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$$ I = \int_{0}^{\infty} \frac{\tanh\left(x\right)\operatorname{sech}\left(x\right)}{x}\,{\rm d}x $$

I started with the substitution $\cosh x=t \implies [0,\infty] \to[1,\infty]$

$$I=\int_1^\infty \frac{1}{t^2 \ln(t+\sqrt{t^2-1})}\,dt$$

then the substitution $t+\sqrt{t^2-1}=z\implies [1,\infty] \to[1,\infty]$

$$I=\int_1^\infty \frac{2(z^2-1)}{(z^2+1)^2\ln(z)}\,dz$$

I am stuck on this integral and I have tried some more substitutions but they seem to make the integral harder to evaluate, I am interested to see how to solve this.

Felix Marin
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Amrut Ayan
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  • It seems no elementary antiderivative exists. – Lucenaposition Jul 16 '24 at 07:18
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    This integral can be done with residue theorem, write $\frac{\tanh(x) \text{sech}(x)}{x}$ as $\frac{\sinh(x)}{x\cosh^{2}(x)}$. Since the integral is even, then you can extend the limit over $R$, choose poles are $(2k+1)\frac{i\pi}{2}$ where $k\ge 0$. The residue is $\frac{-(-1)^k4i}{\pi^2(2k+1)^2}$. Taking sum from $k=0$ to $\infty$ you get $-\frac{4iG}{\pi^2}$, where $G$ is the Catalan constant. The result of the given integral equals $\frac{1}{2}2\pi i \left(-\frac{4iG}{\pi^2}\right)=\frac{4G}{\pi}$ – PelikanSH037 Jul 16 '24 at 08:44

2 Answers2

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Let $J(a)=\int_0^\infty \frac{z^{a+1}}{(1+z^2)^2\ln z}dz$. Then $$J’(a) = \int_0^\infty \frac{t^{a+1}}{(1+t^2)^2} dt\overset{ibp} =-\frac a2 \int_0^\infty \frac{t^{a+1}}{1+t^2}dt =\frac{\pi a}{4}\csc\frac{\pi a}2 $$ and

$$I=\int_1^\infty \frac{2(z^2-1)}{(z^2+1)^2\ln z}\,dz =\int_0^\infty \frac{z^2-1}{(z^2+1)^2\ln z}\,dz$$

$$=\int_{-1}^{1}J’(a)da \overset{t=\tan\frac {\pi a}4} =\frac 4\pi \int_0^{1}\frac {\tan^{-1}t}{t}dt =\frac{4}{\pi}G \\ $$

Quanto
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As @Quanto noticed that $$ I \stackrel{x\mapsto\frac{1}{x}}{=} \int_0^1 \frac{2\left(z^2-1\right)}{\left(1+z^2\right)^2 \ln z} d z, $$ we have $$ I=\int_0^1+\int_1^{\infty}=\int_0^{\infty} \frac{z^2-1}{\left(1+z^2\right)^2 \ln z} d z. $$ Noticing that $$ \frac{z^2-1}{\ln z}=\int_{-1}^1 z^{y+1} d y, $$ we can re-write the integral into a double integral $$ I=\int_{-1}^1 \int_0^{\infty} \frac{z^{y+1}}{\left(1+z^2\right)^2} d z d y. $$ We now let $z=\tan \theta$, then $$ \begin{aligned} \int_0^{\infty} \frac{z^{y+1}}{\left(1+z^2\right)^2} d z =& \int_0^{\frac{\pi}{2}} \sin ^{y+1} \theta \cos ^{1-y} \theta d \theta \\ = & \frac{1}{2} B\left(\frac{y}{2}+1, 1-\frac{y}{2}\right) \\ = & \frac{1}{2} \frac{\Gamma\left(\frac{y}{2}+1\right) \Gamma\left(1-\frac{y}{2}\right)}{\Gamma(2)} \\ = & \frac{1}{2} \cdot \frac{y}{2} \cdot \Gamma\left(\frac{y}{2}\right) \Gamma\left(1-\frac{y}{2}\right) \\ = & \frac{\pi}{4} y \csc \frac{\pi y}{2} \quad \textrm{( By the reflection property of Gamma function)} \end{aligned} $$ Plugging back yields

$$ \begin{aligned} I & =\frac{\pi}{4} \int_{-1}^1 \frac{y}{\sin \left(\frac{\pi y}{2}\right)} d y \\ & =\frac{\pi}{2} \int_0^1 \frac{y}{\sin \left(\frac{\pi y}{2}\right)} d y \\ & =\frac{8}{\pi} \int_0^{\frac{\pi}{4}} \frac{z}{\sin (2 z)} d z \text {, where } z=\frac{\pi y}{4} \\ & =\frac{4}{\pi} \int_0^{\frac{\pi}{4}} z \,d(\ln (\tan z)) \\ & =-\frac{4}{\pi} \int_0^{\frac{\pi}{4}} \ln (\tan z) d z \quad \textrm{ (Via integration by parts)} \\ & =-\frac{4}{\pi}(-G) \cdots (*)\\ & =\frac{4 G}{\pi} \end{aligned} $$

where (*) using the result in the post.

Lai
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