44th Putnam 1983 Problem A6 Let T be the triangle with vertices (0, 0), (a, 0), and (0, a). Find lim a→∞ a 4 exp(-a 3 ) ∫ T exp(x 3 +y 3 ) dx dy. Solution
: 2/9. The corresponding indefinite integral is obviously intractable and the definite integral clearly diverges. But exp(a 3 )/a 4 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=0 a ∫ -s s exp(s 3 /4 + 3s t 2 /4) dt ds. Differentiating wrt a, we get 1/2 ∫ -a a exp( a 3 /4 + 3a/4 t 2 ) dt = 1/2 exp(a 3 /4) ∫ -a a exp( 3a/4 t 2 ) dt. Similarly, differentiating the denominator gives (3/a 2 - 4/a 5 ) exp(a 3 ). We can cancel out the exp(a 3 /4) to get 1/2 ∫ -a a exp( 3a/4 t 2 ) dt and (3/a 2 - 4/a 5 ) exp(3a 3 /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 = a 1/2 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 -1/2 ∫ 0 a√a exp(3s 2 /4) ds. We can cancel the a -1/2 . Now differentiating the numerator gives exp(3a 3 4) 3a 1/2 /2. Differentiating the denominator gives (27a 1/2 /4 + lower powers of a ) exp(3a 3 /4). This evaluates to (3/2)/(27/4) = 2/9 as a tends to infinity. 44th Putnam 1983 © John Scholes jscholes@kalva.demon.co.uk 16 Jan 2001