61st Putnam 2000

Problem A4

Show that limk→∞0k sin x sin x2 dx converges.



We integrate by parts to get ∫0k sin x sin x2 dx = - ∫0k (sin x)/2x d(cos x2) = (cos x2 sin x)/2x |k0 + ∫0k cos x2 ( (cos x)/2x - (sin x)/2x2 ) 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 ∫0k 1/2x2 dx, which converges.

For term 2, we integrate by parts again to get ∫0k (cos x)/4x2 d(sin x2) dx = (sin x2 cos x)/4x2 |0k + ∫0k (sin x2 sin x)/4x2 dx + ∫0k (sin x2 cos x)/8x3 dx.

The first term obviously tends to zero and the second two converge, by comparison of the absolute value of the integrands to 1/x2 and 1/x3.



61st Putnam 2000

© John Scholes
1 Jan 2001