INTEGRAL sinh(ax)/sinh(bx)dx

In this post we will prove 4 beautiful integrals involving the hyperbolic sine function, namely


\begin{aligned} &\int_0^\infty \frac{\sinh(ax)}{\sinh(bx)}\,dx=\frac{\pi}{2b} \tan\left(\frac{a \pi}{2b} \right)\\ &\\ &\int_0^\infty \frac{\sin(ax)}{\sinh(bx)}\,dx=\frac{\pi}{2b} \tanh\left(\frac{a \pi}{2b} \right)\\ & \\ &\int_0^\infty \frac{x}{a^2+x^2}\frac{1}{\sinh(bx)}\,dx=\frac{1}{2} \psi\left(\frac{ab}{2 \pi }+\frac12\right)-\frac12\psi\left(\frac{ab}{2 \pi }\right)-\frac{\pi}{2ab}\\ & \\ &\int_0^\infty \frac{x}{(1+4x^2)^2}\frac{1}{\sinh(\pi x)}\,dx=\frac{\beta(2)}{4}-\frac18 \end{aligned}


\begin{aligned} I&=\int_0^\infty \frac{\sinh(ax)}{\sinh(bx)}\,dx\\ &=\int_0^\infty \frac{e^{ax}-e^{-ax}}{e^{bx}-e^{-bx}}\,dx\\ &=\int_0^\infty \frac{e^{-bx}}{e^{-bx}}\cdot\frac{e^{ax}-e^{-ax}}{e^{bx}-e^{-bx}}\,dx\\ &=\int_0^\infty \frac{e^{-(b-a)x}-e^{-(b+a)x}}{1-e^{-2bx}}\,dx\\ &=\int_0^\infty \left({e^{-(b-a)x}-e^{-(b+a)x} \right)\left(\sum_{k=0}^\infty e^{-2bkx} \right)\,dx\\ &=\sum_{k=0}^\infty \left( \int_0^\infty {e^{-\left((2k+1)b-a\right)x}\,dx-\int_0^\infty e^{-\left((2k+1)b+a\right)x} \,dx\right)\\ &=\sum_{k=1}^\infty \left( \int_0^\infty {e^{-\left((2k-1)b-a\right)x}\,dx-\int_0^\infty e^{-\left((2k-1)b+a\right)x} \,dx\right)\\ &=\sum_{k=1}^\infty \frac{1}{(2k-1)b-a}-\frac{1}{(2k-1)b+a}\\ &=\sum_{k=1}^\infty \frac{2a}{b^2(2k-1)^2-a^2}\\ &=\frac{4\pi^2}{4b^2\pi^2}\sum_{k=1}^\infty \frac{2a}{(2k-1)^2-\frac{a^2}{b^2}}\\ &=\frac{\pi}{2b}\sum_{k=1}^\infty \frac{8\frac{ a \pi}{2b}}{\pi^2(2k-1)^2-\frac{\pi^2a^2}{4b^2}}\\ &=\frac{\pi}{2b} \tan\left(\frac{a \pi}{2b} \right) \qquad \blacksquare \end{aligned}


\boxed{\int_0^\infty \frac{\sinh(ax)}{\sinh(bx)}\,dx=\frac{\pi}{2b} \tan\left(\frac{a \pi}{2b} \right)}(1)


Where We used the partial fraction decomposition of the \tan(x)  letting  x = \frac{a \pi}{2b}  proved here


\tan\,x=\sum_{k=1}^\infty\frac{8x}{\pi^2(2k-1)^2-4x^2}

A second method

\begin{aligned} I&=\int_0^\infty \frac{\sinh(ax)}{\sinh(bx)}\,dx\\ &=\int_0^\infty \frac{e^{ax}-e^{-ax}}{e^{bx}-e^{-bx}}\,dx\\ &=\int_0^\infty \frac{e^{-(b-a)x}-e^{-(b+a)x}}{1-e^{-2bx}}\,dx\\ &=\frac{1}{2b}\int_0^\infty \frac{e^{-\frac{(b-a)}{2b}x}-e^{-\frac{(b+a)}{2b}x}}{1-e^{-x}}\,dx \qquad (2bx \to x)\\ &=\frac{1}{2b}\int_0^\infty \frac{x^{\frac{(b-a)}{2b}x}-x^{\frac{(b+a)}{2b}}}{1-x}\,dx \qquad (e^{-x} \to x)\\ &=\frac{1}{2b}\left(\psi\left(\frac12+\frac{a}{2b} \right)-\psi\left(\frac12-\frac{a}{2b} \right) \right)\\ &=\frac{\pi}{2b}\tan\left(\frac{a \pi}{2b} \right) \qquad \blacksquare \end{aligned}


Where we used the results proved here


\int_0^1\frac{x^{w-1}-x^{z-1}}{1-x}\,dx$=\psi(z)-\psi(w)


and (see appendix below)

\psi\left(\frac{1}{2}+x\right)-\psi\left(\frac{1}{2}-x\right)=\pi \tan(\pi x)

Now, if we let a \to i a in (1) we obtain


\boxed{\int_0^\infty \frac{\sin(ax)}{\sinh(bx)}\,dx=\frac{\pi}{2b} \tanh\left(\frac{a \pi}{2b} \right)}(2)

