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 :)


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 $[0,\infty)$).

  • $\begingroup$ Arg, I wrote the wrong Fourier transform... and I went to the effort of being explicit too - just a constant though $\endgroup$ – Lucas Jun 11 '13 at 2:09
  • $\begingroup$ I get the derivation of the part which is of the same form as an Erlang distribution. But I am having trouble with the partial fractions part. Specifically, I'm wondering if $c_{j\ell}$ is real (it can be), and how to write it explicitly. $\endgroup$ – Lucas Jun 11 '13 at 2:46
  • $\begingroup$ @Lucas I don't really have time to go through the computation right now, but there are a lot of methods to compute the constants $c_{j\ell}$ in the partial fraction decomposition. One way is by residues: If $f(\omega) = \prod_j (1 +i\omega/k_j)^{-g_j}$ and $h_j(\omega) = (1+ i\omega/k_j)^{g_j} f(\omega)$, then the constant $c_{j\ell}$ should be equal to $h_j^{(\ell)}(ik_j)$, for $0\leq \ell\leq g_j-1$. I can add an explicit computation later. $\endgroup$ – Nick Strehlke Jun 11 '13 at 3:22
  • $\begingroup$ Oh, don't worry. I can figure it out. Just not sure it is exactly what I want without going through it. Thanks. $\endgroup$ – Lucas Jun 11 '13 at 4:02
  • $\begingroup$ Just realized that where I said $c_{j\ell} = h_j^{(\ell)}(ik_j)$ for $0\leq \ell \leq g_j - 1$ in the last comment I should have said $c_{j\ell} = h_j^{(g_j-\ell)}(ik_j)/\ell!$ for $1\leq \ell\leq g_j$. Sorry about that. By the way I don't think any formula for the general partial fraction decomposition will be very nice, because you're essentially seeking the most general possible partial fraction decomposition (by extension there won't be a very nice closed form for the inverse Fourier transform since this is equivalent to finding the partial fraction decomposition). $\endgroup$ – Nick Strehlke Jun 11 '13 at 4:24

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