Two Amazing Integrals

Today we will prove the following two amazing integrals


\int_0^\infty \frac{1}{(b^2+x^2)\cosh\left(ax \right)}\,dx=\frac{2 \pi}{b}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2ab+(2k-1)\pi}(1)


\int_0^\infty \frac{\ln(z^2+x^2)}{\left(e^{\pi x}+e^{-\pi x} \right)}\,dx=\ln\Gamma\left( \frac{z }{2}+\frac34\right)-\ln\Gamma\left( \frac{z }{2}+\frac14\right)+\frac{\ln(2)}{2}(2)


In order to prove (1), first recall the following results:


\int_0^\infty e^{-bt}\cos\left(xt \right)\,dt=\frac{b}{b^2+x^2}(3)


\int_0^\infty \frac{\cos\left(xt \right)}{\cosh\left(ax \right)}\,dx=\frac{\pi} {2a \cosh\left(\frac{\pi t}{2a} \right)}(4)

Then,


\begin{aligned} \int_0^\infty \frac{1}{(b^2+x^2)\cosh\left(ax \right)}\,dx&= \frac 1b\int_0^\infty \frac{1}{\cosh\left(ax \right)}\,\int_0^\infty e^{-bt}\cos\left(xt \right)\,dt\,dx\\ &= \frac 1b\int_0^\infty e^{-bt}\int_0^\infty \frac{\cos\left(xt \right)}{\cosh\left(ax \right)}\,dx\,dt\\ &= \frac 1b\int_0^\infty e^{-bt}\left(\frac{\pi} {2a \cosh\left(\frac{\pi t}{2a} \right)} \right)\,dt\\ &= \frac{\pi}{2 a b}\int_0^\infty \frac{ e^{-bt}} {\cosh\left(\frac{\pi t}{2a} \right)} \,dt\\ &= \frac{1}{b}\int_0^\infty \frac{ e^{-\frac{2ab}{\pi}t}} {\cosh\left(t \right)} \,dt \qquad \left(\frac{2ab}{\pi}w=t\right)\\ &= \frac{2}{b}\int_0^\infty \frac{ e^{-ct}} {e^{t}+e^{-t}} \,dt \qquad \left(c=\frac{2ab}{\pi}\right)\\ &= \frac{2}{b}\int_0^\infty \frac{ e^{-ct}e^{-t}} {1+e^{-2t}} \,dt\\ &= \frac{2}{b}\int_0^\infty e^{-ct}e^{-t}\left(\sum_{k=0}^\infty (-1)^{k}e^{-2kt} \right) \,dt\\ &= \frac{2}{b}\int_0^\infty e^{-ct}\left(\sum_{k=0}^\infty (-1)^{k}e^{-(2k+1)t} \right) \,dt\\ &= \frac{2}{b}\int_0^\infty e^{-ct}\left(\sum_{k=1}^\infty (-1)^{k-1}e^{-(2k-1)t} \right) \,dt\\ &= \frac{2}{b}\sum_{k=1}^\infty (-1)^{k-1}\,\int_0^\infty e^{-\left(c+(2k-1)\right)t} \,dt\\ &= \frac{2}{b}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{\frac{2ab}{\pi}+(2k-1)}\\ &=\frac{2 \pi}{b}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2ab+(2k-1)\pi} \qquad \blacksquare \end{aligned}


\boxed{\int_0^\infty \frac{1}{(b^2+x^2)\cosh\left(ax \right)}\,dx=\frac{2 \pi}{b}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2ab+(2k-1)\pi}}(5)

If we let b \to z and a \to \pi in (5), we get


\begin{aligned} \int_0^\infty \frac{1}{(z^2+x^2)\cosh\left(\pi x \right)}\,dx&=\frac{2 \pi}{z}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2z \pi+(2k-1)\pi}\\ &=\frac{2 \pi}{z}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2z \pi+2 \pi k-\pi}\\ &=\frac{2 }{z} \sum_{k=1}^\infty \frac{(-1)^{k-1}}{2z +2 k-1}\\ &=\frac{1 }{z} \sum_{k=1}^\infty \frac{(-1)^{k-1}}{z + k-\frac12}\\ &=\frac{1 }{z} \sum_{k=1}^\infty (-1)^{k-1}\int_0^1x^{z + k+\frac12-1}\,dx\\ &=\frac{1 }{z} \int_0^1x^{z +\frac12}\left(\sum_{k=1}^\infty (-1)^{k-1}x^{k-1}\right)\,dx\\ &=\frac{1 }{z} \int_0^1x^{z +\frac12}\left(\sum_{k=0}^\infty (-1)^{k}x^{k}\right)\,dx\\ &=\frac{1 }{z} \int_0^1\frac{x^{z +\frac12}}{1+x}\,dx\\ &=\frac{1 }{2z}\left(\psi\left( \frac{z +\frac12+1}{2}\right)-\psi\left( \frac{z +\frac12}{2}\right) \right)\\ &=\frac{1 }{2z}\left(\psi\left( \frac{z }{2}+\frac34\right)-\psi\left( \frac{z }{2}+\frac14\right)\right) \qquad \blacksquare \end{aligned}



\boxed{\int_0^\infty \frac{1}{(z^2+x^2)\cosh\left(\pi x \right)}\,dx=\frac{1 }{2z}\left(\psi\left( \frac{z }{2}+\frac34\right)-\psi\left( \frac{z }{2}+\frac14\right)\right)}(6)


Where We used the result


\int_0^1\frac{t^{x-1}}{1+t}dt=\frac{1}{2}\left(\psi\left( \frac{x+1}{2}\right)-\psi\left( \frac{x}{2}\right) \right)

