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.
Answer: (not required) 11131110/107634259.
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.]
© John Scholes
12 Dec 1998