I am in the middle of a problem and I have been working on this integral for a while: $$ \int_0^\infty \frac{x^{2r}}{(1+x^2)^m}\cdot \frac{\Gamma(m)}{\Gamma(1/2) \Gamma(m-1/2)}dx $$ Which I identified the integrand on the right to be the beta function $B(\frac{1}{2},m-\frac{1}{2})=\frac{\Gamma(1/2) \Gamma(m-1/2)}{\Gamma(m)}$.
\begin{align} \frac{1}{B(\frac{1}{2},m-\frac{1}{2})}&\int_0^\infty \frac{x^{2r}}{(1+x^2)^m}dx \\ = \frac{1}{B(\frac{1}{2},m-\frac{1}{2})}&\int^\infty_0 x^r(1-x)^{-m}\cdot x^r(1+x)^{-m} dx \end{align}
In the bottom integral the integrand $x^r(1-x)^{-m}$ is the integrand for the beta function. Anyways, I cannot figure out how to proceed past this, but I am sure I just need to put it in some form of the beta function. Also in this problem, I am supposed to discover that the integral only exists when $2r < 2m -1$.
$$\int_0^{\infty} \dfrac{x^{2r}}{(1+x^2)^{m}} dx = \int_0^\infty \dfrac{t^{r-1/2}}{(1+t)^m} dt = \int_1^{\infty}(z-1)^{r-1/2} \dfrac{1}{z^m} dz$$
Now use $z = \dfrac{1}{x}$