\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(x)}-\frac{a}{e^{2x}}\right)\,\frac{dx}{x}

In this post We will compute the following integral:


\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(x)}-\frac{a}{e^{2x}}\right)\,\frac{dx}{x}=\ln\left(\frac{\pi}{\Gamma^2\left(\frac{1+a}{2} \right)\cos\left(\frac{a \pi }{2} \right) } \right)


\begin{aligned} I(a)&=\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(x)}-\frac{a}{e^{2x}}\right)\frac{dx}{x}\\ & \\ I^\prime(a)&=\int_0^\infty \left(\frac{x\cosh(ax)}{\sinh(x)}-\frac{1}{e^{2x}}\right)\frac{dx}{x}\\ &=\int_0^\infty \left(\frac{\cosh(ax)}{\sinh(x)}-\frac{e^{-2x}}{x}\right)dx\\ &=\int_0^\infty \left(\frac{e^{ax}+e^{-ax}}{e^x-e^{-x}}-\frac{e^{-2x}}{x}\right)dx\\ &=\int_0^1 \left(\frac{x^{a}+x^{-a}}{x^{-1}-x}+\frac{x^2}{\ln(x)}\right)\frac{dx}{x} & (e^{-x} \to x)\\ &=\int_0^1 \left(\frac{x}{x}\cdot\frac{x^{a}+x^{-a}}{x^{-1}-x}+\frac{x^2}{\ln(x)}\right)\frac{dx}{x}\\ &=\int_0^1 \left(\frac{x^{a+1}+x^{1-a}}{1-x^2}+\frac{x^2}{\ln(x)}\right)\frac{dx}{x}\\ &=\frac12\int_0^1 \left(\frac{x^{\frac{a+1}{2}}+x^{\frac{1-a}{2}}}{1-x}+\frac{2x}{\ln(x)}\right)\frac{dx}{\sqrt{x}\sqrt{x}} & (x^2 \to x)\\ &=\frac12\int_0^1 \left(\frac{1}{\ln(x)}+\frac{x^{\frac{a+1}{2}-1}}{1-x}\right)dx+\frac12\int_0^1 \left(\frac{1}{\ln(x)}+\frac{x^{\frac{1-a}{2}-1}}{1-x}\right)dx\\ &=-\frac12\psi\left(\frac{1+a}{2} \right)-\frac12\psi\left(\frac{1-a}{2} \right) \end{aligned}


Then


\begin{aligned} I(a)&=-\frac12\int_0^a\psi\left(\frac{1+u}{2} \right)\,du-\frac12\int_0^a\psi\left(\frac{1-u}{2} \right)\,du\\ &=-\int_{1/2}^{\frac{1+a}{2}}\psi\left(u \right)\,du+\int_{1/2}^{\frac{1-a}{2}}\psi\left(u \right)\,du\\ &=\ln\left(\Gamma\left(\frac{1-a}{2} \right) \right)-\ln\left(\Gamma\left(\frac{1}{2} \right) \right)-\ln\left(\Gamma\left(\frac{1+a}{2} \right) \right)+\ln\left(\Gamma\left(\frac{1}{2} \right) \right)\\ &=\ln\left(\frac{\Gamma\left(\frac{1-a}{2} \right)}{\Gamma\left(\frac{1+a}{2} \right) } \right)\\ &=\ln\left(\frac{\Gamma\left(\frac{1-a}{2} \right)\Gamma\left(\frac{1+a}{2} \right)}{\Gamma\left(\frac{1+a}{2} \right)\Gamma\left(\frac{1+a}{2} \right) } \right)\\ &=\ln\left(\frac{\pi}{\Gamma^2\left(\frac{1+a}{2} \right)\cos\left(\frac{a \pi }{2} \right) } \right) \qquad \blacksquare\\ \end{aligned}


We used that (see here):


\int_0^1 \left(\frac{x^{z-1}}{\ln(x)}+\frac{x^{w-1}}{1-x}\right)dx=\ln(z)-\psi(w)

and (here)

\Gamma\left(\frac{1}{2}-x\right) \Gamma\left(\frac{1}{2}+x\right)=\frac{\pi}{\cos \pi x}

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