Solve the integral $ \int_{-\infty}^{\infty} \frac{\cos(ux)}{(x^{2}+a^{2})^{s}} dx$ Is there any way to compute in closed form (in terms of known functions) the Fourier integral 
$$ \int_{-\infty}^{\infty}  \frac{\cos(ux)}{(x^{2}+a^{2})^{s}} dx$$
where $u$ and $a$ are real positive numbers.
Of course I could evaluate it with residue theorem considering the s-th derivative but i would like to know if exact analytic methods or asymptotic methos are available.
For $s=1$ I know how to do it but i am looking for a more general method, thanks.
 A: Exact solution using residues:
Note that this is the same as the integral of $e^{iuz}/(z^2+a^2)^s$ along the real axis. For $u$ real and positive, $e^{iuz}$ decays as $\mathrm{Im}(z) \to \infty$, so we can deform the contour of integration so that it encircles the single pole in the upper half plane at $z = ia$. (Here I am assuming that $s$ is a positive integer. For other values of $s$ there will be a branch cut and things are more delicate.). Call this contour $\Gamma$. Then we wish to evaluate
$$
\oint_\Gamma \frac{e^{iuz}dz}{(z^2-a^2)^s} 
= \oint_\Gamma \frac{e^{iuz}dz}{(z-ia)^s(z+ia)^s}
= \oint_\Gamma \frac{h(z)dz}{(z-ia)^s}
$$
where $h(z) = e^{iuz}(z+ia)^{-s}$ is holomorphic on the upper half plane. Using the Cauchy integral formula, we find that this integral is equal to
$$
\frac{2\pi i}{(s-1)!} h^{(s-1)}(ia)
$$
If this is not explicit enough, note that
$$
\frac{d^k e^{iuz}}{dz^k}=(iu)^k e^{iuz}
$$
and
\begin{align}
\frac{d^k (z+ia)^{-s}}{dz^k}
&= (-s)(-s+1)\cdots(-s+k-1) (z+ia)^{-s-k}\\
&= (-1)^k \frac{s!}{(s-k)!} (z+ia)^{-s-k}
\end{align}
Plugging these into the generalized Leibniz rule, we have
$$
\frac{d^{s-1} h(z)}{dz^{s-1}} 
= \sum_{k=0}^{s-1} {s-1 \choose k} (-1)^k (iu)^{s-1-k} e^{iuz} \frac{s!}{(s-k)!} (z+ia)^{-s-k}
$$
So putting it altogether, the value of the integral is
$$
\frac{2\pi i}{(s-1)!} e^{-au} \sum_{k=0}^{s-1} {s-1 \choose k} (-1)^k (iu)^{s-1-k} \frac{s!}{(s-k)!} (2ia)^{-s-k}
$$
You could simplify this expression a bit but it is already explicit enough to get it for small values of $s$. For example,
plugging in $s = 1$  we get $\pi e^{-au}/a$. For $s = 2$, we get
$$
\frac{\pi e^{-au}}{2a^3} \left(au + 1 \right)
$$
Note that the result is always $e^{-au}$ times a degree $s-1$ polynomial in $u$. So it seems possible that a clever substitution and repeated integration by parts might also yield the result.
A: You can also get a recursive formula without residues:
Let $I_s(a)=\int \frac{\cos(ux)}{(x^2 + a^2)^s} \ \mathrm{d}x$. Then
$$
\begin{eqnarray}
\frac{\mathrm{d}I_s(a)}{\mathrm{d}a} 
&=& \frac{d}{da}\int_{-\infty}^\infty \frac{\cos(ux)}{(x^2 + a^2)^s} \ \mathrm{d}x \\
&=& -2as \int_{-\infty}^\infty \frac{\cos(ux)}{(x^2 + a^2)^{s+1}} \ \mathrm{d}x \\
&=& -2as I_{s+1}(a)
\end{eqnarray}
$$
This is the recursion. So with $I_1(a) = \pi e^{-au}/a$, we get for $s = 2$
$$
\begin{eqnarray}
I_2(a) &=& \frac{-1}{2as}\pi(-u/a - 1/a^2)e^{-au} \\
&=& \frac{\pi e^{-au}}{2a^3s} (au+1) \\
\end{eqnarray}
$$
and in general 
$$I_{s+1}(a) = \frac{\pi}{2^{s}s!}\left(\frac{-1}{a}\frac{\mathrm{d}}{\mathrm{d}a}\right)^{s}  \ \frac{e^{-au}}a$$

Observation: The operator $\left(\frac{-1}{a}\frac{\mathrm{d}}{\mathrm{d}a}\right)^{s}$ also turns up in the same sort of relation for the Bessel functions $J_n(x)$, where we have
$$J_n(x) = \left(\frac{-1}{x}\frac{\mathrm{d}}{\mathrm{d}x}\right)^{n} J_0(x)$$
for $n\ge 0$.
Additional Question: Is this a coincidence or can the integral somehow be nicely expressed in terms of Bessel functions? Is there a deeper relationship?
Also, the formula derived by Jonathan in the other post
$$
\begin{eqnarray}
I_s(a) &=& \frac{2\pi i}{(s-1)!} e^{-au} \sum_{k=0}^{s-1} {s-1 \choose k} (-1)^k (iu)^{s-1-k} \frac{s!}{(s-k)!} (2ia)^{-s-k} \\
&=&\frac{\pi u^{s-1} e^{-au}}{2^{s-1}a^{s}}  \sum_{k=0}^{s-1} \frac{(-1)^k s! }{(s-k)\, k!\, {(s-1-k)!}^2}\frac{1}{(2au)^k}
\end{eqnarray}
$$
does have some resemblence to the series expressions of Bessel functions.
