The convergence of the improper integral $\int_{0}^\infty\frac{\sin^2(x)}{x^{5/2}}\,dx$

Question

I have to analyse the convergence of $\int_{0}^\infty\frac{\sin^2(x)}{x^{5/2}}\,dx$.

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What I have done is:

First of all, I've divided the integral in two integrals: $\int_{0}^\infty\frac{\sin^2(x)}{x^{5/2}}\,dx=\int_{0}^{1}\frac{\sin^2(x)}{x^{5/2}}\,dx+\int_{1}^\infty\frac{\sin^2(x)}{x^{5/2}}\,dx$

I've analysed the second integral: $\int_{1}^\infty\frac{\sin^2(x)}{x^{5/2}}\,dx\;$ and as $\int_{1}^\infty\frac{1}{x^{5/2}}$ is convergent (owing to the fact that $5/2>1$), by comparison we know that $\frac{\sin^2(x)}{x^{5/2}}$ is convergent.

But now, I have to analyse the first part and I don't know how to do. I want to use the comparison (but I don't know with what to compare it) or the limit theorem (but I neither know how)

Answer 1

We have : $$ \lim_{x\to 0}{\sqrt{x}\frac{\sin^{2}{x}}{x^{\frac{5}{2}}}}=\lim_{x\to 0}{\left(\frac{\sin{x}}{x}\right)^{2}}=1 $$

Thus : $$ \frac{\sin^{2}{x}}{x^{\frac{5}{2}}}\underset{x\to 0}{\sim}\frac{1}{\sqrt{x}} $$

Since $ \int_{0}^{1}{\frac{\mathrm{d}x}{\sqrt{x}}} $ converges, $ \int_{0}^{1}{\frac{\sin^{2}{x}}{x^{\frac{5}{2}}}\,\mathrm{d}x} $ does also converge.

Answer 2

1. Comparison

$$\underbrace{\int_{0}^{\infty} \sin^2(x) x^{-5/2}\,dx}_{I} = \underbrace{\int_{0}^{1} \sin^2(x) x^{-5/2}\,dx}_{I_1} +\underbrace{\int_{1}^{\infty} \sin^2(x) x^{-5/2}\,dx}_{I_2} $$ $I_2$ converges by directly comparing the integrand to $x^{-5/2}$. $I_1$ converges because on $[0,1]$, we have $\sin(x)\leq x$, whence $$ 0\leq I_1\leq \int_0^1 x^2\cdot x^{-5/2}\,dx = \int _0^1 x^{-1/2}\,dx = 2 $$

2. Direct computation

Use integration by parts with $u=\sin^2(x)$: $$ I = \left.- \frac{2}{3} \frac{\sin^2(x)}{x^{3/2}} \right|_{0}^{\infty} + \frac{2}{3} \int _{0}^{\infty} \frac{2\sin(x)\cos(x)}{x^{3/2}}\,dx $$ $$ = \frac{2}{3} \int _{0}^{\infty} \frac{\sin(2x)}{x^{3/2}}\,dx $$ Put $y=2x$: $$ = \frac{4}{3\sqrt{2}} \int _{0}^{\infty} \frac{\sin(y)}{y^{3/2}}\,dy $$ $$ = \frac{4}{3\sqrt{2}}\cdot \Gamma\left(\frac{-1}{2}\right)\sin\left(\frac{-\pi}{4}\right) = \frac{4\sqrt{\pi}}{3} $$

Source: https://dlmf.nist.gov/5.9

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{\sin^{2}\pars{x} \over x^{5/2}}\,\dd x & = \int_{0}^{\infty}\ \overbrace{1 - \cos\pars{2x} \over 2} ^{\ds{\sin^{2}\pars{x}}}\ \overbrace{{1 \over \Gamma\pars{5/2}}\int_{0}^{\infty}t^{3/2}\expo{-xt}\dd t} ^{\ds{1 \over x^{5/2}}}\ \dd x \\[5mm] & = {1 \over 2\,\Gamma\pars{5/2}}\int_{0}^{\infty}t^{3/2}\, \Re\int_{0}^{\infty}\bracks{\expo{-xt} - \expo{-\pars{t - 2\ic}x}}\dd x \,\dd t \\[5mm] & = {2 \over 3\root{\pi}}\int_{0}^{\infty}t^{3/2}\, \Re\pars{{1 \over t} - {1 \over t - 2\ic}}\dd t \\[5mm] & = {2 \over 3\root{\pi}}\int_{0}^{\infty}t^{3/2}\, \pars{{1 \over t} - {t \over t^{2} + 4}}\dd t \\[5mm] & = {8 \over 3\root{\pi}}\int_{0}^{\infty}\,{t^{1/2} \over t^{2} + 4}\,\dd t \\[5mm] & = {8 \over 3\root{\pi}}\,{1 \over 4}\,2\root{2} \int_{0}^{\infty}\,{t^{1/2} \over t^{2} + 1}\,\dd t \\[5mm] & = {4 \over 3}\root{2 \over \pi} \int_{0}^{\infty}\,{t^{1/4} \over t + 1}\,{1 \over 2}\,t^{-1/2}\,\dd t \\[5mm] & = {2 \over 3}\root{2 \over \pi} \int_{1}^{\infty}\,{\pars{t - 1}^{-1/4} \over t}\,\dd t \\[5mm] & = {2 \over 3}\root{2 \over \pi} \int_{1}^{0}\,{\pars{1/t - 1}^{-1/4} \over 1/t} \pars{-\,{\dd t \over t^{2}}} \\[5mm] & = {2 \over 3}\root{2 \over \pi} \int_{0}^{1}t^{-3/4}\pars{1 - t}^{-1/4}\,\dd t = {2 \over 3}\root{2 \over \pi}\,{\Gamma\pars{1/4}\Gamma\pars{3/4} \over \Gamma\pars{1}} \\[5mm] & = {2 \over 3}\root{2 \over \pi}\,{\pi \over \sin\pars{\pi/4}} = \bbx{{4 \over 3}\root{\pi}}\ \approx 2.3633 \end{align}