### Putnam 1993

Problem A5

Let U be the set formed as the union of three open intervals, U = (-100, -10) ∪ (1/101, 1/11) ∪ (101/100, 11/10). Show that ∫U (x2 - x)2/(x3 - 3x + 1)2 dx is rational.

Solution

Moderately hard.

It is not obvious how to integrate the integrand in closed form. So If we sketch the integrand we find that it is positive and close to zero except for poles near -2, 1/2 and 1 1/2. So poles fall between the three ranges of integration. However, it still seems odd that we are given a sum of three ranges. So we look for a substitution that might bring the ranges into coincidence or something like it. A little tinkering suggests y = 1/(1-x). This takes (1/101, 1/11) to (101/100, 11/10), and (101/100, 11/10) to (-100, -10). We also find that it takes (x2 - x)/(x3 - 3x + 1) to (y2 - y)/(y3 - 3y + 1).

Let I(a, b) = ∫ab (x2 - x)2/(x3 - 3x + 1)2 dx. Then we get I(1/101, 1/11) = ∫101/10011/10 (y2 - y)2/(y3- 3y + 1)2 dy/y2. Applying the transformation again gives that I(-100, -10) + I(1/101, 1/11) + I(101/100, 11/10) = ∫-100-10 (z2 - z)2/(z3- 3z + 1)2 (1 + (2z2 - 2z + 1)/( (z - 1)2z2) ) dz = ∫-10-100 (z2 - z + 1)2/(z3 - 3z + 1)2 dz.

At this point we are in a difficulty. It is still not obvious how to evaluate the integral explicitly. But at the same time, there is no other obvious approach. If it does have a closed form, then (az2 + bz + c)/(z3 - 3z + 1) looks a good candidate. Differentiating this, we find that it does in fact work for a = -1, b = 1, c = 0. So the value is K(-10) - K(-100), where K(z) = (z2 - z)/(z3 - 3z + 1). This is obviously rational. [In fact it evaluates to the answer given above with a little more effort.]