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?
$$\int_{0}^{\pi/2}x\sqrt{\tan{x}}\log{\sin{x}}\,\mathrm dx=-\frac{\pi\sqrt{2}}{48}\big(\pi^2+12\pi \log{2}+24\log^2{2}\big) $$
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.
It is worth mentioning that this approach applies to the other question, too.
Substitute $t = \sqrt{\tan x}$
\begin{align} I=&\int_{0}^{\pi/2}x\sqrt{\tan{x}}\ln({\sin x})\ dx\\ = &\int_0^\infty \frac{t^2 \ln\frac{t^4}{1+t^4}\ \tan^{-1}t^2}{1+t^4}dt\\ =&\int_0^\infty \int_0^1 \frac{2y \ t^4 \ln\frac{t^4} {1+t^4}}{(1+t^4)(1+y^4 t^4)}dy \ dt\\ =&\int_0^1 \frac{2y}{1-y^4} \int_0^\infty\bigg(\frac{\ln\frac{t^4} {1+t^4}}{1+y^2t^4}- \frac{\ln\frac{t^4} {1+t^4}}{1+t^4} \bigg)dt \ dy \end{align} Utilize $$J(a)= \int_0^\infty \frac{a \ln\frac{t^4} {1+t^4}}{1+a^4 t^4}dt = -\frac\pi{\sqrt2}\left[\tan^{-1}a+\ln(1+a)+\frac12\ln(1+a^2) \right] $$ to obtain \begin{align} I= &\int_0^1 \frac{2 [J(y)-yJ(1)] }{1-y^4}dy \\ =&\ \sqrt2\pi \int_0^1 \frac{\frac\pi4y-\tan^{-1}y}{1-y^4} +\frac{y\ln2 -\ln(1+y)}{1-y^4} +\frac{y\ln2 -\ln(1+y^2)}{2(1-y^4)}\ dy\\ =&\ \sqrt2\pi \bigg[ \left(G-\frac{\pi^2}{16}-\frac\pi4\ln2\right)+\left( \frac{\pi^2}{12} -\frac\pi4\ln2 -\ln^22\right)\\ &\>\>\>\>\>\>\>\>\>\>\>\>\> +\frac12\left( 2G+ \frac{\pi^2}{8} -\pi\ln2-\ln^22\right)\bigg]\\ =& \ \frac{\pi}{\sqrt2}\left( G+ \frac{\pi^2}{24}-\frac\pi2\ln2 -\frac34\ln^22\right) \end{align}
$$ I=-\frac{\pi}{8}\sqrt{2}(2{\pi}\ln2-4G-\frac{\pi^2}{6}+3\ln^22)$$ Where G is the Catalan's function.
