Prove: $\int_0^\infty \sin (x^2) \, dx$ converges. $\sin x^2$ does not converge as $x \to \infty$, yet its integral from $0$ to $\infty$ does.
I'm trying to understand why and would like some help in working towards a formal proof.
 A: The humps for $x\mapsto \sin(x^2)$ go up and down.  Each has an area smaller than that of the last.  The areas converge to 0 as you progress down the $x$-axis.  By the alternating series test, this converges.
A: $x\mapsto \sin(x^2)$ is integrable on $[0,1]$, so we have to show that $\lim_{A\to +\infty}\int_1^A\sin(x^2)dx$ exists. Make the substitution $t=x^2$, then $x=\sqrt t$ and $dx=\frac{dt}{2\sqrt t}$. We have $$\int_1^A\sin(x^2)dx=\int_1^{A^2}\frac{\sin t}{2\sqrt t}dt=-\frac{\cos A^2}{2\sqrt {A^2}}+\frac{\cos 1}2+\frac 12\int_1^{A^2}\cos t\cdot  t^{-3/2}\frac{-1}2dt,$$
and since $\lim_{A\to +\infty}-\frac{\cos A^2}{2\sqrt {A^2}}+\frac{\cos 1}2=\frac{\cos 1}2$ and the integral $\int_1^{+\infty}t^{-3/2}dt$ exists (is finite), we conclude that $\int_1^{+\infty}\sin(x^2)dx$ and so does $\int_0^{+\infty}\sin(x^2)dx$.
This integral is computable thanks to the residues theorem.
A: There are more involved methods to actually compute this integral. A real variables method is given in this answer and a contour integration method is given in this answer.
Here is a much simpler method to show only the convergence of the integral.
$$
\begin{align}
\int_0^\infty\sin\left(x^2\right)\,\mathrm{d}x
&=\int_0^\infty\frac{\sin(x)}{2\sqrt{x}}\,\mathrm{d}x\tag{1}\\
&=\sum_{k=0}^\infty\int_{2k\pi}^{(2k+2)\pi}\frac{\sin(x)}{2\sqrt{x}}\,\mathrm{d}x\tag{2}\\
&=\sum_{k=0}^\infty\int_{2k\pi}^{(2k+1)\pi}\frac{\sin(x)}2\left(\frac1{\sqrt{x}}-\frac1{\sqrt{x+\pi}}\right)\mathrm{d}x\tag{3}\\
&=\sum_{k=0}^\infty\int_{2k\pi}^{(2k+1)\pi}\frac{\sin(x)}2\frac{\pi}{\sqrt{x}\sqrt{x+\pi}\left(\sqrt{x}+\sqrt{x+\pi}\right)}\mathrm{d}x\tag{4}\\
&\le\int_0^\pi\frac{\sin(x)}2\left(1+\sum_{k=1}^\infty\frac1{4\sqrt{2\pi} k^{3/2}}\right)\mathrm{d}x\tag{5}\\
&=1+\frac{\zeta\!\left(\frac32\right)}{4\sqrt{2\pi}}\tag{6}
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto\sqrt{x}$
$(2)$: break the integral into $2\pi$ segments
$(3)$: $\sin(x+\pi)=-\sin(x)$
$(4)$: algebra
$(5)$: $\frac{\pi}{\sqrt{x}\sqrt{x+\pi}\left(\sqrt{x}+\sqrt{x+\pi}\right)}\le\min\left(1,\frac1{4\sqrt{2\pi} k^{3/2}}\right)$ for $x\ge2k\pi$
$(6)$: evaluate integral and sum
A: I solved this one integral as a particular case of the following formulas:
$$\int\limits_0^\infty  {\sin \left( {a{x^2}} \right)\cos \left( {2bx} \right)dx}  = \sqrt {\frac{\pi }{{8a}}} \left( {\cos \frac{{{b^2}}}{a} - \sin \frac{{{b^2}}}{a}} \right)$$
$$\int\limits_0^\infty  {\cos \left( {a{x^2}} \right)\cos \left( {2bx} \right)dx}  = \sqrt {\frac{\pi }{{8a}}} \left( {\cos \frac{{{b^2}}}{a} + \sin \frac{{{b^2}}}{a}} \right)$$
So you have
$$\int\limits_0^\infty  {\sin \left( {{x^2}} \right)dx}  = \sqrt {\frac{\pi }{8}} $$
A: This is also informative, and works when there is no aspect with closed form. Taking Davide's substitution, define
$$ A_n^+ = \int_{2 \pi n}^{2 \pi n + \pi} \;  \frac{\sin t}{2 \sqrt t} \;  dt \; ,  $$ 
$$ A_n^- = \int_{2 \pi n + \pi}^{2 \pi n + 2 \pi} \;  \frac{\sin t}{2 \sqrt t} \;  dt \; ,  $$ and finally
$$ A_n = A_n^+ + A_n^- = \int_{2 \pi n }^{2 \pi n + 2 \pi} \;  \frac{\sin t}{2 \sqrt t} \;  dt \; ,  $$ 
Next, I just used $\int_{m \pi}^{m \pi + \pi} \sin t dt = \pm 2,$ depending upon the integer $m,$ and took bounds based on the size of the denominators. 
