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Check convergence of integral: $$ \int_{0}^{1} \frac{dx}{|\sin{x}|^{1/2}} $$

My attempt

Dirichlet's test would't help there so I am going to use comparison test:

$$ x \ge \sin{x} \\ \frac{1}{x} \le \frac{1}{\sin{x}} \\ \frac{1}{|x|} \le \frac{1}{|\sin{x}|} \\ \frac{1}{|x|^{1/2}} \le \frac{1}{|\sin{x}|^{1/2}}$$ So

$$ \int_{0}^{1} \frac{dx}{|\sin{x}|^{1/2}} \ge \int_{0}^{1} \frac{1}{|x|^{1/2}} $$ But $\int_{0}^{1} \frac{1}{|x|^{1/2}}$ converges so it doesn't help me.

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  • $\begingroup$ $\sin x \geqslant 2x/\pi$ in that interval $\endgroup$ Commented May 27, 2019 at 21:07
  • $\begingroup$ For small $x$ it's also true that $| \sin x| \gt x- \frac{x^3}{6}$, as you can see from the Taylor series. Does that help? $\endgroup$ Commented May 27, 2019 at 21:07

2 Answers 2

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Instead of using the upper bound $\sin x\le x$, use the to lower bound $\sin x\ge\frac{2x}{\pi}$, which follows from $\sin x$ being concave on $[0,\,\pi/2]$.

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  • $\begingroup$ How you got this smart bound for $\sin{x}$? I know that when I have that $ \sin x\ge\frac{2x}{\pi} $ it is really easy to proof that, but how you got just that idea? $\endgroup$ Commented May 27, 2019 at 21:12
  • $\begingroup$ @Tester1998 Experience, visualising the graph, that sort of thing. Bear in mind $\sin x\sim x$ for small $x$, so we should supply get upper and lower bounds on the integral proportional to the lower bound you obtained, and that's just what we need to prove convergence. $\endgroup$ Commented May 27, 2019 at 21:14
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Near $0$, $|\sin x|\sim |x|$, so $\sqrt{|\sin x|}\sim \sqrt{|x|}$ and $$\frac 1{\sqrt{\sin x}}\sim \frac 1{\sqrt x}\quad \text{ for } x>0.$$

As $\displaystyle\int_0^1\frac{\mathrm d x}{\sqrt x}$ converges, $\;\displaystyle\int_0^1\frac{\mathrm d x}{\sqrt{\sin x}}$ converges too.

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