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

Perform part integration, we have \begin{align} &\int_{0}^{\infty}\frac{\log(\cosh x)}{\cosh x} \ d{x} \\ =&\ 2\tan^{-1}\left(\tanh\frac x2\right)\log(\cosh x)\bigg|_{0}^{\infty} - 2\int_{0}^{\infty}\tan^{-1}\left(\tanh\frac x2\right)\tanh x\ dx \end{align}

But, the first part diverges!

What other path can we take? I thought of expressing the hyperbolic functions in terms of exponentials.

Awaited eagerly for the answer.

Quanto
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Mathsource
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3 Answers3

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

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

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

aleden
  • 4,007
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To avoid divergence in part integration, choose instead $v=\cot^{-1}(\sinh x )$ \begin{align} &\int_{0}^{\infty}\frac{\ln(\cosh x)}{\cosh x}dx =-\int_{0}^{\infty} \ln(\cosh x)\ d(\cot^{-1}\sinh x)\\ \overset{ibp}= &\int_{0}^{\infty}\frac{\cot^{-1}(\sinh x)}{\coth x}dx =\int_{0}^{\infty}\frac1{\coth x}\int_0^1 \frac{\sinh x}{y^2 +\sinh^2 x}dy\ dx\\ =& \int_0^1\frac1{1-y^2}\int_0^\infty \bigg( \frac{1}{1+\sinh^2 x}- \frac{y^2}{y^2+\sinh^2 x}\bigg)\cosh x\ dx \ dy\\ =& \ \frac\pi2 \int_0^1 \frac1{1+y}dy=\frac{\pi}{2}\ln 2 \end{align}

Quanto
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