Evaluate: lim_{k→∞} e^{-k} ∫_{R} (e^{x} - e^{y}) / (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} (e^{x} - e^{y}) / (x - y) dx dy, then the limit is the same as the limit of f'(k)/e^{k}. To work out f'(k), note that f(k + δk) - f(k) = ∫_{δR} (e^{x} - e^{y}) / (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 ∫_{0}^{k} (e^{k} - e^{x}) / (k - x) dx. Hence f'(k)/e^{k} = 2 ∫_{0}^{k} (1 - e^{-(k-x)}) / (k - x) dx. Changing the integration variable to z = k - x, we get 2 ∫_{0}^{k} (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 ∫_{1}^{k} 1/z dz = ln k. This tends to infinity with k.

© John Scholes

jscholes@kalva.demon.co.uk

16 Jan 2001