Integral with Residue Theorem and Jordan's lemma

Question

I'm trying to solve this integral $\int_{-\infty}^{\infty} \frac{dx}{(1+x^2)^{n+1}}$ with the residue theorem Jordan's lemma, I found the single poles $z= \pm i$, so I tried: $$\operatorname{Res}(f(z),z=i) = \lim_{z\to\ i} \frac{1}{n!} \frac{\partial^n}{\partial z^n}\frac{(z-i)^{n+1}}{(z-i)^{n+1}(z+i)^{n+1}}$$ And my answer is $$\operatorname{Res}(f(z),z=i) = \lim_{z\to\ i} -\frac{1}{n!} \frac{(n+1)}{(z+i)^n}= -\frac{(n+1)}{n!(2i)^n} $$ therefore the integral: $$\int_{-\infty}^{\infty} \frac{dx}{(1+x^2)^{n+1}}=2\pi i \operatorname{Res}(f(z),z=i)=-\frac{2\pi i (n+1)}{n!(2i)^n}$$ I don't know if this is right, because because afterwards I have to use this result to prove: $$\text{Integral} =\pi\prod_{k=1}^{n} (2k-1)/ \prod_{k=1}^{n} 2k$$

Answer 1

If $f_n(z)=\dfrac1{(z+i)^{n+1}}$, then $\dfrac1{(z^2+1)^{n+1}}=\dfrac{f_n(z)}{(z-i)^{n+1}}$, and therefore\begin{align}\operatorname{res}\left(i,\frac1{(z^2+1)^{n+1}}\right)&=\frac{f_n^{(n)}(i)}{n!}\\&=(-1)^n\frac{(n+1)(n+2)\ldots(2n)}{n!(2i)^{2n+1}}\\&=-\frac{(n+1)(n+2)\ldots(2n)}{n!2^{2n+1}}i.\end{align}So,$$\int_{-\infty}^\infty\frac1{(x^2+1)^{n+1}}\,\mathrm dx=\pi\frac{(n+1)(n+2)\ldots(2n)}{n!2^{2n}}.$$And, on the other hand,\begin{align}\pi\frac{\prod_{k=1}^n(2k-1)}{\prod_{k=1}^n(2k)}&=\pi\frac{1\times3\times5\times\cdots\times(2n-1)}{2\times4\times6\times\cdots\times(2n)}\\&=\pi\frac{\bigl(1\times3\times5\times\cdots\times(2n-1)\bigr)\times\bigl(2\times4\times6\times\cdots\times(2n)\bigr)}{\bigl(2\times4\times6\times\cdots\times(2n)\bigr)^2}\\&=\pi\frac{(2n)!}{(n!2^n)^2}\\&=\frac{n!(n+1)(n+2)\ldots(2n)}{n!^22^{2n}}\\&=\pi\frac{(n+1)(n+2)\ldots(2n)}{n!2^{2n}}.\end{align}