a contour integral and how to solve it Evaluate: $\displaystyle
\int_{0}^{\infty}\frac{\sin\left(x^{2}\right) + \cos\left(x^{2}\right) - 1}
{x^{2}}\,\mathrm{d}x$.
My problem is that both functions $\sin\left(x^{2}\right)$ and $\cos\left(x^{2}\right)$ are present in the integrand, so taking just the real part or complex part of $\,\mathrm{e}^{\mathrm{i}x^{2}}$ and proceeding the standard evaluation using residue calculus is not going to work.
What possible simplifications could be done to that integrand to solve the problem ?.
 A: By substituting $x=\sqrt{t}$ we are left with
$$ I=\frac{1}{2}\int_{0}^{+\infty}\frac{\sin(t)+\cos(t)-1}{t^{3/2}}\,dt \stackrel{\mathcal{L}}{=}\frac{1}{2}\int_{0}^{+\infty}\frac{s-1}{s+s^3}\cdot\frac{2\sqrt{s}}{\sqrt{\pi}}\,ds$$
by an useful property of the Laplace transform. By setting $s=u^2$ we get:
$$\boxed{ I = \int_{0}^{+\infty}\frac{\sin x^2+\cos x^2-1}{x^2}\,dx = \frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{u^2-1}{1+u^4}\,du = \color{red}{\large 0}}$$
by splitting the integration range as $[0,1]\cup[1,+\infty)$ and applying the substitution $u\mapsto\frac{1}{u}$ on the second "half".
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}} + \cos\pars{x^{2}} - 1 \over x^{2}}\,\dd x
\\[5mm] \stackrel{x^{2}\ \mapsto\ x}{=} &\
{1 \over 2}\int_{0}^{\infty}{\sin\pars{x} + \cos\pars{x} - 1 \over x^{3/2}}
\,\dd x
\,\,\,\stackrel{\mbox{IBP}}{=}\,\,
\int_{0}^{\infty}{\cos\pars{x} - \sin\pars{x} \over x^{1/2}}\,\dd x
\\[5mm] = &\
\Re\int_{0}^{\infty}x^{-1/2}\expo{\ic x}\,\dd x -
\Im\int_{0}^{\infty}x^{-1/2}\expo{\ic x}\,\dd x
\\[5mm] \stackrel{x\ =\ \ic t}{=}\,\,\,&
\Re\int_{0}^{-\infty\ic}\pars{\ic t}^{-1/2}\expo{-t}\ic\,\dd t -
\Im\int_{0}^{-\infty\ic}\pars{\ic t}^{-1/2}\expo{-t}\ic\,\dd x
\\[5mm] = &\
\cos\pars{\pi \over 4}\int_{0}^{\infty}t^{-1/2}\expo{-t}\,\dd t -
\sin\pars{\pi \over 4}\int_{0}^{\infty}t^{-1/2}\expo{-t}\,\dd x = \bbx{\large 0}
\end{align}

By simplicity, I omitted integrals along an arc of radius $\ds{R}$, in the fourth quadrant, which go to zero $\ds{\pars{~\mbox{as}\ {1 \over R^{3/2}}~}}$as the arc radius $\ds{R \to \infty}$. $\ds{z^{-1/2}}$ is the Principal Branch.

