Evaluate $\lim_{a\rightarrow \infty }\frac{1}{a}\int_{0}^{a}\sin(x)\sin(x^2)\,dx$ Calculate $$I=\lim_{a\rightarrow \infty }\frac{1}{a}\int_{0}^{a}\sin(x)\cdot \sin(x^2)\,dx$$
I tried to use the fact that $\lim_{x\rightarrow \infty }\frac{1}{x}\int_{0}^{x}(f(t))\,dt=\frac{1}{T}\int_{0}^{T}f(t)\,dt$ where $T$ is period of the function.
So in my case $T=2\pi$, right? So I have $$I=\lim_{a\rightarrow \infty }\frac{1}{2\pi}\int_{0}^{2\pi}\sin(x)\sin(x^2)\,dx=\lim_{a\rightarrow \infty }\frac{1}{\pi}\int_{0}^{\pi}\sin(x)\sin(x^2)\,dx$$
Now I noted $x=\pi-t$ so $$I=\lim_{a\rightarrow \infty }\frac{1}{\pi}\int_{0}^{\pi}\sin(\pi-t)\sin((\pi-t)^2)\,dt=\lim_{a\rightarrow \infty }\frac{1}{\pi}\int_{0}^{\pi}\sin(t)\sin((\pi-t)^2)\,dt$$
How to continue? Is my method correct?
 A: Note that
$$\frac{d}{dx}\left(\frac{\sin(x)\cos(x^2)}{2x}\right)=\frac{\cos(x)\cos(x^2)}{2x}-\frac{\sin(x)\cos(x^2)}{2x^2}-\sin(x)\sin(x^2)$$
Hence, for $a>1$,
$$\int_{1}^{a}\sin(x)\sin(x^2)\,dx=\left[-\frac{\sin(x)\cos(x^2)}{2x}\right]_1^a+\int_{1}^{a}\frac{\cos(x)\cos(x^2)}{2x}\,dx-\int_{1}^{a}\frac{\sin(x)\cos(x^2)}{2x^2}\,dx$$
and therefore, as $a\to+\infty$,
$$\left|\frac{1}{a}\int_{1}^{a}\sin(x)\sin(x^2)\,dx\right|\leq \frac{1}{a}\left( \frac{1}{2}+\frac{1}{2a}+\int_{1}^{a}\frac{dx}{2x}+\int_{1}^{a}\frac{dx}{2x^2}\right)= \frac{1+\frac{\log(a)}{2}}{a}\to 0$$
Finally
$$\lim_{a\rightarrow +\infty }\frac{\int_{0}^{a}\sin(x)\sin(x^2)\,dx}{a}=\lim_{a\rightarrow \infty }\frac{\int_{0}^{1}\sin(x)\sin(x^2)\,dx}{a}+\lim_{a\rightarrow \infty }\frac{\int_{1}^{a}\sin(x)\sin(x^2)\,dx}{a}=0.$$
Remark. The limit can be evaluated also by using a stronger result: the improper integral $\int_{0}^{+\infty}\sin(x)\sin(x^2)\,dx$ is convergent (see Does the improper integral $\int_0^\infty\sin(x)\sin(x^2)\,\mathrm dx$ converge).
A: You don't really need to compute the integral. If you show that $\int_0^a \sin x \sin (x^2) \,dx$ is bounded (which it is), the limit is zero, since you are multiplying a bounded function by an infinitesimal (as $a \to \infty$). 
A: $$\sin(x^2)$$
is not periodic, so this approach is not working.
Generally,
$$\sin((x+2\pi)^2) = \sin(x^2+4\pi x +4\pi^2) \neq \sin(x^2).$$
Even when $x \in \mathbb N$, when the $4\pi x$-part is a multiple of the period of $\sin(x)$, the $4\pi^2$-part makes the sin-values different.
No other period exits. As $x \to \infty$, the difference in arguments between consecutive maxima (=$1$) and minima (=$-1$) becomes smaller and smaller, as $x^2$ increases in values as well as inclination.
