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.
Best Answer
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.