Show that $\int_{\pi}^{\infty} \frac{1}{x^2 (\sin^2 x)^{1/3}} dx$ is finite. Show that $\int_{\pi}^{\infty} \frac{1}{x^2 (\sin^2 x)^{1/3}} dx$ is finite. I've been trying to use Holder inequality but it seems I can't get the right combination of $p$ and $q$. Maybe I'm on the wrong track?
 A: We have:
$$ \int_{0}^{\pi}\frac{dx}{\left(\sin^2 x\right)^{\frac{1}{3}}}=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}{\Gamma\left(\frac{2}{3}\right)}$$
hence:
$$\begin{eqnarray*} \left|\int_{\pi}^{+\infty}\frac{dx}{x^2\left(\sin^2 x\right)^{\frac{1}{3}}}\,dx\right|&\leq& \pi\left(\frac{1}{\pi^2}+\frac{1}{4\pi^2}+\frac{1}{9\pi^2}\ldots\right)\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}{\Gamma\left(\frac{2}{3}\right)}\\&=&\frac{\pi}{6}\cdot \frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}{\Gamma\left(\frac{2}{3}\right)}.\end{eqnarray*}$$
The exact value of the integral is given by:
$$ I = \frac{1}{\pi}\int_{0}^{1}\frac{\psi'(1+x)\,dx}{\left(\sin^2 (\pi x)\right)^{\frac{1}{3}}}$$
and by computing the Laplace transform of $\frac{1}{(\sin^2 x)^{\frac{1}{3}}}$, the inverse Laplace transform of $\frac{1}{(x+\pi)^2}$ and keeping just the terms $A e^{-\pi s},B s^2 e^{-\pi s}$ of their product we get:
$$ I\leq \frac{\Gamma\left(\frac{1}{6}\right)}{4\pi^{7/2}\Gamma\left(\frac{2}{3}\right)}\left(5\pi^2-\psi'\left(\frac{1}{3}\right)+\psi'\left(\frac{2}{3}\right)\right)\leq\color{red}{\frac{4}{5}}.$$
