Convergence of $\int_{\pi}^{\infty}\frac{dx}{x^2 (\sin^{2}x)^{1/3}}$ [closed]

Question

The title tells the question. I have to show that the improper integral $$\int_{\pi}^{\infty}\frac{dx}{x^2 (\sin^{2}x)^{1/3}}$$ is finite i.e it is convergent.

Any suggestions on how to proceed ? I am having hard time with this. Thank you.

Answer 1

Hint

Take $n \in \mathbb N$ and consider

$$I_n = \int_{\pi}^{n \pi}\frac{dx}{x^2 (\sin^{2}x)^{1/3}}$$

You have

$$ \begin{aligned}0 \le I_n &= \sum_{k=1}^{n-1}\int_{k\pi}^{(k+1)\pi}\frac{dx}{x^2 (\sin^{2}x)^{1/3}}\\ &\le \frac{1}{\pi^2}\sum_{k=1}^{n-1} \frac{1}{k^2}\int_{k\pi}^{(k+1)\pi}\frac{dx}{(\sin^{2}x)^{1/3}}\\ &= \frac{1}{\pi^2}\sum_{k=1}^{n-1} \frac{1}{k^2}\int_{0}^{\pi}\frac{dx}{(\sin^{2}x)^{1/3}} \end{aligned}$$

As $\sum 1/k^2$ converges, we're left to prove that $\int_{0}^{\pi}\frac{dx}{(\sin^{2}x)^{1/3}}=2\int_{0}^{\pi/2}\frac{dx}{(\sin^{2}x)^{1/3}}$ converges. This can be done as $\sin x \simeq x$ around $0$ and $\int_{0}^{\pi/2}\frac{dx}{x^{2/3}}$ converges.

Answer 2

We have $$ \int_{N\pi}^{(N+1)\pi}\frac{d\theta}{\left(\sin^2\theta\right)^{1/3}}=\int_{0}^{\pi}\frac{d\theta}{\left(\sin^2\theta\right)^{1/3}}=\frac{3\,\Gamma\left(\tfrac{1}{3}\right)^3}{2^{4/3}\pi}=\frac{2\pi\cdot 3^{3/4}}{\text{AGM}(2,\sqrt{2+\sqrt{3}})} $$ by Euler's Beta function, the reflection formula for the $\Gamma$ function and the relation between special values of the $\Gamma$ function and the complete elliptic integral of the first kind / the AGM mean. In particular the LHS is less than $\frac{58}{25}\pi$. Our integral equals $$ \int_{0}^{\pi}\frac{1}{\left(\sin^2 \theta\right)^{1/3}}\sum_{n\geq 1}\frac{1}{(\theta+n\pi)^2}\,d\theta=\frac{1}{\pi^2}\int_{0}^{\pi}\frac{\psi'\left(1+\tfrac{\theta}{\pi}\right)}{\left(\sin^2 \theta\right)^{1/3}}\,d\theta $$ or, by the reflection formula for the $\psi'$ function, $$ \int_{0}^{\pi/2}\left[\frac{1}{\sin^2\theta}-\frac{1}{\theta^2}-\frac{1}{(\pi-\theta)^2}\right]\frac{d\theta}{\left(\sin^2\theta\right)^{1/3}}$$ where the term between square brackets is approximately constant on $\left(0,\frac{\pi}{2}\right)$, bounded between $1-\frac{8}{\pi^2}$ and $\frac{1}{3}-\frac{1}{\pi^2}$. This proves that the integral is finite and also allows an approximate evaluation with a relative error $<11\%$. A greater accuracy is achieved by computing the first coefficients of the Fourier series of the term between square brackets, or by performing a Lagrange interpolation.

The reduced form of the original integral exhibits a strong resemblance with the integrals appearing in the computation of some series due to Ramanujan, related to the values of the Riemann $\zeta$ function at rational points.