Question

Converges or Diverges? Help!

Answer 1

\[\int\limits_{0}^{\pi}\frac{ \sin \theta }{ (\pi-\theta)^{\frac{ 2 }{ 7 }} }d \theta \]

Answer 2

Use the fact that sine is bounded:
\[-1\le\sin\theta\le1~~\iff~~-\frac{1}{(\pi-\theta)^{2/7}}\le\frac{\sin\theta}{(\pi-\theta)^{2/7}}\le\frac{1}{(\pi-\theta)^{2/7}}\]
Then you have
\[\large \int_0^\pi\frac{\sin\theta}{(\pi-\theta)^{2/7}}~d\theta\le\int_0^\pi\frac{d\theta}{(\pi-\theta)^{2/7}}\]
A partial flaw to this comparison is that the integral on the right diverges as \(\theta\to\pi\), so judging by this fact alone, we can't say much for the convergence/divergence fro the left integral.

However, we know the flollwing fact:
\[\large\begin{align*}\lim_{\theta\to\pi^-}\frac{\sin\theta}{(\pi-\theta)^{2/7}}&=\lim_{\theta\to\pi^-}\frac{\sin\theta}{(\pi-\theta)^{2/7}}\\\\
&=\lim_{\theta\to\pi^-}\frac{(\pi-\theta)^{5/7}\sin\theta}{\pi-\theta}\\\\
&=\lim_{\theta\to\pi^-}\frac{(\pi-\theta)^{5/7}\sin(\pi-\theta)}{\pi-\theta}\\\\
&=\left(\lim_{\theta\to\pi^-}(\pi-\theta)^{5/7}\right)\left(\lim_{\theta\to\pi^-}\frac{\sin(\pi-\theta)}{\pi-\theta}\right)\\\\
&=0\times1\\\\
&=0\end{align*}\]
This, along with the fact that \(\dfrac{\sin\theta}{(\pi-\theta)^{2/7}}\) is continuous for \(\theta<\pi\), let's us know that the given integral converges.