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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.

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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.

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  • $\begingroup$ Wow ! That's a very good way to solve. Thank you for your time. $\endgroup$ – hiren_garai Jan 3 '19 at 10:56
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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.

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  • $\begingroup$ Thank you for your answer @JackD'Aurizio. But I'm quite new to some of the terms you used in your answer (e.g reflection formula), so basically it will take time to make out. Anyways thank you again for your response. $\endgroup$ – hiren_garai Jan 4 '19 at 2:47

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