Let p(x) be a polynomial with real coefficients of degree 1 or more. Show that there are only finitely many values α such that ∫0α p(x) sin x dx = 0 and ∫0α p(x) cos x dx = 0.
Note that either equation alone has infinitely many zeros.
Let d(x) = p(x) - p''(x) + p(4)(x) - ... . Then we can easily check that ∫ p(x) sin x dx = -d(x) cos x + d'(x) sin x, so the definite integral is -d(α) cos α + d'(α) + d(0). Similarly, the second definite integral is d(α) sin α + d'(α) cos α - d'(0). So any roots α of both equations also satisfy d(α) = d(0) cos α + d'(0) sin α (*). Let k = |d(0)| + |d'(0)|, then |rhs (*)| ≤ k. But d(x) has the same degree as p(x), so |d(x)| > k for all x > some k'. In other words all the roots of (*) must satisfy |α| ≤ k'. By Rolle's theorem, if f(x) has only finitely many zeros, then so does ∫0x f(t) dt. Both p(x) and sin x have only finitely many zeros in any finite range |x| ≤ k', so ∫0α p(x) sin x dx has only finitely many zeros in the range |α| ≤ k'. Hence result.
41st Putnam 1980
© John Scholes
10 Dec 1999
Last corrected/updated 21 Nov 02