Show that $$\int_{0}^{\infty}\frac{\log(\cosh x)}{\cosh x}dx = \frac{\pi}{2}\log 2$$

Using part integration, we have: $$ \int_{0}^{\infty}\frac{\log(\cosh x)}{\cosh x} \ d{x} = 2\arctan\bigl[\tanh(x/2)\bigr]\log(\cosh x)\mid_{0}^{\infty} - 2\int_{0}^{\infty}\arctan\bigl[\tanh(x/2)\bigr]\frac{\sinh x}{\cosh x}dx $$ But the first part diverge. What other path can we use? I thought of expressing the hyperbolic functions in terms of exponentials.

Awaited eagerly for the answer.


$$\int_0^\infty\frac{\ln(\cosh x)}{\cosh x} dx\overset{x=-\ln t}=2 \int_0^1 \frac{\ln\left(\frac{t^2+1}{2t}\right)}{1+t^2} dt\overset{t=\tan \left(\frac{x}{2}\right)}=-\int_0^\frac{\pi}{2}\ln(\sin x)dx=\frac{\pi}{2}\ln 2$$ See here for the last integral.

Alternatively we can combine everything from above into the substitution $e^x=\cot \left(\frac{t}{2}\right)$.

  • 3
    $\begingroup$ You could also write this substitution as $\cosh x=\csc t$, or $t=\frac{\pi}{2}-\operatorname{gd}x$. $\endgroup$ – J.G. Jun 7 at 17:18
  • $\begingroup$ Thanks, the one containing the Gudermannian function is quite fancy (I keep forgetting about that function). Anyway I should've used $\cosh x=\csc x$ from the start. $\endgroup$ – LeBlanc Jun 7 at 17:29
  • 1
    $\begingroup$ It's somehow similar to Weierstrass-Euler Tangent Half-Angle Substitution. Fine job. $\endgroup$ – Felix Marin Jun 9 at 16:38

Another method:

Let $$I(s)=\int_0^\infty \operatorname{sech}^s(x)dx$$ where $$-I'(1)=\int_0^\infty \ln\left(\cosh(x)\right)\operatorname{sech}(x)dx$$ Using my answer from this question we can evaluate $I(s)$ in terms of the Beta Function/Gamma Function.

$$I(s)=\frac{\Gamma(\frac{s}{2})\sqrt{\pi}}{2\Gamma(\frac{s+1}{2})}$$ We proceed to take the derivative: $$I'(s)=\frac{\sqrt{\pi}}{2}\left(\frac{\frac{1}{2}\Gamma'(\frac{s}{2})\Gamma(\frac{s+1}{2})-\frac{1}{2}\Gamma(\frac{s}{2})\Gamma'(\frac{s+1}{2})}{\Gamma^2(\frac{s+1}{2})}\right)$$ $$I'(1)=\frac{\sqrt{\pi}}{4}\left(\Gamma'\left(\frac{1}{2}\right)-\Gamma'(1)\sqrt{\pi}\right)=\frac{\pi}{4}\left(\psi_0\left(\frac{1}{2}\right)+\gamma\right)=\frac{\pi}{4}\left(-2\ln(2)-\gamma+\gamma\right)=-\frac{\pi\ln(2)}{2}$$

So $$-I'(1)=\int_0^\infty \frac{\ln(\cosh(x))}{\cosh(x)}dx=\frac{\pi\ln(2)}{2}$$


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.