We can rewrite (6) as


\int_0^\infty \frac{2z}{(z^2+x^2)\left(e^{\pi x}+e^{-\pi x} \right)}\,dx=\frac{1 }{2}\left(\psi\left( \frac{z }{2}+\frac34\right)-\psi\left( \frac{z }{2}+\frac14\right)\right)
(7)


If we now integrate (7) w.r. to z we have:


\begin{aligned} &\int\int_0^\infty \frac{2z}{(z^2+x^2)\left(e^{\pi x}+e^{-\pi x} \right)}\,dx\,dz=\frac{1 }{2}\int\left(\psi\left( \frac{z }{2}+\frac34\right)-\psi\left( \frac{z }{2}+\frac14\right)\right)\,dz\\ &\int_0^\infty \frac{1}{\left(e^{\pi x}+e^{-\pi x} \right)}\int\frac{2z}{(z^2+x^2)}\,dz\,dx=\frac{1 }{2}\left(2\ln\Gamma\left( \frac{z }{2}+\frac34\right)-2\ln\Gamma\left( \frac{z }{2}+\frac14\right)\right)+C\\ &\int_0^\infty \frac{\ln(z^2+x^2)}{\left(e^{\pi x}+e^{-\pi x} \right)}\,dx=\ln\Gamma\left( \frac{z }{2}+\frac34\right)-\ln\Gamma\left( \frac{z }{2}+\frac14\right)+C\\ \end{aligned}


The evaluation of the constant C is a beautiful exercise per se. Fortunately relying on the previous estabilished Vardi´s integral We may accomplish it easily. Setting z=0 in the last equation, the L.H.S. becomes


\begin{aligned} I&=\int_0^\infty \frac{\ln(x^2)}{e^{\pi x}+e^{-\pi x}}\,dx\\ &=2\int_0^\infty \frac{\ln(x)}{e^{\pi x}+e^{-\pi x}}\,dx\\ &=\int_0^\infty \frac{\ln(x)}{\cosh(\pi x)}\,dx\\ &=\frac{1}{\pi}\int_0^\infty \frac{\ln(x)}{\cosh( x)}\,dx-\frac{\ln(\pi)}{\pi}\int_0^\infty \frac{1}{\cosh( x)}\,dx\\ &=\frac{1}{\pi}\frac{\pi}{2} \ln \left( \frac{4 \pi^3}{\Gamma^4 \left(\frac{1}{4}\right) }\right)-\frac{\ln(\pi)}{\pi}\frac{\pi}{2}\\ &=\frac{1}{2} \ln \left( \frac{4 \pi^3}{\Gamma^4 \left(\frac{1}{4}\right) }\right)-\frac{\ln(\pi)}{2}\\ &=\frac12 \ln(4\pi^3)-\frac12\ln\left(\Gamma^4\left( \frac14\right)\right) -\frac{\ln(\pi)}{2}\\ &=\ln(2)-2\ln\left(\Gamma\left( \frac14\right)\right)+\ln(\pi) \qquad \blacksquare \end{aligned}


Where we used the Vardi´s integral proved here:


\int_0^\infty \frac{\ln(x)}{\cosh( x)}\,dx=\frac{\pi}{2} \ln \left( \frac{4 \pi^3}{\Gamma^4 \left(\frac{1}{4}\right) }\right)


And


\int_0^\infty \frac{1}{\cosh( x)}\,dx=\frac\pi2

The R.H.S. becomes


\begin{aligned} \ln\left(\frac{\Gamma\left( \frac34\right)}{\Gamma\left( \frac14\right)}\right)+C&=\ln\left(\frac{\sqrt{2\pi^2}}{\Gamma^2\left( \frac14\right)}\right)+C\\ &=\frac{\ln\left(2\right)}{2}+\ln(\pi)-2\ln\left(\Gamma\left( \frac14\right) \right)+C \qquad \blacksquare\\ \end{aligned}


Equating L.H.S. and R.H.S. we conclude that


\ln(2)-2\ln\left(\Gamma\left( \frac14\right)\right)+\ln(\pi)=\frac{\ln\left(2\right)}{2}+\ln(\pi)-2\ln\left(\Gamma\left( \frac14\right) \right)+C

C=\frac{\ln(2)}{2} \qquad \blacksquare


And finally


\boxed{\int_0^\infty \frac{\ln(z^2+x^2)}{\left(e^{\pi x}+e^{-\pi x} \right)}\,dx=\ln\Gamma\left( \frac{z }{2}+\frac34\right)-\ln\Gamma\left( \frac{z }{2}+\frac14\right)+\frac{\ln(2)}{2}}(8)


Or


\boxed{\int_0^\infty \frac{\ln(z^2+x^2)}{\cosh(\pi x)}\,dx=2\ln\Gamma\left( \frac{z }{2}+\frac34\right)-2\ln\Gamma\left( \frac{z }{2}+\frac14\right)+\ln(2)}(9)

Appendix

Recall Legendre Duplication Formula for the Gamma Function


\Gamma\left( 2x\right)=\frac{2^{2x-1}\Gamma\left( x\right)\Gamma\left( x+\frac12\right)}{\sqrt{\pi}}

Letting x \to \frac14

\begin{aligned} \Gamma\left( \frac12\right)=\frac{2^{-\frac12}\Gamma\left( \frac14\right)\Gamma\left( \frac34\right)}{\sqrt{\pi}}\\ \sqrt{\pi}=\frac{\Gamma\left( \frac14\right)\Gamma\left( \frac34\right)}{\sqrt{2\pi}}\\ \Gamma\left( \frac34\right)=\frac{\sqrt{2\pi^2}}{\Gamma\left( \frac14\right)}\\ \end{aligned}

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