Calculate $k\int_{-\infty}^{\infty}\frac{x^r}{(1+x^2)^m}dx$ Calculate $$\mu_r = k\int_{-\infty}^{\infty}\frac{x^r}{(1+x^2)^m}dx$$
What I have tried:
For $m>1$, making the substitution $z = \frac{1}{1+x^2} \implies x = z^{-\frac{1}{2}}(1-z)^{\frac{1}{2}} \implies dx = -\frac{1}{2z^{\frac{3}{2}}(1-z)^{\frac{1}{2}}}$
Plugging these in
$$ \begin{align} \mu_{2r} = k\int_{-\infty}^{\infty}\left(\frac{x^{r/m}}{1+x^2}\right)^mdx &= k\int_0^{1}\left(z^{-\frac{r}{2m}+m}(1-z)^{\frac{2}{2m}}\right)^m\left(-\frac{1}{2z^{\frac{3}{2}}(1-z^{\frac{1}{2}})}\right)dz \\ &= -k\int_0^1\frac{1}{2}(1-z)^{\frac{r}{2}-\frac{1}{2}}z^{m-\frac{r}{2}-\frac{3}{2}}dz \end{align}$$
However, when working through the book it shows instead
$$k\int_0^1z^{m-r-3/2}(1-z)^{r-1/2}dz$$.
How do I get it in that form but also where do the integral bounds $\int_0^1$ come from? I understand that the $r/2$ cancels when we have $2r$ which gets the bit inside the integral. However, I do not know how to rid $-1/2$, I'm guessing it had something to do with the integral bounds.
I know that I can do
$$\int_{\infty}^{\infty} = \int_{-\infty}^1+\int^\infty_1$$
and that $\frac{1}{1+\infty^2} \to 0$
So something like
$$-2\int^0_1=2\int_0^1 \implies k\int_0^1z^{m-r-3/2}(1-z)^{r-1/2}dz$$
Which would remove $-1/2$. However, what's stopping me from using $\int_{-\infty}^1$? Does it become an odd function with this integral, and hence equal to 0?
It becomes a beta distribution with $B(m-r-1/2, r+1/2)$, where $k$ is the result from this question of mine
 A: If $r\not\in\mathbb{Z}$, $x^r$ is not well-defined for $x\lt0$. If $r$ is odd, then by symmetry the integral is $0$. So, let us assume that $r\in2\mathbb{Z}$.
$$
\begin{align}
k\int_{-\infty}^\infty\frac{x^r}{\left(1+x^2\right)^m}\,\mathrm{d}x
&=2k\int_0^\infty\frac{x^r}{\left(1+x^2\right)^m}\,\mathrm{d}x\tag{1a}\\
&=k\int_0^\infty\frac{x^{\frac{r-1}2}}{(1+x)^m}\,\mathrm{d}x\tag{1b}\\[3pt]
&=k\frac{\Gamma\!\left(\frac{r+1}2\right)\Gamma\!\left(m-\frac{r+1}2\right)}{\Gamma(m)}\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: the integrand is even, so we can fold the domain
$\text{(1b)}$: substitute $x\mapsto x^{1/2}$
$\text{(1c)}$: apply the Beta integral
Another approach to step $(3)$ using $u=\frac x{1+x}$ and $x=\frac u{1-u}$ and $\mathrm{d}x=\frac{\mathrm{d}u}{(1-u)^2}$. For $x\in[0,\infty)$ we have $u\in[0,1)$.
$$
\begin{align}
k\int_0^\infty\frac{x^{\frac{r-1}2}}{(1+x)^m}\,\mathrm{d}x
&=k\int_0^1\frac{u^{\frac{r-1}2}}{(1-u)^{\frac{r-1}2}}(1-u)^{m-2}\,\mathrm{d}u\tag{2a}\\
&=k\int_0^1u^{\frac{r+1}2-1}(1-u)^{m-\frac{r+1}2-1}\,\mathrm{d}u\tag{2b}\\
&=k\frac{\Gamma\!\left(\frac{r+1}2\right)\Gamma\!\left(m-\frac{r+1}2\right)}{\Gamma(m)}\tag{2c}
\end{align}
$$
Explanation:
$\text{(2a)}$: apply the substitution given
$\text{(2b)}$: simplify
$\text{(2c)}$: apply the Beta integral
If you restrict the domain of integration, the restriction on $r$ can be relaxed to $r\in\mathbb{R}$ and $r\gt-1$, and dividing the integral given above by $2$.
