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

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)
 A: 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.
A: *

*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
$$
0\leq I_1\leq \int_0^1 x^2\cdot x^{-5/2}\,dx = \int _0^1 x^{-1/2}\,dx = 2
$$


*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
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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}
