2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ do you want to use Dominated convergence theorem ? $\endgroup$ – Khosrotash Jul 22 '15 at 21:32
  • $\begingroup$ I do believe things are set up for a simple DCT argument. $\endgroup$ – zhw. Jul 22 '15 at 21:34
2
$\begingroup$

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: $$ \lim_{n\to +\infty}\int_{0}^{+\infty}\frac{e^{-x/n} dx}{1+(x-n)^2} = \color{red}{\frac{\pi}{e}}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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