Integral $\int_0^{\infty} \sin(x^2)/x^2\,dx$ Does anyone have  a proof for $$\int_0^{\infty}\frac{\sin(x^2)}{x^2}\,dx=\sqrt{\frac{\pi}{2}}.$$
I tried to get it from contour integrating $$\frac{e^{iz^2}-1}{z^2},$$ but failed.
Thanks.
 A: I'll extend @sos440's answer a bit.
$$
I = \int_o^\infty \frac {\sin x^2}{x^2}dx = -\left. \frac{\sin x^2}x \right|_0^\infty + 2\int_0^\infty \cos x^2 dx \\
\left .\frac{\sin x^2}x \right |_0^\infty = \lim_{x \rightarrow \infty} \frac {\sin x^2}x - \lim_{x \rightarrow 0} \frac {\sin x^2}x = 0-\lim \frac 1x \left ( x^2 + O(x^4)\right ) = -\lim_{x \rightarrow 0}\left( x + O(x^3)\right ) = 0 \\
$$
To take Fresnel's integral $C(x) = \int_0^x \cos x^2 dx$ at $\infty$ just take a contour like here, and you're done.
PS: $I = 2\lim_{x \rightarrow \infty} C(x) = 2 \sqrt{\frac \pi 8} = \sqrt{\frac \pi 2}$
A: Another solution. (which does not use complex analysis)
Substitute $u=x^2$, then the integral becomes
$$A:=\int_{0}^{\infty}\frac{\sin (x^2)}{x^2}dx=\frac{1}{2}\int_{0}^{\infty}u^{-3/2}\sin u du$$
Now we'll consider more general one;
$$f(p):=\int_{0}^{\infty}\frac{\sin u}{u^p}du\phantom{a} (p\in (1, 2))$$
From the definition of the gamma function, we can easily deduce that
$$\frac{\Gamma(p)}{u^p}=\int_{0}^{\infty}v^{p-1}e^{-uv}dv$$
This formula gives
$$f(p)\Gamma(p)=\int_{0}^{\infty}\sin u\left(\int_{0}^{\infty}v^{p-1}e^{-uv} dv\right)du$$
By Fubini's theorem,
$$f(p)\Gamma(p)=\int_{0}^{\infty}\int_{0}^{\infty}v^{p-1}e^{-uv}\sin u dvdu=\int_{0}^{\infty}\int_{0}^{\infty}v^{p-1}e^{-uv}\sin u du dv$$
So
$$f(p)\Gamma(p)=\int_{0}^{\infty}v^{p-1}\left(\int_{0}^{\infty}e^{-uv}\sin udu\right)dv=\int_{0}^{\infty}\frac{v^{p-1}}{v^2+1}dv$$
Substitute $v=\tan \theta$, then
$$\begin{aligned}f(p)\Gamma(p)&=\int_{0}^{\pi /2}\tan^{p-1}\theta d\theta=\int_{0}^{\pi /2}\cos^{2\cdot(1-p/2)-1}\theta\sin^{2\cdot p/2-1}\theta d\theta\\&=\frac{1}{2}\operatorname{B}\left(1-\frac{p}{2},\frac{p}{2}\right)=\frac{1}{2}\Gamma\left(1-\frac{p}{2}\right)\Gamma\left(\frac{p}{2}\right)\end{aligned}$$
($\operatorname{B}(\phantom{x},\phantom{y})$ denotes Beta function.)
By Euler's reflection formula,
$$f(p)\Gamma(p)=\frac{1}{2}\Gamma\left(1-\frac{p}{2}\right)\Gamma\left(\frac{p}{2}\right)=\frac{1}{2}\frac{\pi}{\sin(\pi p/2)}$$
Therefore
$$f(p)=\int_{0}^{\infty}\frac{\sin u}{u^p}du=\frac{1}{2}\frac{\pi}{\Gamma(p) \sin(\pi p/2)}$$
Substituting $\displaystyle p=\frac{3}{2}$, we obtain
$$A=\frac{1}{2}f\left(\frac{3}{2}\right)=\frac{1}{2}\cdot\frac{1}{2}\frac{\pi}{\sqrt{\pi}/2 \cdot \sqrt{2}/2}=\frac{\sqrt{\pi}}{\sqrt{2}}$$
A: Since $\mathcal{L}(\sin x)(s)=\frac{1}{1+s^2}$ and $\mathcal{L}^{-1}\left(\frac{1}{x\sqrt{x}}\right)=\frac{2}{\sqrt{\pi}}\sqrt{s}$ we have
$$ \int_{0}^{+\infty}\frac{\sin(x^2)}{x^2}\,dx = \frac{1}{2}\int_{0}^{+\infty}\frac{\sin x}{x\sqrt{x}}\,dx = \frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{\sqrt{s}}{1+s^2}\,ds=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}\frac{u^2 du}{1+u^4} $$
and by the residue theorem $\int_{-\infty}^{+\infty}\frac{u^2 du}{1+u^4}\,du=\frac{\pi}{\sqrt{2}}$, so the original integral equals $\sqrt{\frac{\pi}{2}}$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{\infty}{\sin\pars{x^{2}} \over x^{2}}\,\dd x & =
\int_{0}^{\infty}{1 \over 2}\int_{-1}^{1}\expo{\ic k x^{2}}\dd k\,\dd x =
{1 \over 2}\int_{-1}^{1}\
\overbrace{\int_{0}^{\infty}\expo{\ic k x^{2}}\dd x}
^{\ds{\begin{array}{c}
      Fresnel\ Integral:
      \\[1mm]
      \ds{=\ {1 \over 2}\,\root{\pi \over 2}\,{\ic k + \verts{k} \over \verts{k}^{3/2}}}
      \end{array}}}\,\ \dd k
\\[5mm] & =
{1 \over 2}\bracks{{1 \over 2}\root{\pi \over 2}\pars{2\int_{0}^{1}k^{-1/2}
\,\dd k}} = \bbx{\root{\pi \over 2}} \approx 1.2533
\end{align}

See the Fresnel Integral.

