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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.

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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 $e^x=\cot \left(\frac{t}{2}\right)$.

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    $\begingroup$ 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. $\endgroup$ – Zacky Jun 7 at 17:29
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    $\begingroup$ It's somehow similar to Weierstrass-Euler Tangent Half-Angle Substitution. Fine job. $\endgroup$ – Felix Marin Jun 9 at 16:38
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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 $$-I'(1)=\int_0^\infty \frac{\ln(\cosh(x))}{\cosh(x)}dx=\frac{\pi\ln(2)}{2}$$

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