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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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4 Answers 4

up vote 10 down vote accepted

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
2  
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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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{t \equiv \sin\pars{x}\quad\iff x = \arcsin\pars{t}}$: \begin{align} \color{#00f}{\large\int_{0}^{1}{\dd x \over \root{\sin\pars{x}}}}&= \int_{0}^{\sin\pars{1}}t^{-1/2}\,{\dd t \over \root{1 - t^{2}}} = \int_{0}^{\root{\sin\pars{1}}}t^{-1/4}\pars{1 - t}^{-1/2}\half\,t^{-1/2}\,\dd t \\[3mm]&= \half\int_{0}^{\root{\sin\pars{1}}}t^{-3/4}\pars{1 - t}^{-1/2}\,\dd t =\color{#00f}{\large\half\,{\rm B}_{\sin^{1/2}\pars{1}}\pars{{1 \over 4},\half}} \\[3mm]&=\color{#00f}{\large2\sin^{1/8}\pars{1}\ _{2}{\rm F}_{1}\pars{{1 \over 4},\half;{3 \over 4};\sin^{1/8}\pars{1}}} \approx 2.3283 \end{align} ${\rm B}_{x}\pars{p,q}$ and $_{2}{\rm F}_{1}$ are the Incomplete Beta and the Hypergeometric functions, respectively.

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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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