Difficult limit $\lim_{x\to\infty} e^{-x}\int_0^x \int_0^x \frac{e^u-e^v}{u-v} \,\mathrm{d}u\mathrm{d}v$ I need to calculate this limit:
$$\lim_{x\to\infty} e^{-x}\int_0^x \int_0^x \frac{e^u-e^v}{u-v} \,\mathrm{d}u\mathrm{d}v.$$
How do I do it?
There's a hint that I should use de l'Hospital's rule.
 A: The integral over the region $0\leq u\leq v\leq x$ equals the integral over the region $0\leq v\leq u\leq x$ by symmetry, hence the integral can be written as:
$$ \iint_{(0,x)^2}\frac{e^u-e^v}{u-v}\,du\,dv = 2\int_{0}^{x}\int_{0}^{a}\frac{e^a-e^b}{a-b}\,db\,da = 2\int_{0}^{x}\int_{0}^{1}\frac{e^{a}-e^{ca}}{1-c}\,dc\,da$$
or as:
$$ 2\int_{0}^{1}\frac{(1-e^{ux})+u(e^x-1)}{u(1-u)}\,du =2\int_{0}^{1}\frac{1+e^x-e^{ux}-e^{(1-u)x}}{u}\,du.$$
The last integral can be written in terms of the incomplete $\Gamma$ function, and by standard asymptotics we get
$$ 2e^{-x}\int_{0}^{1}\frac{1+e^x-e^{ux}-e^{(1-u)x}}{u}\,du \gg \log(x) $$
from which it follows that the wanted limit is $\color{red}{\large +\infty}$. The same can be deduced from Frullani's integral and the dominated convergence theorem.
A: Expand the exponentials in their standard Taylor series to see the integrand equals
$$\sum_{n=0}^{\infty}\frac{1}{n!}\frac{u^n-v^n}{u-v}= \sum_{n=1}^{\infty}\frac{1}{n!}(u^{n-1} + u^{n-2}v + \cdots + uv^{n-2} + v^{n-1}).$$
For each $n$ we have
$$\int_0^x\int_0^x (u^{n-1} + u^{n-2}v + \cdots + uv^{n-2} + v^{n-1})\,dv\,du = x^{n+1}\left ( \frac{1}{n\cdot 1} + \frac{1}{(n-1)\cdot 2} + \frac{1}{(n-2)\cdot 3} +\cdots + \frac{1}{2\cdot (n-1)} + \frac{1}{1\cdot n}\right) \ge x^{n+1}\frac{1}{n}\left ( \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n}\right ) \ge x^{n+1}\frac{\ln n}{n}.$$
Thus the full expression of interest is at least
$$e^{-x}\sum_{n=1}^{\infty} \frac{\ln n}{n\cdot n!}x^{n+1} \ge e^{-x}\sum_{n=1}^{\infty} \ln n\frac{x^{n+1}}{(n+1)!}.$$
If we had just $e^{-x}\sum_{n=1}^{\infty} \frac{x^{n+1}}{(n+1)!}$ on the right, the limit on the right would be $1.$ But we have the extra $\ln n$ in front of the standard coefficients, hence the limit on the right is $\infty.$ Therefore the limit on the left is $\infty$ and we're done.
