[Math] Does the improper integral $\int_0^\infty\sin(x)\sin(x^2)\,\mathrm dx$ converge

Question

Does the following improper integral converge?

$$\lim_{B \to \infty}\int_0^B\sin(x)\sin(x^2)\,\mathrm dx$$

Answer 1

Notice $x^2 + x = (x+1)x = (x+1)^2 - (x+1)$, we have: $$\begin{align}\int_0^B \sin(x) \sin(x^2) dx &= \frac12\int_0^B(\cos(x^2-x)-\cos(x^2+x)) dx\\ &= \frac12\left(\int_0^B - \int_1^{B+1}\right)\cos(x^2-x)dx\\ &= \frac12\left(\int_0^1 - \int_B^{B+1}\right)\cos(x^2-x)dx \end{align}$$ Integrate by parts. For large $B$, we have: $$\begin{align} &\int_B^{B+1} \cos(x^2 - x)dx = \left[ \frac{\sin(x^2 - x)}{2x - 1} \right]_B^{B+1} + \int_B^{B+1}\frac{2\sin(x^2 - x)dx}{(2x - 1)^2}\\ \implies & \left|\int_B^{B+1} \cos(x^2 - x)dx\right| \le \frac{2}{2B-1} + \frac{2}{(2B-1)^2}\\ \implies & \lim_{B\to\infty} \int_B^{B+1} \cos(x^2 - x)dx = 0\\ \implies & \lim_{B\to\infty} \int_0^B \sin(x) \sin(x^2) dx = \frac12 \int_0^1 \cos(x^2 - x) dx. \end{align}$$