1981 Putnam problem A-3 : Differentiating under a double integral?

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.

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G(t) is an integral over a square [0, t] x [0, t]. So G(t+dt) is an integral over a slightly larger square, which can be decomposed into the original square plus two strips of width dt and a square of area dt^2 which can be neglected in the limit. The expression for G'(t) is the sum of the integrals over those strips, divided by dt, in the limit as dt approaches zero. – Qiaochu Yuan Jan 3 '11 at 21:59
Thanks Qiaochu. I think I understand now. – AnonymousCoward Jan 3 '11 at 22:05
You might want to use $$..math...$$ instead of $\displaystyle...math...$. The double dollar sign puts you in displaystyle, and it renders better. It also means you don't have to leave the blank lines to interrupt a paragraph. – Arturo Magidin Jan 3 '11 at 22:20

Let $$F(y,t)=\int_0^t \frac{e^x-e^y}{x-y} dx.$$ Then your $G(t)$ is $$G(t)=\int_0^t F(y,t) dy.$$
Then $$G'(t)=F(t,t)+\int_0^t \frac{\partial F}{\partial t} (y,t) dy$$ $$=\int_0^t \frac{e^x-e^t}{x-t} dx + \int_0^t\frac{e^t -e^y}{t-y} dy .$$