### 41st Putnam 1980

**Problem A5**

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 ∫_{0}^{α} p(x) sin x dx = 0 and ∫_{0}^{α} p(x) cos x dx = 0.

**Solution**

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 ∫_{0}^{x} f(t) dt. Both p(x) and sin x have only finitely many zeros in any finite range |x| ≤ k', so ∫_{0}^{α} p(x) sin x dx has only finitely many zeros in the range |α| ≤ k'. Hence result.

41st Putnam 1980

© John Scholes

jscholes@kalva.demon.co.uk

10 Dec 1999

Last corrected/updated 21 Nov 02