A trigonometric-integral inequality This problem comes from a discussion with one of my friends: 

Prove that: $$\displaystyle \lim_{n \to \infty}\int_{1}^{n}{\sin (x)\sin(x^2)}\,{\mathrm dx}< \lim_{n \to \infty}\int_{1}^{n}{\sin(x^2)}\,{\mathrm dx}$$

I wonder if there is a reasonable way to somehow solve such a problem. Thanks. 
 A: The idea goes as follows. Consider your inequality
$$
\displaystyle \lim_{n \to \infty}\int_{1}^{n}{\sin (x)\sin(x^2)}\,{\mathrm dx}< \lim_{n \to \infty}\int_{1}^{n}{\sin(x^2)}\,{\mathrm dx}
$$
but both integrals are meaningful in the given limit being standard values of Fresnel functions. So, you have
$$
\int_{1}^{\infty}{\sin (x)\sin(x^2)}\,{\mathrm dx}< \int_{1}^{\infty}{\sin(x^2)}\,{\mathrm dx}.
$$
The rhs can be immediately evaluated to give
$$
  \int_{1}^{\infty}{\sin(x^2)}\,{\mathrm dx}=\frac{1}{2}\sqrt{\frac{\pi}{2}}\left[1-2S\left(\sqrt{2}{\pi}\right)\right].
$$
I have introduced here a Fresnel function. These are defined in the following way
$$
   S(x)=\int_0^x dt\sin\left(\frac{\pi}{2}t^2\right)\qquad C(x)=\int_0^x dt\cos\left(\frac{\pi}{2}t^2\right)
$$
and $\frac{\pi}{2}$ factor that you see in the integrals is the reason why your integral is evaluated at that value for $S$. For your case is $S\left(\sqrt{2}{\pi}\right)\sim 0.247558$ and so the rhs evaluates to 0.316389.
On the lhs you have to do some more work. You will have
$$
  \sin(x)\sin(x^2)=\frac{1}{2}\left[\cos(x-x^2)-\cos(x+x^2)\right]
$$
and so you have to evaluate
$$
  \int_{1}^{\infty}{\sin (x)\sin(x^2)}\,{\mathrm dx}=\frac{1}{2}\int_{1}^{\infty}\left[\cos(x-x^2)-\cos(x+x^2)\right]\,{\mathrm dx}.
$$
In order to evaluate these integrals we firstly note that $x^2+x=\left(x+\frac{1}{2}\right)^2-\frac{1}{4}$ and  $x^2-x=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}$ and so
$$
  \frac{1}{2}\int_{1}^{\infty}\left[\cos(x-x^2)-\cos(x+x^2)\right]\,{\mathrm dx}=
   \frac{1}{2}\int_{\frac{1}{2}}^{\infty}\cos\left(y^2-\frac{1}{4}\right)\,{\mathrm dy}-\frac{1}{2}\int_{\frac{3}{2}}^{\infty}\cos\left(y^2-\frac{1}{4}\right)\,{\mathrm dy}
$$
after an obvious change of variables in the integrals. But we recognize immediately that
$$
   \int_{1}^{\infty}{\sin (x)\sin(x^2)}\,{\mathrm dx}= \frac{1}{2}\int_{\frac{1}{2}}^{\frac{3}{2}}\cos\left(y^2-\frac{1}{4}\right)\,{\mathrm dy}
$$
and expanding the cosine we realize that
$$
   \int_{\frac{1}{2}}^{\frac{3}{2}}\cos\left(y^2-\frac{1}{4}\right)\,{\mathrm dy}=
\int_{\frac{1}{2}}^{\frac{3}{2}}\cos(y^2)\cos\left(\frac{1}{4}\right)\,{\mathrm dy}+\int_{\frac{1}{2}}^{\frac{3}{2}}\sin(y^2)\sin\left(\frac{1}{4}\right)\,{\mathrm dy}
$$
and we are back to Fresnel functions. Already at this stage, you should be able to realize that the inequality holds as you are evaluating Frensel functions on a smaller interval and these are also multiplied by factors that are lesser than unity. But we can go on with numerical evaluation obtaining 0.286035 that is lower than 0.316389 as it should.
A: May be there is none, general at least, since if to change the lower limits of integration from $1$ to $a$ the inequality wouldn't hold for all $a$. Here is a plot of the limits for different values of $a$, lhs in red and rhs in blue: 

