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

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},} newcommand{braces}[1]{leftlbrace,{#1},rightrbrace} newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack} newcommand{dd}{mathrm{d}} newcommand{ds}[1]{displaystyle{#1}} newcommand{expo}[1]{,mathrm{e}^{#1},} newcommand{ic}{mathrm{i}} newcommand{mc}[1]{mathcal{#1}} newcommand{mrm}[1]{mathrm{#1}} newcommand{pars}[1]{left(,{#1},right)} newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}} newcommand{root}[2][]{,sqrt[#1]{,{#2},},} newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}} newcommand{verts}[1]{leftvert,{#1},rightvert}$ begin{align} int_{0}^{infty}{sin^{2}pars{x} over x^{5/2}},dd x & = int_{0}^{infty} overbrace{1 - cospars{2x} over 2} ^{ds{sin^{2}pars{x}}} overbrace{{1 over Gammapars{5/2}}int_{0}^{infty}t^{3/2}expo{-xt}dd t} ^{ds{1 over x^{5/2}}} dd x \[5mm] & = {1 over 2,Gammapars{5/2}}int_{0}^{infty}t^{3/2}, Reint_{0}^{infty}bracks{expo{-xt} - expo{-pars{t - 2ic}x}}dd x ,dd t \[5mm] & = {2 over 3root{pi}}int_{0}^{infty}t^{3/2}, Repars{{1 over t} - {1 over t - 2ic}}dd t \[5mm] & = {2 over 3root{pi}}int_{0}^{infty}t^{3/2}, pars{{1 over t} - {t over t^{2} + 4}}dd t \[5mm] & = {8 over 3root{pi}}int_{0}^{infty},{t^{1/2} over t^{2} + 4},dd t \[5mm] & = {8 over 3root{pi}},{1 over 4},2root{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},{Gammapars{1/4}Gammapars{3/4} over Gammapars{1}} \[5mm] & = {2 over 3}root{2 over pi},{pi over sinpars{pi/4}} = bbx{{4 over 3}root{pi}} approx 2.3633 end{align}

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 $$ 0leq I_1leq int_0^1 x^2cdot x^{-5/2},dx = int _0^1 x^{-1/2},dx = 2 $$

2. Direct computation

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

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

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

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

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.