Problem - S. Stewart (Saudi Arabia)
Evaluate
\[I=\int_0^\infty\frac{\ln\left(\cos^2(x)\right)\sin^3(x)}{x^3\left(1+2\cos^2(x)\right)}\,\mathrm dx.\]Solution
We show that
\[I=-\frac{\pi}{4}\ln(2)-\frac{\pi}{4\sqrt{3}}\ln(1+\sqrt{3}).\]Start by rewriting the integral as a sum
\[\begin{align}I &= \sum_{n=0}^\infty\int_{2\pi n}^{2\pi(n+1)}\frac{\ln\left(\cos^2(x)\right)\sin^3(x)}{x^3\left(1+2\cos^2(x)\right)}\,\mathrm dx \nonumber \\&= \int_0^{2\pi}\frac{\ln\left(\cos^2(x)\right)\sin^3(x)}{1+2\cos^2(x)}\sum_{n=0}^\infty\frac{1}{(x+2\pi n)^3}\,\mathrm dx,\,x\mapsto x+2\pi n \nonumber \\&= -\frac{1}{16\pi^3}\int_0^{2\pi}\frac{\ln\left(\cos^2(x)\right)\sin^3(x)}{1+2\cos^2(x)}\psi''\left(\frac{x}{2\pi}\right)\,\mathrm dx\tag{1}\end{align}\]where $\psi^{\prime \prime}(z)=\frac{\mathrm d^3}{\mathrm dz^3}\ln(\Gamma(z)).$ Making the substitution $x\mapsto 2\pi-x$ in $(1)$ shows
\[\int_0^{2\pi}\frac{\ln\left(\cos^2(x)\right)\sin^3(x)}{1+2\cos^2(x)}\psi''\left(\frac{x}{2\pi}\right)\,\mathrm dx=-\int_0^{2\pi}\frac{\ln\left(\cos^2(x)\right)\sin^3(x)}{1+2\cos^2(x)}\psi''\left(1-\frac{x}{2\pi}\right)\,\mathrm dx\]and by averaging these integrals we find that
\[\begin{align} I &= \frac{1}{32\pi^3}\int_0^{2\pi}\frac{\ln\left(\cos^2(x)\right)\sin^3(x)}{1+2\cos^2(x)}\left(\psi''\left(1-\frac{x}{2\pi}\right)-\psi''\left(\frac{x}{2\pi}\right)\right)\,\mathrm dx \nonumber \\ &= \frac{1}{32}\int_0^{2\pi}\frac{\ln\left(\cos^2(x)\right)\sin^3(x)}{1+2\cos^2(x)}\sin(x)\csc^4\left(\frac{x}{2}\right)\,\mathrm dx.\tag{2}\end{align}\]The last line in the above can be obtained by computing the third derivative of the logarithm of the reflection formula, $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$ to acquire the identity
\[\psi''(1-z)-\psi''(z)=2\pi^3\cos(\pi z)\csc^3(\pi z),\]or
\[\psi''\left(1-\frac{z}{2\pi}\right)-\psi''\left(\frac{z}{2\pi}\right)=\pi^3\sin(z)\csc^4\left(\frac{z}{2}\right)\]upon appropriate substitution.
