Show that lim_{k→∞} ∫_{0}^{k} sin x sin x^{2} dx converges.

**Solution**

We integrate by parts to get ∫_{0}^{k} sin x sin x^{2} dx = - ∫_{0}^{k} (sin x)/2x d(cos x^{2}) = (cos x^{2} sin x)/2x |_{k}^{0} + ∫_{0}^{k} cos x^{2} ( (cos x)/2x - (sin x)/2x^{2} ) dx = term 1 + term 2 - term 3.

term 1 obviously converges to 1/2.

For term 3 the integral of the absolute value is less than ∫_{0}^{k} 1/2x^{2} dx, which converges.

For term 2, we integrate by parts again to get ∫_{0}^{k} (cos x)/4x^{2} d(sin x^{2}) dx = (sin x^{2} cos x)/4x^{2} |_{0}^{k} + ∫_{0}^{k} (sin x^{2} sin x)/4x^{2} dx + ∫_{0}^{k} (sin x^{2} cos x)/8x^{3} dx.

The first term obviously tends to zero and the second two converge, by comparison of the absolute value of the integrands to 1/x^{2} and 1/x^{3}.

© John Scholes

jscholes@kalva.demon.co.uk

1 Jan 2001