Limit of integral over $[0,1]$ of $\frac{ne^{-x}}{1+nx}$

Question

I need some help to calculate the following limit (in measure theory):

$$\lim_{n \to \infty} \int_{0}^{1} \frac{ne^{-x}}{1+nx}dx$$

My first idea was to use either the monotone convergence theorem or the dominated convergence theorem. So before trying anything, I just tried to take the limit of the term in the integral and integrate it, like follows:

$$\lim_{n \to \infty} \frac{ne^{-x}}{1+nx} = \frac{e^{-x}}{x}$$

$$\int_{0}^{1} \frac{e^{-x}}{x}dx$$

And that's where I got stuck. I checked in Wolfram Alpha for the result, so I may have an idea on how to proceed, but it says:

$$\int \frac{e^{-x}}{x}dx = Ei(-x)$$

which doesn't help me. So I guess that's not the way to proceed. I thought that maybe I need to upper and lower bound it by something that converges to the same value, but even then I have to idea how. Any hints?

Answer 1

Thanks to John Dawkins' hint I've got the answer:

$$\frac{ne^{-1}}{nx+1} \leq \frac{ne^{-x}}{nx+1}$$

and

$$\lim_{n \to \infty} \uparrow \frac{ne^{-1}}{nx+1} = \frac{e^{-1}}{x}$$

$$\int_{0}^{1} \frac{e^{-1}}{x}dx = +\infty$$

$$\Longrightarrow \int_{0}^{1} \lim_{n \to \infty} \frac{ne^{-x}}{nx+1}dx = \int_{0}^{1} \frac{e^{-x}}{x}dx = +\infty$$

Then either use the Fatou lemma or the monotone convergence theorem.

Answer 2

All we need is to recognize that

$$\int_0^1\frac{ne^{-x}}{1+nx}\,dx\ge \int_0^1\frac{n/e}{1+nx}\,dx=(n/e)\log(1+n)\to \infty$$

And we are done.