Resuming at $(2)$,
\[\begin{align} I &= \frac{1}{2}\int_0^{2\pi}\frac{\ln\left(\cos^2(x)\right)}{1+2\cos^2(x)}\cos^4\left(\frac{x}{2}\right)\,\mathrm dx \nonumber \\ &= \frac{1}{2}\int_0^{2\pi}\frac{\ln\left(\cos^2(x)\right)}{1+2\cos^2(x)}\left(\frac{1+\cos(x)}{2}\right)^2\,\mathrm dx \nonumber \\ &= \int_{-1}^1\frac{\ln\left(u^2\right)}{1+2u^2}\left(\frac{1+u}{2}\right)^2\frac{\mathrm du}{\sqrt{1-u^2}},\,u=\cos(x) \nonumber \\ &= \int_{0}^1\frac{\ln\left(u^2\right)}{1+2u^2}\left(\frac{1+u}{2}\right)^2\frac{\mathrm du}{\sqrt{1-u^2}}-\int_0^{-1}\frac{\ln\left(u^2\right)}{1+2u^2}\left(\frac{1+u}{2}\right)^2\frac{\mathrm du}{\sqrt{1-u^2}} \nonumber \\ &= \int_{0}^1\frac{\ln\left(u^2\right)}{1+2u^2}\left(\frac{1+u}{2}\right)^2\frac{\mathrm du}{\sqrt{1-u^2}}+\int_0^1\frac{\ln\left(u^2\right)}{1+2u^2}\left(\frac{1-u}{2}\right)^2\frac{\mathrm du}{\sqrt{1-u^2}} \nonumber \\ &= \frac{1}{2}\int_0^1\frac{\ln\left(u^2\right)}{1+2u^2}\cdot\frac{1+u^2}{\sqrt{1-u^2}}\,\mathrm du \nonumber \\ &= -\frac{1}{4}\int_0^\infty\frac{\ln(1+t)(2+t)}{(1+t)(3+t)\sqrt{t}} \,\mathrm dt,\,u=\frac{1}{\sqrt{1+t}} \nonumber \\ &=-\frac{1}{8}\int_0^\infty\frac{\ln(1+t)}{(1+t)\sqrt{t}}\,\mathrm dt-\frac{1}{8}\int_0^\infty\frac{\ln(1+t)}{(3+t)\sqrt{t}}\,\mathrm dt.\tag{3}\end{align}\]Seeking a closed form for these two integrals, we examine the more general integral
\[J(\alpha,\,\beta)=\int_0^\infty\frac{\ln(1+\beta t)}{(\alpha+t)\sqrt{t}}\,\mathrm dt,\]which we can approach with the Feynman trick.
\[\begin{align*} \frac{\partial}{\partial\beta}J(\alpha,\,\beta) &= \int_0^\infty\frac{\sqrt{t}}{(1+\beta t)(\alpha+t)}\,\mathrm dt \\ &= \frac{1}{\alpha\beta-1}\int_0^\infty\left(\frac{\beta\sqrt{t}}{1+\beta t}-\frac{\sqrt{t}}{\alpha+t}\right)\,\mathrm dt \\ &= \frac{2}{\alpha\beta-1}\left[\sqrt{\alpha}\arctan\left(\sqrt{\frac{t}{\alpha}}\right)-\frac{1}{\sqrt{\beta}}\arctan(\sqrt{\beta t})\right]\Bigg\vert_0^\infty \\ &= \frac{\pi}{(1+\sqrt{\alpha\beta})\sqrt{\beta}}.\end{align*}\]The computation of the last integral shown above may be done by letting $t=v^2/\beta$ in the first term and $t=\alpha w^2$ in the second term. Now,
\[\begin{align*} J(\alpha,\,1) &= J(\alpha,\,1)-J(\alpha,\,0) \\ &= \pi\int_0^1\frac{\mathrm d\beta}{(1+\sqrt{\alpha\beta})\sqrt{\beta}} \\ &= \frac{2\pi}{\sqrt{\alpha}}\int_0^{\sqrt{\alpha}}\frac{\mathrm ds}{1+s},\,\beta=s^2/\alpha \\ &= \frac{2\pi}{\sqrt{\alpha}}\ln\left(1+\sqrt{\alpha}\right).\end{align*}\]Applying this result to $(3)$ gives
\[\begin{align*} I &= -\frac{1}{8}J(1,\,1)-\frac{1}{8}J(3,\,1) \\ &= -\frac{\pi}{4}\ln(2)-\frac{\pi}{4\sqrt{3}}\ln(1+\sqrt{3})\end{align*}\]as claimed.