44th Putnam 1983

Problem A6

Let T be the triangle with vertices (0, 0), (a, 0), and (0, a). Find lima→∞ a4exp(-a3) ∫T exp(x3+y3) dx dy.



Answer: 2/9.

The corresponding indefinite integral is obviously intractable and the definite integral clearly diverges. But exp(a3)/a4 also diverges. So we have the ratio of two divergent quantities. This calls for l'Hôpital's rule: to evaluate lim f(a)/g(a) we take lim f'(a)/g'(a). The first step is to rearrange the integral so that a only occurs as an integration limit for one of the variables (thus making it easier to differentiate with respect to it).

After a little experimentation we take s = x + y, t = x - y. The Jacobian is 1/2 and so we get 1/2 ∫s=0a-ss exp(s3/4 + 3s t2/4) dt ds. Differentiating wrt a, we get 1/2 ∫-aa exp( a3/4 + 3a/4 t2) dt = 1/2 exp(a3/4) ∫-aa exp( 3a/4 t2) dt. Similarly, differentiating the denominator gives (3/a2 - 4/a5) exp(a3). We can cancel out the exp(a3/4) to get 1/2 ∫-aa exp( 3a/4 t2) dt and (3/a2 - 4/a5) exp(3a3/4), but both these still diverge. Accordingly, we must apply the rule again.

We would like to eliminate the a in the integrand to make differentiation simpler. This can be achieved by setting s = a1/2 t. Notice that the integrand has the same value for t and -t (or s and -s) so we can further simplify by taking the integration from 0 to a and doubling. Thus we get for the numerator a-1/20a√a exp(3s2/4) ds. We can cancel the a-1/2. Now differentiating the numerator gives exp(3a34) 3a1/2/2. Differentiating the denominator gives (27a1/2/4 + lower powers of a ) exp(3a3/4). This evaluates to (3/2)/(27/4) = 2/9 as a tends to infinity.



44th Putnam 1983

© John Scholes
16 Jan 2001