Putnam 2000, Problem A-4

Show that the improper integral
$$\lim_{B \to \infty} \int_0^B \sin(x) \sin(x^2) \, dx$$
converges.

Solution:
To avoid some improper integrals at \(0\), we may as well replace the left endpoint of integration by some \(\epsilon > 0\). We now use integration by parts:  
\[
\int_\epsilon^B \sin x \sin x^2 dx = \int_\epsilon^B \frac{\sin x}{2x} \sin x^2 (2x dx) = -\frac{\sin x}{2x} \cos x^2 \bigg|_\epsilon^B + \int_\epsilon^B \left( \frac{\cos x}{2x} - \frac{\sin x}{2x^2} \right) \cos x^2 dx.
\]  

Now \(\frac{\sin x}{2x} \cos x^2\) tends to \(0\) as \(B \to \infty\), and the integral of \(\frac{\sin x}{2x^2} \cos x^2\) converges absolutely by comparison with \(1/x^2\). Thus it suffices to note that  
\[
\int_\epsilon^B \frac{\cos x}{2x} \cos x^2 dx = \int_\epsilon^B \frac{\cos x}{4x^2} \cos x^2 (2x dx) = \frac{\cos x}{4x^2} \sin x^2 \bigg|_\epsilon^B - \int_\epsilon^B \frac{2x \cos x - \sin x}{4x^3} \sin x^2 dx,
\]  
and that the final integral converges absolutely by comparison to \(1/x^3\).  

An alternate approach is to first rewrite \(\sin x \sin x^2\) as \(\frac{1}{2}(\cos(x^2 - x) - \cos(x^2 + x))\). Then  
\[
\int_\epsilon^B \cos(x^2 + x) dx = -\frac{\sin(x^2 + x)}{2x + 1} \bigg|_\epsilon^B - \int_\epsilon^B \frac{2 \sin(x^2 + x)}{(2x + 1)^2} dx
\]