An Irresistible integral: $\int_0^\infty \frac{x^{2m}}{(ax^2+b)^n}\mathrm dx=\frac{\pi}{2a^mb^{n-m-1}\sqrt{ab}}\frac{(2m-1)!!(2n-2m-3)!!}{(2m-2)!!}$ According to the author of Irresistible Integrals,
$$I(n,m;a,b)=\int_0^\infty \frac{x^{2m}\mathrm dx}{(ax^2+b)^n}=\frac{\pi}{2a^mb^{n-m-1}\sqrt{ab}}\frac{(2m-1)!!(2n-2m-3)!!}{(2m-2)!!}$$
Which I am attempting to prove. My progress:
$$J(n;a,b)=I(n,0;a,b)=\int_0^\infty \frac{\mathrm dx}{(ax^2+b)^n}$$
As I have shown in other posts of mine, this integral satisfies the recurrence
$$J(n;a,b)=\frac{2n-3}{2bn}J(n-1;a,b)$$
And has the base case $$J(1;a,b)=\frac\pi{2\sqrt{ab}}$$
So 
$$J(n;a,b)=\frac{\pi}{2^{2n-1}b^{n-1}\sqrt{ab}}{2n-2\choose n-1}$$
Then I noticed that 
$$\frac{\partial}{\partial a}J(n;a,b)=-nI(n+1,1;a,b)$$
$$\frac{\partial^2}{\partial a^2}J(n;a,b)=n(n+1)I(n+2,2;a,b)$$
$$...$$
$$\frac{\partial^k}{\partial a^k}J(n;a,b)=(-1)^k(n)_kI(n+k,k;a,b)=\frac{\pi\sqrt\pi}{2^{2n-1}b^{n-1}a^{k}\sqrt{ab}\,\Gamma(\frac{1-2k}2)}{2n-2\choose n-1}$$
Here $(n)_k=\frac{\Gamma(n+k)}{\Gamma(n)}$. Yet I am confused as to how to proceed. Could I have some help?
 A: You may substitute $ax^2 + b = b/u$ to obtain
\begin{align*}
\int_{0}^{\infty} \frac{x^{2m}}{(ax^2+b)^n} \, \mathrm{d}x
&= \int_{0}^{1} \frac{\left( \frac{b}{a}(\frac{1}{u}-1) \right)^m}{\left(\frac{b}{u}\right)^n} \, \frac{\sqrt{b/a}}{2u^{3/2}(1-u)^{1/2}} \mathrm{d}u \\
&= \frac{(b/a)^{m+1/2}}{2 b^n} \int_{0}^{1} u^{n-m-3/2}(1-u)^{m-1/2} \, \mathrm{d}u \\
&= \frac{(b/a)^{m+1/2}}{2 b^n} \mathrm{B}\left(n-m-\tfrac{1}{2}, m + \tfrac{1}{2}\right).
\end{align*}
Hope the computation is correct as I hurried a bit when writing things down...
A: Here is an alternative method where I will make use of a so-called Schwinger parametrisation. Such a parametrisation is particularly suited to integrals where a polynomial raised to a power appears in the denominator of the integrand. 
For a positive, continuous function $\beta (x)$ observe that for $p > 0$
$$\frac{1}{\beta^p (x)} = \frac{1}{\Gamma (p)} \int_0^\infty u^{p - 1} e^{-u \beta(x)} \, du.$$
It is this observation that is known as the Schwinger parametrisation.
I will assume $a,b > 0$ and $m,n \in \mathbb{N}$ such that $n > m$. Choosing $\beta (x) = ax^2 + b$ and $p = n$ we have
$$\frac{1}{(ax^2 + b)^n} = \frac{1}{\Gamma (n)} \int_0^\infty u^{n - 1} e^{-u(ax^2 + b)} \,du.$$
The integral $I(m, n; a, b)$ can thus be rewritten as
\begin{align}
I(m,n; a,b) &= \frac{1}{\Gamma (n)} \int_0^\infty \int_0^\infty x^{2m} u^{n - 1} e^{-u(ax^2 + b)} \, du \, dx\\
&= \frac{1}{\Gamma (n)} \int_0^\infty u^{n - 1} e^{-ub} \int_0^\infty x^{2m} e^{-uax^2} \, dx \, du,
\end{align}
after the order of integration has been changed.
Enforcing a substitution of $x \mapsto \sqrt{\dfrac{x}{ua}}$ leads to
\begin{align}
I(m,n;a,b) &= \frac{1}{2 \Gamma (n) a^{m + 1/2}} \int_0^\infty u^{n - m - 3/2} e^{-ub} \int_0^\infty x^{m - 1/2} e^{-x} \, dx\\
&= \frac{\Gamma (m + 1/2)}{2 \Gamma (n) a^{m + 1/2}} \int_0^\infty u^{n - m -3/2} e^{-ub} \, du.
\end{align}
Next, enforcing a substitution of $u \mapsto u/b$ yields 
\begin{align}
I(m,n;a,b) &= \frac{\Gamma (m + 1/2)}{2 \Gamma (n) b^{n - m - 1/2} a^{m + 1/2}} \int_0^\infty u^{n - m -3/2} e^{-u} \, du\\
&= \frac{\Gamma (m + 1/2) \Gamma (n - m - 1/2)}{2 \Gamma (n) b^{n - m - 1/2} a^{m + 1/2}},
\end{align}
or in terms of the beta function
$$I(m,n;a,b) = \frac{1}{2b^n} \left (\frac{b}{a} \right )^{m + \frac{1}{2}} \operatorname{B} \left (n - m - \frac{1}{2}, m + \frac{1}{2} \right ),$$
in agreement with the result given by @Sangchul Lee. Granted, the substitution Sangchul Lee uses is pretty slick, and gets one to this point a lot quicker than making use of a Schwinger parametrisation, but it at least shows you an alternative approach to reaching the same point.
Note that in terms of central binomial coefficients the result can be expressed as
$$I(m,n; a, b) = \frac{n \pi}{(2n - 2m - 1) 2^{2n} a^m b^{n - m - 1} \sqrt{ab}} \frac{\binom{2m}{m} \binom{2n - 2m}{n - m}}{\binom{n}{m}}.$$
