Solving this Fourier transform?

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.

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Is $s$ a natural number? And how did you find the $s=1$ case? – Ragib Zaman Nov 16 '11 at 0:44
Is there any reason you don't consider residues to be an exact analytic method? – Jonathan Nov 16 '11 at 0:52

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.

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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.

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I haven't thought about it too much, but a quick look at the wikipedia page suggests that these functions might be closely related to the Laguerre polynomials, and the Laguerre polynomials are indeed related to Bessel functions. In any case I'm sure these functions are listed in tables of Fourier transforms, so they probably have somebody's name attached to them. – Jonathan Nov 16 '11 at 23:09