How to get $\int_0^∞\frac{dx}{(x^2+1)^n}=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}\overset{?}=\frac{\sqrt{\pi}}{2}\frac{Γ(n-\frac{1}{2})}{Γ(n)}$? I recently solved the following integral using a recursive formula and integration by parts.
$$\int_0^\infty\frac{dx}{(x^2+1)^n}=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}$$
Where $(n)!!$ represents the double factorial, not to be confused with $(n!)!$.
But when I plug this same integral into WolframAlpha I get this:
$$\int_0^\infty\frac{dx}{(x^2+1)^n}=\frac{\sqrt{\pi}}{2}\frac{\Gamma(n-\frac{1}{2})}{\Gamma(n)}$$
How do I prove that these two results are equivalent?
 A: Recursion gives
$$\newcommand{\g}[1]{\Gamma \left( #1 \right)}
\newcommand{\para}[1]{\left( #1 \right)}
\begin{align*}
\g{ n - \frac 1 2 } &= \para{ n -1 - \frac 1 2 }\cdot   \g{(n-1) - \frac 1 2}\\
\g{ n - \frac 1 2 } &= \para{ n -1 - \frac 1 2 }\para{ n -2 - \frac 1 2 }\cdot  \g{(n-2) - \frac 1 2}\\
&= \vdots \\
&=\para{ n -1 - \frac 1 2 }\para{ n -2 - \frac 1 2 }\cdot \cdots \cdot \para{ \frac 3 2 } \para{ \frac 1 2 }\cdot  \g{ \frac 1 2 } \\
&=\para{ \frac{2n-3}{2} }\para{ \frac{2n-5}{2} }\para{ \frac{2n-7}{2} } \cdots\para{ \frac{1}{2} } \sqrt \pi \\
&= \frac{(2n-3)!!}{2^{n-1}} \sqrt \pi
\end{align*}$$
Of course,
$$\g{n} = (n-1)!$$
Then their ratio is
$$\frac{\g{n-1/2}}{\g{n}} = \frac{\sqrt \pi}{2^{n-1}} \frac{(2n-3)!!}{(n-1)!}$$
By a property of the double factorial,
$$2^{n-1}(n-1)! = (2n-2)!!$$
so
$$\frac{\g{n-1/2}}{\g{n}} = \sqrt{\pi} \frac{(2n-3)!!}{(2n-2)!!}$$
The desired result follows.
A: It is not difficult to show that the product of the even integers from $ \ (2n - 2) \ $ down to $ \ 2 \ $ is
$$ (2n - 2)·(2n - 4)· \ldots · 2 \ \ = \ \ 2^{n - 1} · \Gamma(n) \ \ . $$
Starting with $ \ \Gamma\left(\frac12\right) \ = \ \sqrt{\pi} \ \ , $ and the recursion to obtain larger half-integer values of the gamma function, we can then find the product of the odd integers from $ \ (2n - 1) \ $ down to $ \ 1 \ $ as
$$ (2n - 1)·(2n - 3)· \ldots · 1 \ \ = \ \ \frac{2^{n}}{\sqrt{\pi}} · \Gamma\left(n+\frac12\right) \ \ . $$
You'll need just a little modification to produce the relation you're looking for.
ADDENDUM --
It might be mentioned that you may also see this result expressed as
$$\int_0^\infty\frac{dx}{(x^2+1)^n}=\frac12· \mathbf{B} \left(n - \frac12 \ , \ \frac12 \right) \ \ , $$
with the beta function $ \ \mathbf{B}(x \ , \ y) \ = \ \frac{\Gamma(x)·\Gamma(y)}{\Gamma(x + y)} \ \ $ being used.  Such ratios of gamma functions, in connection with the beta function, turn up in quite a few improper and other definite integrals; see, as just one example, Compute the integral $\displaystyle{\int_0^1 \dfrac{x^a}{\sqrt{1-x}} dx}$ .
