Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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
add comment

1 Answer 1

up vote 7 down vote accepted

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

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.