# Convergence of the Integral $\int_0^1 \frac{1}{\sqrt{\sin x}} \, dx$

Here is an old question from my real analysis exam. It has been bugging me for the good part of a year. Does the following integral converge?

$$\int_0^1 \frac{1}{\sqrt{\sin x}} \, dx$$

I'm pretty sure the comparison test is the way to go. Any insight would be greatly appreciated. Thanks.

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HINT:

You're right: the comparison test is the way to go. In the domain of integration, i.e., the interval $[0,1]$, the only point that causes concern is $0$. As $x \to 0$, note that the integrand grows unbounded. Therefore, to decide the convergence or divergence of the integral, we need to bound the growth of the integrand near $0$. This is the idea behind the (limit) comparison test.

To implement the above idea, we could use the standard fact $\sin x \sim x$ for $x$ close to $0$ (i.e., as $x \to 0$). Therefore, our integral $\int_0^1 \frac{1}{\sqrt{\sin x}} ~\mathrm dx$ converges if and only if $\int_0^1 \frac{1}{\sqrt{x}} ~\mathrm dx$ (the integral of the test function) converges. Do you know how to establish the convergence (or divergence) of the latter integral?

Convergence of the test integral: The integral $\int_0^1 \frac{1}{\sqrt{x}} ~\mathrm dx$ in fact converges (and so does our original integral). To see this, note that $$\int_{\delta}^{1} \frac{1}{\sqrt{x}} ~\mathrm dx = \left. 2 \sqrt{x} \right|_{\delta}^{1} = 2 - 2 \sqrt{\delta} \to 2,$$ as $\delta \to 0$.

In fact, one could similarly see that the integral $\int_0^1 x^p ~\mathrm dx$ converges if and only if $p \gt -1$.

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Using the change of variables $x\mapsto 1/x$ changes the limits and seems to allow for the integral test. Am I right to assume we can then use the comparison test with the general harmonic series? –  Henry Shearman Dec 6 '11 at 1:07
Why changing variables? Using Will Jagy's hint, you can bound your integral above by the integral of $1/\sqrt{x/2}$ between $0$ and $1$, which is finite, hence you're done. –  Patrick Da Silva Dec 6 '11 at 1:27
Yeah that works. Just needed to use the integral $$\lim_{t\to 0}\,\, \int_t^1 \frac{1}{\sqrt{x/2}}\, dx.$$ –  Henry Shearman Dec 6 '11 at 1:35
Or you can use the fact that $\int_0^1 1/x^pdx$ is a convergent integral if and only if $p<1$, which follows by direct computation of the integral. –  a.r. Dec 6 '11 at 4:16

For $0 \leq x \leq 1,$ we get $$\frac{x}{2} \leq \; \sin x \; \leq x$$

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A possible solution only with the comparison test:

It suffices to show that $\displaystyle{\int_{0}^{\pi /2}\dfrac{\mathrm{d}x}{\sqrt{\sin x}}}$ converges, since the integrand is positive, therefore $\displaystyle{\int_{0}^{1}\dfrac{\mathrm{d}x}{\sqrt{\sin x}}} \le \int_{0}^{\pi /2}\dfrac{\mathrm{d}x}{\sqrt{\sin x}}$.

Set $x = \arcsin y$ and $\mathrm{d}x = \dfrac{\mathrm{d}y}{\sqrt{1-y^2}}$. Then, \begin{aligned} \int_{0}^{\pi /2}\dfrac{\mathrm{d}x}{\sqrt{\sin x}}&=\int_{0}^{1}\dfrac{\mathrm{d}y}{\sqrt{y}\sqrt{1-y^2}} \\ &=\int_{0}^{1} \dfrac{\mathrm{d}y}{\sqrt{y}\sqrt{1-y}\sqrt{1+y}} \\ &\le \int_{0}^{1}\dfrac{\mathrm{d}y}{\sqrt{y}\sqrt{1-y}} \\ &=2\int_{0}^{1}\dfrac{\mathrm{d}y}{\sqrt{1-y^2}}\\ &\le2\int_{0}^{1}\dfrac{\mathrm{d}y}{\sqrt{1-y}} \\ &=4. \end{aligned}

So, the integral is convergent and $$\int_{0}^{1}\dfrac{\mathrm{d}x}{\sqrt{\sin x}} \le 4.$$

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