Recall

\int_0^\infty e^{-at}\sin(xt)\,dt=\frac{x}{x^2+a^2}


\int_{0}^{1} \frac{x^{m}}{1+x} d x=\frac{1}{2 }\left(\psi\left(\frac{m+2}{2 }\right)-\psi\left(\frac{m+1}{2 }\right)\right)


\begin{aligned} I&=\int_0^\infty \frac{x}{a^2+x^2}\frac{1}{\sinh(bx)}\,dx\\ &=\int_0^\infty \int_0^\infty \frac{e^{-at}\sin(xt)}{\sinh(bx)}\,dt\,dx\\ &=\int_0^\infty e^{-at}\int_0^\infty \frac{\sin(tx)}{\sinh(bx)}\,dx\,dt\\ &=\frac{\pi}{2b}\int_0^\infty e^{-at} \tanh\left(\frac{t \pi}{2b} \right)\,dt\\ &=\int_0^\infty e^{-\frac{2ab}{\pi}t} \tanh\left(t \right)\,dt \qquad \left( \frac{t \pi}{2b} \to t \right) \\ &=\int_0^\infty e^{-kt} \frac{e^{t}-e^{-t}}{e^{t}+e^{-t}}\,dt \qquad \left( \frac{2ab}{\pi} = k \right) \\ &=\int_0^\infty e^{-kt} \frac{1-e^{-2t}}{1+e^{-2t}}\,dt \\ &=\frac12 \int_0^\infty e^{-\frac{k}{2}t} \frac{1-e^{-t}}{1+e^{-t}}\,dt \\ &=\frac12 \int_0^1 \frac{t^{\frac{k}{2}-1}-t^{\frac{k}{2}}}{1+t}\,dt \\ &=\frac12 \int_0^1 \frac{t^{\frac{k}{2}-1}}{1+t}\,dt-\frac12 \int_0^1 \frac{t^{\frac{k}{2}}}{1+t}\,dt \\ &=\frac{1}{4 }\left(\psi\left(\frac{k}{4 }+\frac12\right)-\psi\left(\frac{k}{4 }\right)\right)-\frac{1}{4 }\left(\psi\left(\frac{k}{4 }+1\right)-\psi\left(\frac{k}{4 }+\frac12\right)\right)\\ &=\frac{1}{2} \psi\left(\frac{k}{4 }+\frac12\right)-\frac{1}{4 }\left(\psi\left(\frac{k}{4 }+1\right)+\psi\left(\frac{k}{4 }\right)\right)\\ &=\frac{1}{2} \psi\left(\frac{k}{4 }+\frac12\right)-\frac{1}{4 }\left(\frac{4}{k}+2\psi\left(\frac{k}{4 }\right)\right)\\ &=\frac{1}{2} \psi\left(\frac{k}{4 }+\frac12\right)-\frac12\psi\left(\frac{k}{4 }\right)-\frac1k\\ &=\frac{1}{2} \psi\left(\frac{ab}{2 \pi }+\frac12\right)-\frac12\psi\left(\frac{ab}{2 \pi }\right)-\frac{\pi}{2ab} \qquad \blacksquare\\ \end{aligned}


\int_0^\infty \frac{x}{a^2+x^2}\frac{1}{\sinh(bx)}\,dx=\frac{1}{2} \psi\left(\frac{ab}{2 \pi }+\frac12\right)-\frac12\psi\left(\frac{ab}{2 \pi }\right)-\frac{\pi}{2ab}(3)


Differentiating both sides of (3) w.r. to a


-\int_0^\infty \frac{2ax}{(a^2+x^2)^2}\frac{1}{\sinh(bx)}\,dx=\frac{b}{4 \pi} \psi^\prime\left(\frac{ab}{2 \pi }+\frac12\right)-\frac{b}{4 \pi}\psi^\prime\left(\frac{ab}{2 \pi }\right)+\frac{\pi}{2ba^2}


\int_0^\infty \frac{x}{(1+\frac{x^2}{a^2})^2}\frac{1}{\sinh(bx)}\,dx=\frac{a^3b}{8 \pi}\psi^\prime\left(\frac{ab}{2 \pi }\right)-\frac{a^3b}{8 \pi} \psi^\prime\left(\frac{ab}{2 \pi }+\frac12\right)-\frac{a\pi}{4b}

setting a=\frac12 and b=\pi


\begin{aligned} \int_0^\infty \frac{x}{(1+4x^2)^2}\frac{1}{\sinh(\pi x)}\,dx&=\frac{1}{64}\psi^\prime\left(\frac{1}{4}\right)-\frac{1}{64} \psi^\prime\left(\frac{3}{4 }\right)-\frac{1}{8}\\ &=\frac{1}{64}\left(\pi^2+8\beta(2)-\pi^2+8\beta(2) \right)-\frac18\\ &=\frac{\beta(2)}{4}-\frac18 \qquad \blacksquare \end{aligned}


Where we used the results proved here

\psi^\prime\left(\frac{1}{4}\right)=\pi^2+8\beta(2)

and

\psi^\prime\left(\frac{3}{4}\right)=\pi^2-8\beta(2)

Appendix

Recall the reflection formula proved here:

\Gamma(a) \Gamma(1-a)=\frac{\pi}{\sin (a \pi)}

Letting x \longrightarrow \frac{1}{2}-x we get


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


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


Taking logs on both sides and differentiating w.t. to x


\boxed{\psi\left(\frac{1}{2}+x\right)-\psi\left(\frac{1}{2}-x\right)=\pi \tan(\pi x)}

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