Compute $\lim_{n\to\infty} \int_0^{\infty}\frac{e^{\frac{-x}{n}}}{1+(x-n)^2}dx$

Question

Compute $\lim_{n\to\infty} \int_0^{\infty}\frac{e^{\frac{-x}{n}}}{1+(x-n)^2}dx$. I'm considering to use Dominated convergence theorem but the hint of this problem is the limit is non-zero. Any hints will be appreciated. Thank you.

Answer 1

Hint: If you let $x=y+n,$ you get

$$\int_{-n}^\infty \frac{\exp (-(y+n)/n)}{1+y^2}dy=\frac{1}{e}\int_{-n}^\infty \frac{\exp (-y/n)}{1+y^2}dy$$

Answer 2

We have: $$ \int_{0}^{+\infty}\frac{e^{-x/n} dx}{1+(x-n)^2}=\int_{0}^{+\infty}\frac{e^{-x}\,dx}{\frac{1}{n}+n(x-1)^2}$$ but since: $$ \int_{0}^{+\infty}\frac{dx}{1+(x-n)^2} = \frac{\pi}{2}+\arctan(n)=\pi-O\left(\frac{1}{n}\right) $$ we have that $\frac{1}{\frac{1}{n}+n(x-1)^2}$ converges in distribution to $\pi\cdot\delta(x-1)$ and: $$ \lim_{n\to +\infty}\int_{0}^{+\infty}\frac{e^{-x/n} dx}{1+(x-n)^2} = \color{red}{\frac{\pi}{e}}.$$