I was trying some old problems and got stuck on this one. Then when I looked at the answer there was a step I could not understand. Perhaps you can explain it to me.
A-3 Find
$ \displaystyle \lim_{t \to \infty}\left[ e^{-t}\int_0^t \int_0^t \frac{e^x - e^y}{x - y}dx dy \right]$
or show that the limit does not exist.
Solution Let $G(t)$ be the double integral. Then \lim\limits_{t \to \infty}\frac{G(t)}{e^t} = \lim\limits_{t \to \infty}\frac{G'(t)}{e^t} by L'Hopital. Then G'(t) = \int_0^t \frac{e^x-e^t}{x-t}dx+ \int_0^t \frac{e^y-e^t}{y-t}dy so G'(t) = 2\int_0^t \frac{e^x-e^t}{x-t}dx.
Then, the answer continues to show that \lim_{t \to \infty}\frac{G'(t)}{e^t}=\infty since, \frac{G'(t)}{2e^t} = \int_0^t \frac{e^{x-t}-1}{x-t}dx = \int_0^t\frac{1-e^{-y}}{y}dy > \int_1^t\frac{1-e^{-y}}{y}dy > \left(1-e^{-1}\right)log\,t. My question is how G'(t) was found. I understand the rest of the solution. I understand differentiating under the integral in the one dimensional case, but I do not understand how it works in the case of a double integral (which I assume is what is being done here), and I couldn't produce the answer's result.