{n = 1}^\infty {\sin \left( {\frac{{2n\pi }}{3}} \right)} \int\limits_0^\infty {x{e^{ - nx}}dx} \ = \frac{2}{{\sqrt 3 }}\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}\sin \left( {\frac{{2n\pi }}{3}} \right)} = \frac{2}{{\sqrt 3 }}{\rm{C}}{{\rm{l}}_2}\left( {\frac{{2\pi }}{3}} \right) \end{array}$ – OnTheWay Oct 01 '25 at 08:42
  • @Claude Leibovici, Happy to see you again after my 13-day journey. – Lai Oct 01 '25 at 08:43
  • ${\text{C}}{{\text{l}}_2}\left( z \right){\text{ is order 2 Clausen function, and can be expressed as trigamma function}}$https://en.wikipedia.org/wiki/Clausen_function#Relation_to_the_polygamma_function – OnTheWay Oct 01 '25 at 08:43
  • @OnTheWay, thank you for your interesting solution. But your answer is -G, instead of 0.7813. – Lai Oct 01 '25 at 09:11
  • @Lai It's not : https://www.wolframalpha.com/input?i=%282%2FSqrt%5B3%5D%29Sum%5BSin%5B2nPi%2F3%5D%2Fn%5E2%2C%7Bn%2C1%2CInfinity%7D%5D – OnTheWay Oct 01 '25 at 09:14
  • Answers don't lie. If the integral value is expressed in terms of $\psi'$, it has to be so. Unless $\psi'$ enjoys a special property that allows to simplify the expression, there is no escape. –  Oct 01 '25 at 09:27
  • @OnTheWay, your answer is correct. Please post your answer as a solution and explain on how to get the first nice expansion. – Lai Oct 01 '25 at 12:34

3 Answers3

4

$$I=\int \frac{x}{1+a\cosh x}\,dx=\int \frac{2 \log (y)}{a y^2+2 y+a}\,dy=\frac 2a \int \frac{ \log (y)} {(y-\alpha)(y-\beta) }\,dy$$ where $$(\alpha,\beta)=-\frac{1\pm\sqrt{1-a^2}}{a}$$ Now $$\frac{1} {(y-\alpha)(y-\beta) }=\frac{1}{\alpha -\beta }\Big(\frac 1 {y-\alpha}- \frac 1 {y-\beta}\Big)$$ Using twice $$\int \frac {\log(y)}{y-\gamma}\,dy=\text{Li}_2\left(\frac{y}{\gamma }\right)+\log (y) \log \left(1-\frac{y}{\gamma }\right)$$

Recombine the logarithms and use the bounds.

Then, $$J=\int_0^\infty \frac{x}{1+a\cosh x}\,dx$$ $$J=-\frac{1}{\sqrt{1-a^2}}\Bigg(\text{Li}_2\left(\frac{a}{\sqrt{1-a^2}- 1}\right)+ \text{Li}_2\left(-\frac{a} {\sqrt{1-a^2}+1}\right) \Bigg)$$

Claude Leibovici
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In general \begin{align} I(a)=&\int _0^{\infty }\frac{x}{1+\sec a\cosh x}dx\\ =&\ \frac2{\tan a } \Im \int_0^\infty \frac x{1+e^{x-i a}}dx=- \frac2{\tan a }\Im \text{Li}_2(-e^{i a}) \end{align} and, in particular $$\int _0^{\infty }\frac{x}{1+2\cosh x}dx=I(\frac\pi3) = -\frac2{\sqrt3}\Im \text{Li}_2(-e^{i \frac\pi3})$$

Quanto
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  • Always impressed by your solutions and out of curiosity, what did make you thinking about $\sec(a)$ instead of my dumb $a$ ? – Claude Leibovici Oct 01 '25 at 14:10
  • @ClaudeLeibovici $\sec a$ makes it easier to partially fractionalize the integrand. – Quanto Oct 01 '25 at 14:14
  • @Quanto, your solution is wonderful. But by WA, http://www.wolframalpha.com/input/?i=imaginary%20part%20of%202%2F%E2%88%9A3Li2%28-e%5E%28i%CF%80%2F3%29, your answer is -0.7813 instead of 0.7813. – Lai Oct 02 '25 at 00:27
  • @Quanto, I had just tried and found that your first step is so wonderful that gave the final answer as $\frac2{\tan a }\Im \text{Li}_2(-e^{i a})$ missing a negative sign, which should be $-\frac2{\tan a }\Im \text{Li}_2(-e^{i a})$. So $$\int _0^{\infty }\frac{x}{1+2\cosh x}dx=I(\frac\pi3) = -\frac2{\sqrt3}\Im \text{Li}_2(-e^{i \frac\pi3}).$$ Am I right? – Lai Oct 02 '25 at 01:14
  • @Lai Thanks for the catch. Will correct – Quanto Oct 02 '25 at 01:22
  • I am glad to have such a magical solution to the integral in general. Admiration always drives me to learn. – Lai Oct 02 '25 at 01:30
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The simplest way to deduce this result is to look the integral up in formula 3.531.2 of the Gradsteyn-Ryzhik tables using the substitution $x=2t$ and then $u=\pi/6$.
$$ \int_0^\infty \frac{t}{\cosh(2t)+\cos(2u)}dt = \frac{u\ln 2 -L(u)}{\sin(2u)} $$ [Eventually the Lobachevsky function $L$ turns into the trigamma functions, so there no simplification beyond what's already been posted here.]

R. J. Mathar
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