I have been messing around with this integral that has some particular special values$$I(a)=\int_0^\infty \ln\left(\tanh(ax)\right)dx$$

I found that $$I(1)=-\frac{\pi^2}{8}$$ $$I\left(\frac{1}{2}\right)=-\frac{\pi^2}{4}$$ $$I\left(\frac{1}{4}\right)=-\frac{\pi^2}{2}$$ $$...$$ and so on. It appears that $I(2^{-n})=-2^{n-3}\pi^2$. Can anyone explain how to derive a general closed form for $I(a)$ or at least why $I(2^{-n})$ takes on the particular values above?


We have $$I^\prime(a)=\int_0^\infty\frac{x\operatorname{sech}^2 ax dx}{\tanh ax}=\int_0^\infty\frac{2x dx}{\sinh 2ax}=\frac{1}{2a^2}\int_0^\infty\frac{y dy}{\sinh y},$$so constants $A,\,B$ exist with$$I(a)=A-\frac{B}{a}.$$From your results we can infer $A=0,\,B=\frac{\pi^2}{8}$.

Edit: a slicker way is to write $$\int_0^\infty\ln\frac{1-e^{-2ax}}{1+e^{-2ax}}dx=-2\int_0^\infty\sum_{n=0}^\infty\frac{1}{2n+1}e^{-(4n+2)ax} dx=-\frac{\pi^2}{8a}.$$

  • $\begingroup$ Wow, this is surprisingly simple! I did not consider differentiating. I also believe that the value of the last integral can be calculated by utilizing the geometric series and the value for $\zeta(2)$. Thank you so much! $\endgroup$ – aleden May 13 at 20:40
  • $\begingroup$ @aleden It can, yes, or from the results you'd already obtained. (I'd be interested to know your original approach to them.) $\endgroup$ – J.G. May 13 at 20:42
  • $\begingroup$ I used wolfram alpha to calculate those values to see if there was any sort of pattern. Now I can see that $I(a)$ is proportionate to the constant $\frac{\pi^2}{8}$. $\endgroup$ – aleden May 13 at 20:45
  • $\begingroup$ @J.G. What's the point in even differentiating? You can make the substitution $u=ax$ in $I(a)$ to conclude that $I(a)=\frac{k}a$ for some $k\in\mathbb{C}$. Then using $I(1)=k=-\frac{\pi^2}8$ gives the final result. $\endgroup$ – Peter Foreman May 13 at 21:37
  • 1
    $\begingroup$ @PeterForeman Sorry, force of logarithm-encountering habit. $\endgroup$ – J.G. May 13 at 22:23

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.