Prove that $\lim_{t \to \infty} \int_1^t \sin(x)\sin(x^2)\,dx$ converges Question_

Prove that $$\lim_{t \to \infty} \int_1^t \sin(x) \sin(x^2) \, dx$$
  converges.

I think the indefinite integration of $\sin(x)\sin(x^2)$ is impossible. Besides, I've wondered whether the definite integration of it is possible or not. 
I've tried to use the condition that $t \to \infty$. The one that came up to my mind is to use partial integration. When using it, we can have:
$$\int_{1}^{t}\sin(x)\sin(x^2) \, dx=-\left[ \sin(x^2) \cos(x) \right]_1^t + \int_1^t 2x\cos(x^2)\sin(x) \, dx$$
However, since $\sin(t^2)\cos(t)$ diverges as $t \to \infty$, I couldn't determine whether the given integration diverges or not. Due to this, I re-tried to have partial integration in a quite different way:
$$\int_1^t \sin(x)\sin(x^2) \, dx=\int_1^t \frac{\sin(x)}{2x}(2x\sin(x^2)) \, dx=-\left[\frac{\sin(x)}{2x} \cos(x^2)\right]_1^t + \int_1^t \frac{x\cos(x)-\sin(x)}{2x^2} \cos(x^2) \, dx$$
In this case, $\left[\sin(t)\cos(t^2)/2t\right]$ goes to $0$ as $t \to \infty$.  Therefore, it is enough to see the integration part only. Unfortunately, I'm stuck here. Could you give me some key ideas that can investigate whether $$\int_{1}^{t}\frac{x\cos(x)-\sin(x)}{2x^2}\cos(x^2)dx$$ converges or not?
The other way of solution is also welcome! Thanks for your advice.
 A: This is an old Putnam problem [2000, A4]: show that $\displaystyle{\lim_{B\to \infty}\int _0^B \sin(x)\sin(x^2)\,dx}$ exists. 
Since we are interested in the limit, we can assume $B>1$. To simplify matters, we introduce a factor of 4.
$$
\lim_{B\to \infty}2\int _0^B \sin(x)\cdot 2\sin(x^2)\,dx
$$$\sin(x^2)$ doesn't have an elementary antiderivative, but $2x\sin(x^2)$ does. We multiply and divide by $x$ to introduce this factor
$$
\lim_{B\to \infty}2\int _0^B\frac{ \sin(x)}{x}\cdot2 x\sin(x^2)\,dx
$$Let's use IBP to trade one integral for another. Let $u=\frac{\sin(x)}{x}$,  $dv=2x\sin(x^2)dx$:
$$
2\int _0^B\frac{ \sin(x)}{x}\cdot2 x\sin(x^2)\,dx = \left.-2\cos(x^2)\frac{\sin(x)}{x}\right|_0^B + \int _0^B 2\cos(x^2)\cdot \frac{x \cos (x)-\sin (x)}{x^2}\,dx
$$The boundary term evaluates to 2 as $B\to\infty$. Now we use that $B>1$ to split up the new integral:
$$
=2 + 2\int _0^1 \cos(x^2)\cdot \frac{x \cos (x)-\sin (x)}{x^2}\,dx+ \int _1^B 2\cos(x^2)\cdot \frac{x \cos (x)-\sin (x)}{x^2}\,dx
$$For the first integral, note that cosine is positive and continuous on $[0,1]$. Then by the Mean Value Theorem for Integrals, for some $\xi\in(0,1)$ we have
$$
\int _0^1 2\cos(x^2)\cdot \frac{x \cos (x)-\sin (x)}{x^2}\,dx = 2\cos(\xi^2)\int _0^1  \frac{x \cos (x)-\sin (x)}{x^2}\,dx = 2\cos(\xi^2) \left.\frac{\sin(x)}{x}\right|_0^1
$$
$$
= 2\cos(\xi^2)\left(\frac{\sin(1)}{1}-1\right);
$$this is a finite number. The presence of $x^2$ in the denominator of the second integral is a good sign: we know that $x^{-2}$ is improperly integrable on $[1,\infty).$ We have to tweak things a little but the goal is to get a function comparable with $x^{-2}$ on $[1,B]$. Now we reintroduce a factor of $x/x$ as earlier and use IBP again:
$$
\int _1^B 2x\cos(x^2)\cdot \frac{x \cos (x)-\sin (x)}{x^3}\,dx
$$
$$
= \left.\sin(x^2)\cdot \frac{x \cos (x)-\sin (x)}{x^3} \right|_1^B - \int_1^B \sin(x^2)\cdot \frac{-x^2\sin (x)+3 \sin (x)-3 x \cos (x)}{x^4}\,dx
$$The boundary terms are finite in the limit. This new integral has the form we want: it is roughly of the form $x^{-2}$. We now use several basic comparisons to show it converges as $B\to \infty$. First, take absolute values:
$$
\left|\int_1^B \sin(x^2)\cdot \frac{-x^2\sin (x)+3 \sin (x)-3 x \cos (x)}{x^4}\,dx\right|
$$
$$
\leq \int_1^B\left| \sin(x^2)\cdot \frac{-x^2\sin (x)+3 \sin (x)-3 x \cos (x)}{x^4}\right|\,dx
$$
$|\sin(\theta)|\leq1$ for any real $\theta$, and dividing the numerator and denominator by $x^2$ gives
$$
\leq \int_1^B1\cdot\left| \frac{-x^2\sin (x)+3 \sin (x)-3 x \cos (x)}{x^4}\right|\,dx
$$
$$
= \int_1^B\frac{\left|-\sin (x)+3 \sin (x)/x^2-3 \cos (x)/x\right|}{x^2}\,dx
$$
$$
\leq \int_1^B\frac{1+3+3}{x^2}\,dx=\int_1^B\frac{7}{x^2}\,dx
$$This last integral is finite. The original integral is bounded by a number of finite pieces, hence is finite.
A: You can write
$$ \sin(x) \sin(x^2) = \frac{1}{2} \left[\cos(x^2-x) - \cos(x^2+x)\right]$$
and let $u = x^2-x$, $v=x^2+x$ to obtain
\begin{align}
\int \limits_1^t \sin(x) \sin(x^2) \, \mathrm{d} x &= \frac{1}{2} \int \limits_1^t \left[\cos(x^2-x) - \cos(x^2+x)\right] \, \mathrm{d} x  \\
&= \frac{1}{4} \left[\int \limits_0^{t^2-t} \frac{\cos(u)}{\sqrt{u + \frac{1}{4}}} \, \mathrm{d}u - \int \limits_2^{t^2+t} \frac{\cos(v)}{\sqrt{v + \frac{1}{4}}} \, \mathrm{d}v\right] .
\end{align}
The convergence of this expression as $t \to \infty$ is ensured by Dirichlet's test for integrals or integration by parts.