I suppose we need to start with $n \geq 1.$ With that,
$$ \frac{1}{\sqrt{2 \pi n + \pi}} \leq  A_n^+ \leq \frac{1}{\sqrt{2 \pi n}},  $$
$$ \frac{-1}{\sqrt{2 \pi n + \pi}} \leq  A_n^- \leq \frac{-1}{\sqrt{2 \pi n + 2 \pi}},  $$ and 
$$ 0 \leq  A_n \leq \frac{1}{\sqrt{ 8 \pi} \; \; n^{3/2}}.$$
What does this say about convergence? The integral is $$ \sum_{n = 0}^\infty A_n.   $$ Convergence does not depend on the initial terms, so we may start at the more convenient $n=1.$ From the $3/2$ exponent in the estimate of $A_n,$ we see that the sum is a finite constant. We do see modest oscillation in the indefinite integral, however the $\sqrt n$ terms in the denominators of $A_n^+$ and $A_n^-$ tell us that eventually the indefinite integral stays within any desired distance of the infinite integral.
This idea, cancellation of alternating contributions, can be used with far worse integrands, $\sin (x^5 - x - 1)$  comes to mind.
A: Just with elementary tools: see Here,How to prove only by Transformation that: $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $

In fact we have: \begin{split} \int_0^\infty \sin(x^2) dx=\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx<\infty\end{split}
  See below

Employing the change of variables $2u =x^2$ after integration by parts we get  \begin{split} \int_0^\infty \sin(x^2) dx&=&\frac{1}{\sqrt{2}}\int^\infty_0\frac{\sin(2x)}{\sqrt{x}}\,dx\\& =&\frac{1}{\sqrt{2}} \int^\infty_0\frac{\sin(2x)}{x^{1/2}}\,dx\\&=&\frac{1}{\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx \\&= &\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx\end{split}
Given that $ \sin 2x  =(\sin^2x)'$ and $$\lim_{x\to 0}\frac{\sin x}{x}=1$$
However, $$ \int^\infty_1\frac{\sin^2 x}{x^{3/2}}\,dx\le \int^\infty_1\frac{1}{x^{3/2}}\,dx<\infty$$
since $|\sin x|\le |x|$ we have, 
$$\int^1_0\frac{\sin^2 x}{x^{3/2}}\,dx \le \int^1_0\frac{\ x^2}{x^{3/2}}\,dx = \int^1_0\sqrt x\,dx<\infty$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\sin\pars{x^{2}}\,\dd x} =
{1 \over 4}\int_{0}^{\infty}x^{\color{red}{3/4} - 1}\,\,\,
{\sin\pars{\root{x}} \over \root{x}}\,\dd x
\end{align}
Note that
$\ds{{\sin\pars{\root{x}} \over \root{x}} =
\sum_{k = 0}^{\infty}\,\,\color{red}{\Gamma\pars{1 + k} \over \Gamma\pars{2 + 2k}}
\,{\pars{-x}^{k} \over k!}}$.
With Ramanujan-MT:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\sin\pars{x^{2}}\,\dd x} =
{1 \over 4}\Gamma\pars{\color{red}{3 \over 4}}\,
{\Gamma\pars{1 - \color{red}{3/4}} \over
\Gamma\pars{2 - 2\bracks{\color{red}{3/4}}}} =
{1 \over 4\,\Gamma\pars{1/2}}\,{\pi \over \sin\pars{\pi/4}}
\\[5mm] = &\
{1 \over 4\root{\pi}}\,{\pi \over 1/\root{2}} =
\bbx{{\root{2} \over 4}\,\root{\pi}} \approx 0.6267 \\
\end{align}
