Help with $\int _0^{\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x$

Solution 1:

I present another approach, utilising a well-known property of the Laplace Transform i.e. $$\int_{0}^{+\infty} f\left(x\right) g\left(x\right) \, dx = \int_{0}^{+\infty} \left(\mathcal{L} f\right)\left(y\right)\left(\mathcal{L}^{-1} g\right)\left( y\right) \, dy$$ Letting $f\left(x\right) = \tanh \left(x\right) \text{sech} \left(x\right)$ and $g(x) = \frac{1}{x}$: $$\left( \mathcal{L} f \right) (y) = 1 + \frac{1}{2} y \left( \psi^{(0)} \left(\frac{1+y}{4}\right)- \psi^{(0)} \left(\frac{3+y}{4}\right)\right)$$ $$\left( \mathcal{L}^{-1} g \right) \left(y \right) = 1$$ Where $\psi$ represents the polygamma function.

$$\implies I = \int_{0}^{\infty} \left(1 + \frac{1}{2} y \left( \psi^{(0)} \left(\frac{1+y}{4}\right)- \psi^{(0)} \left(\frac{3+y}{4}\right)\right)\right) \, dy$$ I will now proceed to integrate indefinitely, then take limits at the end.

$$\int 1 \, dy + \frac{1}{2} \int y \, \psi^{(0)} \left(\frac{1+y}{4}\right) \, dy - \frac{1}{2} \int y \, \psi^{(0)} \left(\frac{3+y}{4}\right) \, dy$$ We proceed with integrating by parts. $$=y + \frac{1}{2} \left( 4y \ln \Gamma \left( \frac{1+y}{4} \right) - 4 \int \ln \Gamma \left( \frac{1+y}{4} \right) \, dy \right)-\frac{1}{2} \left( 4y \ln \Gamma \left( \frac{3+y}{4} \right) - 4 \int \ln \Gamma \left( \frac{3+y}{4} \right) \, dy \right)$$ $$=y + 2 y \ln \Gamma \left( \frac{1+y}{4}\right) - 8 \psi^{(-2)} \left(\frac{1+y}{4}\right) - 2y \ln \Gamma \left( \frac{3+y}{4} \right) + 8 \psi^{(-2)} \left( \frac{3+y}{4} \right)$$ Taking limits to $\infty$ and $0$, and then subtracting gives us: $$2\ln (2\pi) + 8 \psi^{(-2)} \left( \frac{1}{4} \right)-8 \psi^{(-2)} \left( \frac{3}{4} \right)$$ $$=\boxed{\frac{4G}{\pi}}$$ (the equality was checked with WolframAlpha) as required.

Solution 2:

Let $I(a)=\int _0^{\infty }\frac{\sinh (ax)}{x\cosh ^2x}{d}x$. Then $$I’(a)= \int _{-\infty}^{\infty }\frac{\cosh (ax)}{2\cosh ^2x}{d}x \overset{t=e^{2x}}=\int_0^\infty \frac{t^{-\frac1a}+t^{\frac1a}}{2(1+t)^2}dt =\frac{\pi a}{2\sin\frac{\pi a}2} $$ and $$\int _0^{\infty }\frac{\sinh x}{x\cosh ^2x}{d}x =\int_0^1 I’(a)da \overset{s=\tan\frac {\pi a}4} =\frac 4\pi \int_0^{1}\frac {\tan^{-1}s}{s}ds =\frac{4}{\pi}G \\ $$ where the Maclaurin series for the inverse tangent function leads to $ G $.