Proving $I_{n+1,a}= \frac{2n-1}{2n} \cdot I_{n,a}$ given $I_{n,a}=\int_{-\infty} ^{\infty} \frac{dx}{(1+\frac{x^2}{a})^n}$

Question

For $n \geq 1$, $a>0$ denote: $$I_{n,a}=\int_{-\infty} ^\infty \frac{dx}{(1+\frac{x^2}{a})^n}$$

Prove that the improper integral $I_{n,a}$ converges and satisfies: $$I_{n+1,a}= \frac{2n-1}{2n} \cdot I_{n,a}$$

I managed to prove that $I_{n,a}$ converges but I'm having trouble with the second part.

Answer 1

hint

For the recursive formula,

write the numerator of $I_{n+1,a} $ as

$$1=1+\frac {x^2}{a}-\frac{x^2}{a} .$$

You will find that $$I_{n+1,a}=$$ $$=I_{n,a}-\frac {1}{2}\int_{-\infty}^{+\infty}x\frac {\frac {2x}{a}}{(1+\frac {x^2}{a})^{n+1}}dx $$ $$=I_{n,a}-\frac {1}{2}\int_{-\infty}^{+\infty}u (x)v'(x)dx $$

with $$u (x)=x $$ and $$v (x)=-\frac {1}{n}\frac {1}{(1+\frac {x^2}{a})^n}.$$

To finish, use by parts integration.