(n-1) derivative in residue theorem $$
\int _0^{+\infty }\frac{\mathrm{d}x}{\left(x^2+a^2\right)^n}
$$
This integral has been addressed for $n = 2$, let's attempt to solve it for all $n$.
The residue theorem has led me to:
$$
\frac{2\pi i}{\left(n-1\right)!}\frac{\mathrm{d}^{\:n-1}}{\mathrm{d}z^{\:n-1}}\frac{1}{\left(z + \mathrm{i}a\right)^{\,n}}
$$
I could use some help with that derivative, thank you.
 A: To Compute Your Derivative
Apply $\frac{\mathrm{d}}{\mathrm{d}z}(z+ia)^k=k(z+ia)^{k-1}$, iteratively, $n-1$ times ($k=-n$ to $k=-2n+2$): 
$$
\begin{align}
\frac1{(n-1)!}\frac{\mathrm{d}^{n-1}}{\mathrm{d}z^{n-1}} (z+ia)^{-n}
&=(-1)^{n-1}\frac1{(n-1)!}n(n+1)(n+2)\cdots(2n-2)(z+ia)^{-2n+1}\\
&=(-1)^{n-1}\binom{2n-2}{n-1}(z+ia)^{-2n+1}
\end{align}
$$
Now plug in $z=ia$.

A Different Approach
Assume $a\gt0$. Using the binomial theorem, we can compute the residue at $z=ia$:
$$
\begin{align}
\left[(z-ia)^{-1}\right]\frac1{(z-ia)^n(z+ia)^n}
&=\left[(z-ia)^{-1}\right](z-ia)^{-n}{\underbrace{(z-ia+2ia)}_{z+ia}}^{-n}\tag1\\
&=\left[(z-ia)^{n-1}\right](z-ia+2ia)^{-n}\tag2\\[6pt]
&=(2ia)^{-n}\left[(z-ia)^{n-1}\right]\left(1+\frac{z-ia}{2ia}\right)^{-n}\tag3\\[3pt]
&=(2ia)^{-n}\binom{-n}{n-1}(2ia)^{1-n}\tag4\\
&=(2ia)^{1-2n}(-1)^{n-1}\binom{2n-2}{n-1}\tag5\\
&=-i\frac1{(2a)^{2n-1}}\binom{2n-2}{n-1}\tag6
\end{align}
$$
Explanation:
$(1)$: write $z+ia=z-ia+2ia$
$(2)$: $\left[(z-ia)^{-1}\right](z-ia)^{-n}f(z)=\left[(z-ia)^{n-1}\right]f(z)$
$(3)$: pull $(2ia)^{-n}$ out front
$(4)$: apply the "coefficient of" operator to the binomial expansion
$(5)$: compute the negative binomial coefficient
$(6)$: collect terms
Since the integrand is even, we take half the integral over the whole real line, which is $\pi i$ times the residue at $z=ia$.
$$
\int_0^\infty\frac{\mathrm{d}x}{\left(x^2+a^2\right)^n}=\frac{\pi}{(2a)^{2n-1}}\binom{2n-2}{n-1}
$$
A: I would start by writing the following:
$$\frac{d^{n-1}}{dz^{n-1}} \frac{1}{(z+ia)^n} = \frac{d^{n-1}}{dz^{n-1}} (z+ia)^{-n} $$
Then each derivative uses the fact that $\frac{d}{dz}(u^\alpha ) = \alpha u^{\alpha-1} \frac{d}{dz}(u) $, and here $u = z+ia \Rightarrow \frac{d}{dz}u = 1$.
I think with this you can solve it, if I didn't miss anything.
