For $n\ge1,a>0$ we define: $$I_{n,a}=\int_{-\infty}^{\infty}\frac{dx}{(1+x^2/a)^n}$$
This question comes in two parts:
Proving that this integral converges, and that: $$I_{n+1,a}=\frac{2n+1}{2n}I_{n,a}$$
and the second part is proving that $$\lim_{n\to \infty}I_{n,n}=\int_{-\infty}^{\infty}e^{-x^2}dx$$
Now, using integration by parts gave me this: $$I_{n,a}=\int_{-\infty}^{\infty}\frac{\frac{2x^2}{a}n}{(1+x^2/a)^{n+1}}dx$$ which is where I got stuck.
For the second part it is easy to see that $\lim_{n\to \infty}\frac{1}{(1+x^2/n)^n}=e^{-x^2}$ but how do I justify the limit of the whole integral?
Thanks.
You have found that partial integration yields the result:
$$I_{n,a}=\int_{-\infty}^{\infty}\frac{\frac{2x^2}{a}n}{(1+x^2/a)^{n+1}}dx$$
You can then write this as:
$$I_{n,a}=2n\int_{-\infty}^{\infty}\frac{\frac{x^2}{a}+1-1}{(1+x^2/a)^{n+1}}dx=2n\left(I_{n,a}-I_{n+1,a}\right)$$
Solving for $I_{n+1,a}$ yields:
$$I_{n+1,a} = \frac{2n-1}{2n} I_{n,a}$$
Then to prove that you can take the limit inside the integral, you can use theorems such as the dominated convergence theorem. Since by the binomial theorem:
$$\left(1+\frac{x^2}{n}\right)^n = 1 + x^2 + \frac{n-1}{2} x^4\cdots \geq1+x^2 $$
we have:
$$f_n(x) = \frac{1}{\left(1+\frac{x^2}{n}\right)^n}\leq \frac{1}{1+x^2}$$
The condition in Lebesgue's Dominated Convergence Theorem that $f_n(x)$ be dominated by an integrable function is then met, therefore the limit of the integral of $f_n(x)$ is given by the integral of the limit function.