Problem - S. Stewart (Saudi Arabia)
Prove
\[\int_0^{\pi/2}\frac{\sin(4x)}{\ln(\tan(x))}\,\mathrm dx=-\frac{14\zeta(3)}{\pi^2}.\]Solution
We define
\[I(s)=\int_0^{\pi/2}\frac{\sin(4x)\tan^s(x)}{\ln(\tan(x))}\,\mathrm dx\]and investigate its derivative,
\[\begin{align*} I'(s) &= \int_0^{\pi/2}\sin(4x)\tan^s(x)\,\mathrm dx \\ &= 4\int_0^{\pi/2}\left(\sin(x)\cos^3(x)-\sin^3(x)\cos(x)\right)\tan^s(x)\,\mathrm dx \\ &= 2\left(\text{B}\left(1+\frac{s}{2},\,2-\frac{s}{2}\right)-\text{B}\left(2+\frac{s}{2},\,1-\frac{s}{2}\right)\right), \\ &\qquad\text{used }\text{B}(x,\,y)=2\int_0^{\pi/2}\sin^{2x-1}(\theta)\cos^{2y-1}(\theta)\,\mathrm d\theta \\ &= 2\left(\frac{\Gamma\left(1+\frac{s}{2}\right)\Gamma\left(2-\frac{s}{2}\right)}{\Gamma(3)}-\frac{\Gamma\left(2+\frac{s}{2}\right)\Gamma\left(1-\frac{s}{2}\right)}{\Gamma(3)}\right) \\ &= \frac{\pi s(2-s)}{4}\csc\left(\frac{\pi s}{2}\right)-\frac{\pi s(2+s)}{4}\csc\left(\frac{\pi s}{2}\right), \\ &\qquad\text{by the reflection formula }\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin(\pi s)} \\ &= -\frac{\pi s^2}{2}\csc\left(\frac{\pi s}{2}\right).\end{align*}\]Integrating along the non-negative portion of the imaginary axis gives
\[\begin{align*} -\frac{\pi}{2}\int_0^{i\infty}s^2\csc\left(\frac{\pi s}{2}\right)\,\mathrm ds &= \frac{4}{\pi^2}\int_0^\infty s^2\text{csch}(s)\,\mathrm ds,\,s\mapsto \frac{2is}{\pi} \\ &= \frac{8}{\pi^2}\int_0^\infty\frac{s^2}{e^s-e^{-s}}\,\mathrm ds \\ &= \frac{8}{\pi^2}\int_0^\infty\frac{e^{-s}s^2}{1-e^{-2s}}\,\mathrm ds \\ &= \frac{8}{\pi^2}\int_0^\infty\sum_{n=0}^\infty e^{-(2n+1)s}s^2\,\mathrm ds \\ &= \frac{8}{\pi^2}\sum_{n=0}^\infty\int_0^\infty e^{-(2n+1)s}s^2\,\mathrm ds, \\ &\qquad\text{monotone convergence on }(0,\,\infty) \\ &= \frac{8}{\pi^2}\sum_{n=0}^\infty\frac{1}{(2n+1)^3}\int_0^\infty e^{-s}s^2\,\mathrm ds,\,s\mapsto\frac{s}{2n+1} \\ &= \frac{8\Gamma(3)}{\pi^2}\left(\sum_{n=1}^\infty\frac{1}{n^3}-\sum_{n=1}^\infty\frac{1}{(2n)^3}\right) \\ &= \frac{16}{\pi^2}\left(\zeta(3)-\frac{\zeta(3)}{8}\right) \\ &= \frac{14\zeta(3)}{\pi^2}.\end{align*}\]From the fundamental theorem of calculus we also have,
\[-\frac{\pi}{2}\int_0^{i\infty}s^2\csc\left(\frac{\pi s}{2}\right)\,\mathrm ds=\lim_{s\to \infty}I(is)-I(0),\]but the above limit is $0$,
\[\begin{align*} \lim_{s\to\infty}\left|\int_0^{\frac{\pi}{2}}\frac{\sin(4x)e^{is\ln(\tan(x))}}{\ln(\tan(x))}\,\mathrm dx\right| &\leq \underset{x\in[0,\,\pi/2]}{\text{max}}\left|\frac{\sin(4x)}{\ln(\tan(x))}\right|\lim_{s\to\infty}\left|\int_0^{\pi/2}e^{is\ln(\tan(x))}\,\mathrm dx\right| \\ &= 2\lim_{s\to\infty}\left|\int_0^{\pi/2}e^{is\ln(\tan(x))}\,\mathrm dx\right| \\ &= 2\lim_{s\to\infty}\left|\int_0^\infty\frac{e^u}{e^{2u}+1}e^{isu}\,\mathrm du\right|,\,u=\ln(\tan(x)) \\ &= 2\lim_{s\to\infty}\left|\int_0^\infty\frac{(e^{2u}-1)e^u}{(e^{2u}-1)(e^{2u}+1)}e^{isu}\,\mathrm du\right| \\ &= 2\lim_{s\to\infty}\left|\int_0^\infty\left(\frac{e^{(3+is)u}}{e^{4u}-1}-\frac{e^{(1+is)u}}{e^{4u}-1}\right)\,\mathrm du\right|\\ &= 2\lim_{s\to\infty}\left|\int_0^\infty\left(\frac{e^{(-1+is)u}}{1-e^{-4u}}-\frac{e^{(-3+is)u}}{1-e^{-4u}}\right)\,\mathrm du\right|\\ &= \frac{1}{2}\lim_{s\to\infty}\left|\int_0^\infty\left(\frac{e^{-\left(\frac{1}{4}-\frac{is}{4}\right)t}}{1-e^{-t}}-\frac{e^{-\left(\frac{3}{4}-\frac{is}{4}\right)t}}{1-e^{-t}}\right)\,\mathrm dt\right|,\;u=\frac{t}{4} \\ &= \frac{1}{2}\lim_{s\to\infty}\left|\int_0^\infty\left(\frac{e^{-t}}{t}-\frac{e^{-\left(\frac{3}{4}-\frac{is}{4}\right)t}}{1-e^{-t}}\right)\,\mathrm dt-\int_0^\infty\left(\frac{e^{-t}}{t}-\frac{e^{-\left(\frac{1}{4}-\frac{is}{4}\right)t}}{1-e^{-t}}\right)\,\mathrm dt\right| \\ &= \frac{1}{2}\lim_{s\to\infty}\left|\psi\left(\frac{3}{4}-\frac{is}{4}\right)-\psi\left(\frac{1}{4}-\frac{is}{4}\right)\right| \\ &=0.\end{align*}\]That the last limit goes to $0$ can be seen from Binet’s second integral for the digamma function,
\[\psi(z)=\ln(z)-\frac{1}{2z}-2\int_0^\infty\frac{t}{(t^2+z^2)(e^{2\pi t}-1)}\,\mathrm dt.\]This means
\[\frac{\pi}{2}\int_0^{i\infty}s^2\csc\left(\frac{\pi s}{2}\right)\,\mathrm ds=I(0),\] \[\therefore\int_0^{\pi/2}\frac{\sin(4x)}{\ln(\tan(x))}\,\mathrm dx=-\frac{14\zeta(3)}{\pi^2}.\]