I seek to prove convergence of the following integral
$$\int_{\pi}^{\infty} \frac{dx}{x^2 (\sin(x))^{\frac{2}{3}}}$$
My attempted solution is as follows.
I use the identity
$$\int_0^{\frac{\pi}{2}} \sin^{2x-1} (u) \cos^{2y – 1}(u) du = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x + y)}$$
Choose $x = \frac{1}{6}$ and $y = \frac{1}{2}$.
$$
\begin{align}
\int_{\pi}^{\infty} \frac{dx}{x^2 (\sin(x))^{\frac{2}{3}}} & \stackrel{?}{=} \sum_{n=1}^{\infty} \int_{n \pi}^{(n + 1) \pi} \frac{dx}{x^2(\sin(x))^{\frac{2}{3}}} \\
& \leq \sum_{n=1}^{\infty} \frac{2}{n^2 \pi^2} \frac{\Gamma\left(-\frac{2}{3}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma\left(-\frac{1}{6}\right)} \\
& = \frac{\Gamma\left(-\frac{2}{3}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma\left(-\frac{1}{6}\right)} \frac{2}{\pi^2} \frac{\pi^2}{6} \\
& = \frac{1}{3} \frac{\Gamma\left(-\frac{2}{3}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma\left(-\frac{1}{6}\right)} < \infty \\
\end{align}
$$
Question: I know that special care must be given to justifying the interchange of an infinite sum and an integral. Is my first equality my first equality (marked by the ?) correct? If so, I would be very appreciative to see other peoples alternative methods of solving this problem.
Best Answer
I assume that you have some consistent way to define $\sin(x)^{-2/3}$ on all of $[\pi,\infty)$ other than the singularities. It doesn't really matter what that is, as long as it has absolute value $|\sin(x)|^{-2/3}$.
With that in mind, you have
$$\int_\pi^\infty x^{-2} |\sin(x)|^{-2/3} dx \leq \sum_{n=1}^\infty ((2n-1)\pi)^{-2} \int_{(2n-1)\pi}^{(2n+1)\pi} |\sin(x)|^{-2/3} dx \leq C \sum_{n=1}^\infty n^{-2}<\infty.$$