Inverse Fourier Transform of $\prod_{j=1}^n \frac{k_j}{k_j+i\omega}$ I'm trying to work out the inverse Fourier transform of
$$F(\omega)=\prod_{j=1}^n \frac{k_j}{k_j+i\omega}$$
with $k_j \in \mathbb{R}^+$ and using the definition of the Fourier transform where
$$f(x) = \mathcal{F^{-1}}\left[F\right](x) = \int_{-\infty}^\infty F(\omega) e^{2\pi i \omega x} d\omega$$
I only care about $x>0$.
So far...
To explain my difficulty, $F$ can be written as:
$$F(\omega)=\left(\prod_{j=1}^n i k_j\right) {\huge/} \left(\prod_{j=1}^n (ik_j - \omega)\right)$$
The polynomial on the right has roots $ik_j$, and the nature of these roots determines the form of the resulting transform. Consider the example where there are two roots, if they are unique:
$$f_2(x) = \frac{k_1k_2}{k_2-k_1}\left(e^{-k_1x}-e^{-k_2 x}\right)$$
(using subscripted $f$ to distinguish possible solutions). But if they are degenerate and equal to $k$
$$f_1(x) = k^2 x e^{-k x}$$
which can be obtained from the first by letting $k_1=k$ and $k_2=k+\Delta k$ and looking at the limit $\lim_{k_2\to k_1} f_2(x)$
$$\lim_{\Delta k\to0} k(k+\Delta k)\frac{e^{-kx}-e^{-(k+\Delta k) x}}{\Delta k} = -k^2 \frac{d}{dk}e^{-kx} = f_1(x)$$
To make the degeneracy of the roots clearer I have written it as:
$$F(\omega)=\prod_{j=1}^n \left(\frac{k_j}{k_j+i\omega}\right)^{g_j}$$
with each $k_i$ being different (not the same $n$). And I have formula for the non-degenerate case ($g_j = 1$):
$$f(x)=\left( \prod_{i=1}^n k_i \right) \sum_{i=1}^n \frac{e^{-k_i t}}{\prod_{j\neq i} k_j - k_i}$$
but I am struggling with a closed form for cases where $g_j > 1$. Any suggestions?
Thanks :)
 A: Here's an approach to this problem. Using partial fraction decomposition you can write your second product $\prod_j (1+i\omega/k_j)^{-g_j}$ as a sum $\sum_j\sum_\ell c_{j,\ell}(1+i\omega/k_j)^{-\ell}$, so it's enough to compute the inverse Fourier transform of $(1+i\omega/k)^{-\ell}$ whenever $\ell$ is a positive integer and $k>0$.
To do this, write $(1+i\omega/k)^{-1}$ as the Fourier transform of $2\pi k\,\mathbb 1_{[0,\infty)}(x)e^{-2\pi k x}$:
$$
2\pi k\int_0^\infty  e^{-2\pi k x}e^{-2\pi i \omega x}\,dx = {1\over 1 + i\omega/k}.
$$
Now differentiate both sides $\ell-1$ times with respect to $\omega$, passing the differentiation operator under the sign of integration on the left side; the result is
$$
2\pi k\int_0^\infty  (-2\pi i x)^{\ell-1}  e^{-2\pi k x}e^{-2\pi i \omega x}\,dx = {(-i/k)^{\ell-1}(\ell-1)!\over (1+i\omega/k)^\ell}.
$$
It follows at once that
$$
{(2\pi k)^\ell \over (\ell-1)!}\mathbb{1}_{[0,\infty)}(x)x^{\ell-1}e^{-2\pi kx}
$$
is the function with Fourier transform $(1+i\omega/k)^{-\ell}$ (where $\mathbb 1_{[0,\infty)}$ is the characteristic function of $[0,\infty)$).
