Substituting into an integral to create a Beta function

Question

My probability textbook has a solution that turns a certain density function into something more familiar. Starting with,

$\int_{-\infty}^{\infty} \frac{dx}{(1+x^2)^m}$

We substitute in:

$v = (1 + x^2)^{-1}$

to get:

$\int_{-\infty}^{\infty} \frac{dx}{(1+x^2)^m}$ = $\int_{0}^{1} v^{m-\frac{3}{2}}(1-v)^{-\frac{1}{2}} dv$

I'm not quite sure how they got from the left side of that to the right...

Answer 1

First note that $$ \int_{-\infty}^{\infty}(1+x^2)^{-m}\;dx=2\int_0^{\infty}(1+x^2)^{-m}\;dx$$ since the integrand is even, so we can assume that $x\geq 0$. Then if $v=\frac{1}{1+x^2}$ then $$ x=\sqrt{\frac{1}{v}-1}=\frac{(1-v)^{\frac{1}{2}}}{v^{\frac{1}{2}}} $$ and $$ \frac{dv}{dx}=-\frac{2x}{(1+x^2)^2}=-2\cdot\frac{(1-v)^{\frac{1}{2}}}{v^{\frac{1}{2}}}\cdot v^2=-2(1-v)^{\frac{1}{2}}v^{\frac{3}{2}} $$

Moreover, if $x=0$ then $v=1$, while $v\to0$ as $x\to\infty$. Therefore $$ 2\int_0^{\infty}(1+x^2)^{-m}\;dx=\int_0^12v^m\cdot[2(1-v)^{\frac{1}{2}}v^{\frac{3}{2}}]^{-1}\;dv=\int_0^1v^{m-\frac{3}{2}}(1-v)^{-\frac{1}{2}}\;dv$$ as claimed.

Answer 2

You have $$ \int_{-\infty}^{\infty} \frac{dx}{(1+x^2)} = 2 \int_{-\infty}^0 \frac{dx}{(1+x^2)}$$

using the substitution, $x=- \frac{(1-v)^{1/2}}{v^{1/2}}$ and $dv=\frac{-2x}{(1+x^2)^2}dx$.

From here, the equality follows.