Problem - F. Holland (Ireland)
Prove
\[\int_0^{\infty}\left(\frac{\text{tanh}(x)}{x}\right)^2\,\mathrm dx=\frac{14\zeta(3)}{\pi^2}.\]Solution
Consider the contour integral
\[I_k=\oint_{C_k}\left(\frac{\text{tanh}(z)}{z}\right)^2\,\mathrm dz\]where $C_k$ is a positively oriented rectangular contour connecting the points $-k\pi$, ${k\pi}$, $(1+i){k\pi}$, and $(-1+i)k\pi$.
Breaking the contour up yields four integrals,
\[\begin{align*} I_k &= \left(\int_{k\pi}^{(1+i){k\pi}}+\int_{(1+i){k\pi}}^{(i-1){k\pi}}+\int_{(i-1){k\pi}}^{-{k\pi}}+\int_{-{k\pi}}^{k\pi}\right)\left(\frac{\text{tanh}(z)}{z}\right)^2\,\mathrm dz.\end{align*}\]The integral along the right edge satisfies
\[\begin{align*} \left|\int_{k\pi}^{(1+i){k\pi}}\left(\frac{\text{tanh}(z)}{z}\right)^2\,\mathrm dz\right| &\leq \underset{y\in[0,\,k\pi]}{\text{max}}\left|\frac{\text{tanh}^2(k\pi+iy)}{(k\pi+iy)^2}\right|\cdot\left|\int_{k\pi}^{(1+i)k\pi}\,\mathrm dz\right| \\ &= \frac{\text{tanh}^2(k\pi)}{k\pi} \\ &< \frac{1}{k\pi}\end{align*}\]and the same argument shows that
\[\begin{align*} \left|\int_{(i-1){k\pi}}^{-{k\pi}}\left(\frac{\text{tanh}(z)}{z}\right)^2\,\mathrm dz\right|<\frac{1}{k\pi}\end{align*}\]for the left edge.
This leaves the top integral,
\[\begin{align*} \left|\int_{(1+i){k\pi}}^{(i-1){k\pi}}\left(\frac{\text{tanh}(z)}{z}\right)^2\,\mathrm dz\right| &\leq\underset{x\in[-k\pi,\,k\pi]}{\text{max}}\left|\frac{\text{tanh}^2(k\pi i+x)}{(k\pi i+x)^2}\right|\cdot\left|\int_{(1+i){k\pi}}^{(i-1){k\pi}}\,\mathrm dz\right| \\ &< 2k\pi\underset{x\in[-k\pi,\,k\pi]}{\text{max}}\left|\frac{1}{(k\pi i+x)^2}\right| \\ &< \frac{2}{k\pi}.\end{align*}\]So
\[\begin{align*} \lim_{k\to\infty}\oint_{C_k}\left(\frac{\text{tanh}(z)}{z}\right)^2\,\mathrm dz &=\int_{-\infty}^{\infty}\left(\frac{\text{tanh}(x)}{x}\right)^2\,\mathrm dx \\ &= 2\int_0^{\infty}\left(\frac{\text{tanh}(x)}{x}\right)^2\,\mathrm dx.\end{align*}\]We can also evaluate the contour integral using the residue theorem,
\[\begin{align*} \lim_{k\to\infty}I_k &= 2\pi i\sum_{n=0}^\infty\underset{z=\frac{\pi i}{2}+n\pi i}{\text{Res}}\left(\frac{\text{tanh}(z)}{z}\right)^2 \\ &= 2\pi i\sum_{n=0}^\infty\lim_{z\to \frac{\pi i}{2}+n\pi i}\frac{\mathrm d}{\mathrm dz}\left(\frac{\left(z-\left(\frac{\pi i}{2}+n\pi i\right)\right)\text{tanh}(z)}{z}\right)^2 \\ &= 32\pi\sum_{n=0}^\infty\frac{1}{(2n\pi+\pi)^3} \\ &= \frac{32}{\pi^2}\left(\zeta(3)-\frac{\zeta(3)}{8}\right) \\ &= \frac{28\zeta(3)}{\pi^2}.\end{align*}\]But this is twice the integral of interest, therefore
\[\int_0^{\infty}\left(\frac{\text{tanh}(x)}{x}\right)^2\,\mathrm dx=\frac{14\zeta(3)}{\pi^2}\]as desired.
Similar integrals
Using the same method we can also evaluate similar integrals involving higher powers of the integrand.
\[\int_0^{\infty}\left(\frac{\text{tanh}(x)}{x}\right)^4\,\mathrm dx=\frac{2540\zeta(7)}{\pi^6}-\frac{496\zeta(5)}{3\pi^4}\] \[\int_0^{\infty}\left(\frac{\text{tanh}(x)}{x}\right)^6\,\mathrm dx=\frac{5842\zeta(7)}{5\pi^6}-\frac{57232\zeta(9)}{\pi^8}+\frac{515844\zeta(11)}{\pi^{10}}\] \[\int_0^{\infty}\left(\frac{\text{tanh}(x)}{x}\right)^8\,\mathrm dx = \frac{112456344\zeta(15)}{\pi^{14}}-\frac{17299392\zeta(13)}{\pi^{12}}+\frac{720544\zeta(11)}{\pi^{10}}-\frac{102784\zeta(9)}{15\pi^8}\]