Convergence of $ I=\int_0^\infty \sin x\sin(x^2)\mathrm{d}x$ 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. :)
 A: We shall use only substitution and integration by parts to show that the integral of interest, $\int_1^L \sin(x)\sin(x^2)\,dx$, is convergent.
First, enforcing the substitution $x\to \sqrt{x}$ reveals
$$\int_1^L \sin(x)\sin(x^2)\,dx=\frac12\int_1^L \frac{\sin(\sqrt{x})\sin(x)}{\sqrt{x}}\,dx \tag 1$$

Second, integrating by parts the integral on the right-hand side of $(1)$ with $u=\frac{\sin(\sqrt{x})}{\sqrt{x}}$ and $v=-\cos(x)$ yields
$$\begin{align}
\int_1^L \sin(x)\sin(x^2)\,dx&=\frac12\left.\left(-\frac{\sin(\sqrt{x})\cos(x)}{\sqrt{x}}\right)\right|_{x=1}^{x=L}\\\\
&+\frac14 \int_1^L \left(\frac{\cos(\sqrt{x})\cos(x)}{x}-\frac{\sin(\sqrt{x})\cos(x)}{x^{3/2}}\right)\,dx\tag2
\end{align}$$

Third, integrating by parts the first term in the integral on the right-hand side of $(2)$ with $u=\frac{\cos(\sqrt{x})}{x}$ and $v=\sin(x)$, we obtain
$$\begin{align}\int_1^L \frac{\cos(\sqrt{x})\cos(x)}{x}\,&=\left.\left(\frac{\cos(\sqrt{x})\sin(x)}{x}\right)\right|_{x=1}^{x=L}\\\\
&- \int_1^L \left(\frac{\sin(\sqrt{x})\sin(x)}{2x^{3/2}}+\frac{\cos(\sqrt{x})\sin(x)}{x^2}\right)\,dx\tag 3
\end{align}$$

Substituting $(3)$ into $(2)$ shows that all integrals involved are of the forms
$$I_1=\int_1^L \frac{\sin(\sqrt{x})\cos(x)}{x^{3/2}}\,dx$$
and
$$I_2=\int_1^L \frac{\cos(\sqrt{x})\sin(x)}{x^2}\,dx$$
Both $I_1$ and $I_2$ are absolutely convergent as $L\to\infty$ since $\int_1^\infty \frac{1}{x^{3/2}}\,dx<\infty$ and $\int_1^L \frac{1}{x^2}\,dx<\infty$.
Therefore, the integral of interest converges as was to be shown!.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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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.

$$
\int_{0}^{\infty}\sin\pars{x}\sin\pars{x^{2}}\,\dd x =
\bbx{\int_{0}^{1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x}
\quad\mbox{which is clearly}\ convergent.
$$
A: It is enough to show that both the integrals
$$ I_1 = \int_{0}^{+\infty}\cos(x+x^2)\,dx,\qquad I_2=\int_{0}^{+\infty}\cos(x^2-x)\,dx $$
are (conditionally) convergent. $f(x)=x+x^2$ is an increasing function on $\mathbb{R}^+$ and by setting $x=f^{-1}(t)$ we get:
$$ I_1 = \int_{0}^{+\infty}\frac{\cos t}{\sqrt{1+4t}}\,dt $$
that clearly is a convergent integral by Dirichlet's test: $\cos t$ has a bounded primitive and $\frac{1}{\sqrt{1+4t}}$ is decreasing towards zero on $\mathbb{R}^+$. A similar argument proves the convergence of $I_2$: $g(x)=x^2-x$ is increasing on $\left(\frac{1}{2},+\infty\right)$ and the integral $\int_{0}^{1/2}\cos(x^2-x)\,dx$ is clearly bounded by $\frac{1}{2}$ in absolute value.
The Laplace transform gives a way for computing a good numerical approximation of $I_1$. We have:
$$ 0\leq I_1 = \frac{1}{2\sqrt{\pi}}\int_{0}^{+\infty}\frac{\sqrt{s}e^{-s/4}}{1+s^2}\,ds \stackrel{\text{Cauchy-Schwarz}}{\leq}\frac{1}{2\sqrt{\pi}}\sqrt{\int_{0}^{+\infty}e^{-s/2}\,ds\int_{0}^{+\infty}\frac{s\,ds}{(1+s^2)^2}}$$
hence $0\leq I_1\leq \frac{1}{2\sqrt{\pi}}$.
A: You could just keep going by explicitly taking the derivative:
$$\left(\sin t\over2t\right)'={\cos t\over 2t}-{\sin t\over2t^2}$$
which now gives two integrals to be shown convergent:
$$\int_{\pi/2}^\infty{\cos t\cos t^2\over2t}dt\qquad\text{and}\qquad\int_{\pi/2}^\infty{\sin t\cos t^2\over2t^2}dt$$
The second of these is clearly convergent, since $\left|\sin t\cos t^2\over2t^2\right|\le{1\over2t^2}$ and $\int_1^\infty{dx\over x^2}=1$.  For the first integral, do integration by parts again, using
$$u={\cos t\over4t^2}\qquad\text{and}\qquad dv=2t\cos t^2\,dt$$
so that
$$\int_{\pi/2}^\infty{\cos t\cos t^2\over2t}dt={\cos t\sin t^2\over4t^2}\Big|_{\pi/2}^\infty+\int_{\pi/2}^\infty\left({\sin\over4t^2}+{\cos t\over8t^3}\right)\sin t^2\,dt$$
and the remaining integral is again clearly convergent.
