Proving $\int_{0}^{\pi/2}x\sqrt{\tan{x}}\log{\sin{x}}\,\mathrm dx=-\frac{\pi\sqrt{2}}{48}(\pi^2+12\pi \log{2}+24\log^2{2})$

When trying to solve this problem: How to Integrate $\int^{\pi/2}_{0} x \ln(\cos x) \sqrt{\tan x}\,dx$

I found his sister integral has an interesting closed form provided my calculation is correct. I use an ugly series to find it. Can you use other methods to evaluate the integral? Such as Gamma function or residue method?

$$\int_{0}^{\pi/2}x\sqrt{\tan{x}}\log{\sin{x}}\,\mathrm dx=-\frac{\pi\sqrt{2}}{48}\big(\pi^2+12\pi \log{2}+24\log^2{2}\big)$$

• A critique. You didn't solve this integral. You collaborated with multiple people, and acted as a simplify and generalize button at the end of someone else's hard work. It's also worth mentioning that the series from the first question is hard or else you wouldn't had to change from sin to cos. In other words if you look at this answer you'll see that you just need Catalan's constant to evaluate. – Zach466920 Jun 10 '15 at 15:02
• I managed to prove through differentiation under the integral sign and the residue theorem that you integral just depends on the values of the digamma and trigamma functions at $z=\frac{1}{4}$. The Catalan constant arises in $\psi'\left(\frac{1}{4}\right)$. – Jack D'Aurizio Jun 10 '15 at 15:22
• @Zach466920,Catalan's constant is cancel out – math110 Jun 10 '15 at 15:40
• @Zach466920, you also use jack reslut understand $\ln{\sin{x}}=\ln{\cos{x}}+\ln{\tan{x}}$,if before can't evaluate not wrong,then the Catalan's constant is cancel out. – math110 Jun 10 '15 at 15:50
• trying contour integration on this problem causes a headache...^^ – tired Jun 10 '15 at 17:19

For first: $$\int_{0}^{\pi/2}x\sqrt{\tan x}\log\sin x\,dx = \frac{1}{2}\int_{0}^{+\infty}\frac{\sqrt{t}\arctan t\log\frac{t^2}{1+t^2}}{1+t^2}\,dt.$$ Now differentiation under the integral plus the residue theorem give: $$\begin{eqnarray*}\int_{0}^{+\infty}\frac{t^\alpha \arctan t}{1+t^2}\,dt &=&\int_{0}^{1}\int_{0}^{+\infty}\frac{t^{\alpha+1}}{(1+t^2)(1+\beta^2 t^2)}\,dt\,d\beta\\&=&\frac{\pi}{2\sin\left(\frac{\pi\alpha}{2}\right)}\int_{0}^{1}\frac{1-\beta^{-\alpha}}{\beta^2-1}d\beta\\&=&-\frac{\pi}{2\sin\left(\frac{\pi\alpha}{2}\right)}\left(\log 2+\frac{1}{2} H_{-\frac{\alpha+1}{2}}\right)\end{eqnarray*}$$ so $\int_{0}^{+\infty}\frac{\sqrt{t}\arctan t\log t}{1+t^2}\,dt$, that is the derivative of the previous expression at $\alpha=\frac{1}{2}$, just depends on the values of $\psi(z)$ and $\psi'(z)$ at $z=\frac{1}{4}$. Luckily, they are both not so difficult to compute: $$\psi\left(\frac{1}{4}\right)=-\frac{\pi}{2}-3\log 2-\gamma,\qquad \psi'\left(\frac{1}{4}\right)= \pi^2+8K,$$ where $K$ is the Catalan constant.
The other piece, $\int_{0}^{+\infty}\frac{\sqrt{t}\arctan t\log(1+t^2)}{1+t^2}\,dt$, can be computed in a similar way.
$$I=-\frac{\pi}{8}\sqrt{2}(2{\pi}\ln2-4G-\frac{\pi^2}{6}+3\ln^22)$$ Where G is the Catalan's function.