TITLE: Convergence of $ I=\int_0^\infty \sin x\sin(x^2)\mathrm{d}x$
QUESTION [10 upvotes]: I am trying to prove that the improper integral
$$ I=\int_0^\infty \sin x\sin(x^2)\mathrm{d}x$$
converges.
Here's my work:
It suffices to show that
$$\int_\frac{\pi}{2}^\infty \sin (x) \sin(x^2)\mathrm{d}x$$
converges. Using integration by parts,
\begin{align*}
\int_\frac{\pi}{2}^\infty \sin (t) \sin(t^2)\mathrm{d}t&=\int_\frac{\pi}{2}^\infty \frac{\sin (t)}{2t}\cdot 2t\sin(t^2)\mathrm{d}t\\
&=\underline{\bigg[ -\frac{\sin (t)}{2t}\cos(t^2)\bigg]_\frac{\pi}{2}^\infty}+{\int_\frac{\pi}{2}^\infty \left(\frac{\sin (t)}{2t}\right)'\cos(t^2)\mathrm{d}t}
\end{align*}
The underlined part is a constant...
Then I got stuck. I'd like to use "sandwich rule" using the fact that $-1\leq \cos(t^2)\leq 1$, but I can't find a way to apply it properly.
How can I proceed from here? Any correction and/or help would be appreciated. :)

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\Lambda > 0}$:

\begin{align}
\int_{0}^{\Lambda}\sin\pars{x}\sin\pars{x^{2}}\,\dd x & =
{1 \over 2}\int_{0}^{\Lambda}\cos\pars{x^{2} - x}\,\dd x -
{1 \over 2}\int_{0}^{\Lambda}\cos\pars{x^{2} + x}\,\dd x
\\[5mm] & =
{1 \over 2}\int_{1/2}^{\Lambda + 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x -
{1 \over 2}\int_{-1/2}^{\Lambda - 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x
\\[1cm] & =
{1 \over 2}\int_{1/2}^{\Lambda - 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x +
{1 \over 2}\int_{\Lambda - 1/2}^{\Lambda + 1/2}
\cos\pars{x^{2} - {1 \over 4}}\,\dd x
\\[2mm] & -
{1 \over 2}\int_{-1/2}^{1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x -
{1 \over 2}\int_{1/2}^{\Lambda - 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x
\\[1cm] & =
{1 \over 2}\int_{-1/2}^{1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x
\\[2mm] & +
\bracks{%
{1 \over 2}\int_{1/2}^{\Lambda + 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x -
{1 \over 2}\int_{1/2}^{\Lambda - 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x}
\end{align}

As $\ds{\Lambda \to \infty}$, the last two integrals converge since they can be reduced to convergent
  Fresnel Integrals.

$$
\int_{0}^{\infty}\sin\pars{x}\sin\pars{x^{2}}\,\dd x =
\bbx{\int_{0}^{1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x}
\quad\mbox{which is clearly}\ convergent.
$$