Does the improper integral $\int_0^\infty\sin(x)\sin(x^2)\,\mathrm dx$ converge Does the following improper integral converge?
$$\lim_{B \to \infty}\int_0^B\sin(x)\sin(x^2)\,\mathrm dx$$

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:
$$\begin{align}
&\int_B^{B+1} \cos(x^2 - x)dx = \left[ \frac{\sin(x^2 - x)}{2x - 1} \right]_B^{B+1} + \int_B^{B+1}\frac{2\sin(x^2 - x)dx}{(2x - 1)^2}\\
\implies & \left|\int_B^{B+1} \cos(x^2 - x)dx\right| \le \frac{2}{2B-1} + \frac{2}{(2B-1)^2}\\
\implies & \lim_{B\to\infty} \int_B^{B+1} \cos(x^2 - x)dx = 0\\
\implies & \lim_{B\to\infty} \int_0^B \sin(x) \sin(x^2) dx = \frac12 \int_0^1 \cos(x^2 - x) dx. 
\end{align}$$
