Evaluate: limk→∞ e-k ∫R (ex - ey) / (x - y) dx dy, where R is the rectangle 0 ≤ x, y, ≤ k.
Solution
Answer: It diverges to plus infinity.
We use L'Hôpital's rule. Let f(k) = ∫R (ex - ey) / (x - y) dx dy, then the limit is the same as the limit of f'(k)/ek. To work out f'(k), note that f(k + δk) - f(k) = ∫δR (ex - ey) / (x - y) dx dy, where δR is the strip from x = 0 to x = k at y = k, width δk and the strip from y = 0 to k at x = k width δk. The integral over the two strips is obviously the same, so we get f'(k) = 2 ∫0k (ek - ex) / (k - x) dx. Hence f'(k)/ek = 2 ∫0k (1 - e-(k-x)) / (k - x) dx. Changing the integration variable to z = k - x, we get 2 ∫0k (1 - e-z) / z dz. The integrand is always positive and for z >= 1 is at least 1 - 1/e > 1/2. So the integral is at least ∫1k 1/z dz = ln k. This tends to infinity with k.
© John Scholes
jscholes@kalva.demon.co.uk
16 Jan 2001