AMM12199

Problem - S. Sharma (India)

Show that

$${\int }_0^{\mathrm{\infty }}\frac{x\sinh (x)}{3+4{\sinh }^2(x)}\mathrm{d}x=\frac{{\pi }^2}{24}.$$

Solution

Call the integral in question $I$. Let $x\mapsto \ln (x)$, so that $\mathrm{d}x\mapsto \frac{\mathrm{d}x}{x}$ and

$$\begin{array}{rl}I& ={\int }_1^{\mathrm{\infty }}\frac{\ln (x)\left(\frac{x-\frac{1}{x}}{2}\right)}{3+4{\left(\frac{x-\frac{1}{x}}{2}\right)}^2}\cdot \frac{\mathrm{d}x}{x}\\ & ={\int }_1^{\mathrm{\infty }}\frac{\ln (x)(x^2-1)}{2(x^4+x^2+1)}\mathrm{d}x\\ & =-\frac{1}{2}{\int }_1^{\mathrm{\infty }}\frac{\ln (x)(1-x^2{)}^2}{1-x^6}\mathrm{d}x.\end{array}$$

Now if we make the change of variables $x\mapsto \frac{1}{x}$, so that $\mathrm{d}x\mapsto -\frac{\mathrm{d}x}{x^2}$, we get

$$\begin{array}{rl}I& =-\frac{1}{2}{\int }_0^1\frac{\ln (x)(1-x^2{)}^2}{1-x^6}\mathrm{d}x\\ & =-\frac{1}{2}{\int }_0^1\frac{\ln (x)(1-2x^2+x^4)}{1-x^6}\mathrm{d}x\\ & =-\frac{1}{2}{\int }_0^1\ln (x)\sum _{n=0}^{\mathrm{\infty }}(x^{6n}-2x^{6n+2}+x^{6n+4})\mathrm{d}x\\ & =\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}(\frac{1}{(6n+1{)}^2}-\frac{2}{(6n+3{)}^2}+\frac{1}{(6n+5{)}^2}).\end{array}$$

The above interchange of integration and summation is justified, as the series converges uniformly on $[0,1]$. And the integrals of the form ${\int }_0^1\ln (x)x^{\alpha }\mathrm{d}x$ may be calculated by applying differentiation under the integral sign to ${\int }_0^1{x}^{\alpha }\mathrm{d}x$.

Evaluating the above infinite series is relatively straightforward, as it reduces to a sum of trigamma functions and a well known series,

$$\begin{array}{rl}I& =\frac{1}{2}(\frac{{\psi }^{\prime }\left(\frac{1}{6}\right)}{36}+\frac{{\psi }^{\prime }\left(\frac{5}{6}\right)}{36}-\frac{2}{9}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2}).\end{array}$$

By the reflection formula for the trigamma function we have ${\psi }^{\prime }(1/6)+{\psi }^{\prime }(5/6)={\pi }^2{\csc }^2(\pi /6)$, and it is easily derived from $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}={\pi }^2/6$ that $\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2}={\pi }^2/8$. So it follows that

$$\begin{array}{rl}I& =\frac{1}{2}(\frac{{\pi }^2}{36}{\csc }^2\left(\frac{\pi }{6}\right)-\frac{2}{9}\cdot \frac{{\pi }^2}{8})\\ & =\frac{1}{2}(\frac{{\pi }^2}{9}-\frac{{\pi }^2}{36})\\ & =\frac{{\pi }^2}{24}.\end{array}$$