TITLE: Evaluting a limit of integral
QUESTION [5 upvotes]: I am working on the following problem 
$$\lim_{n\rightarrow\infty} \sqrt{n}2^n\int_0^\infty x^n(1+x^2)^{-n} \ dx$$
I can rewrite the integral as $$\lim_{n\rightarrow\infty}\int_0^\infty \sqrt{n}\left(\frac{2x}{1+x^2}\right)^n dx$$ 
Next I want to use the Gaussian integral to evaluate this. I tried a substution of $u=\sqrt{n}x$ but didnt work well. Any other suggestions?

REPLY [4 votes]: Hoping that I did not make any mistake :  for $n>1$, $$\int_0^\infty x^n(1+x^2)^{-n} \ dx=\frac{\sqrt{\pi } 2^{-n} \Gamma \left(\frac{n-1}{2}\right)}{\Gamma
   \left(\frac{n}{2}\right)}$$ So, $$A= \sqrt{n}2^n\int_0^\infty x^n(1+x^2)^{-n} \ dx=\sqrt{\pi n}\frac{ \Gamma \left(\frac{n-1}{2}\right)}{\Gamma
   \left(\frac{n}{2}\right)}$$ For large values of $n$, asymptotic expansion of the $\Gamma$ function gives $$A=\sqrt{2 \pi }+\frac{3 \sqrt{\frac{\pi }{2}}}{2 n}+\frac{25 \sqrt{\frac{\pi }{2}}}{16
   n^2}+O\left(\left(\frac{1}{n}\right)^{5/2}\right)$$