Inverse Fourier transform of $F(w)=\frac{1-\sum_{i=1}^n{a_ie^{-{r_k}{b_i}|w|}}}{|w|}$

Question

I encountered this problem in my research. I don't know to calculate the inverse Fourier transform (IFT) of this function: $$F(w)=\prod_{k=1}^m \frac{1-\sum_{i=1}^n{a_ie^{-{r_k}{b_i}|w|}}}{|w|}$$ where $a_i \in \mathbb{R}$, $b_i \in \mathbb{R^+} (i=1,...,n)$, $r_k \in \mathbb{R^+} (k=1,...,m)$ are known real parameters, $\sum_{i=1}^n{a_i} =1$ (so that $|F(0)|<+\infty$), and $m=1,2$ or $3$.

If the explicit expression can't be derived, then ideas for solving the problem will be great as well.

Answer 1

I shall follow the convention that IFT of $F$ is $\int_{\mathbb R}F(w)e^{2\pi i w t}dw$, take a scaling if necessary. First of all, a crucial lemma $$\int_0^{\infty } \frac{e^{-aw} \cos (b w)-e^{-cw} \cos (d w)}{w} \, dw=\frac{1}{2} \log \left(\frac{c^2+d^2}{a^2+b^2}\right)$$ Which is direct by applying Feynman's trick on either $a$ or $c$. Now, for case $m=1$ one have the IFT equals $$I_1=\int_{\mathbb R} \frac{1-\sum_i a_i e^{-r_1 b_i |w|}}{|w|}e^{2\pi i w t}dw=2\int_{\mathbb R^+} \frac{1-\sum_i a_i e^{-r_1 b_i w}}{w}\cos(2\pi t w)dw=\sum_i a_i K_1(r_1 b_i)$$ Where we have took a reflection on part $\mathbb R^-$, applied Euler formula and rewritten $1$ as $\sum_i a_i$. Now the lemma gives $$K_1(s)=\int_{\mathbb R^+}\frac{1-e^{-s w}}{w}\cos(2\pi t w)dw=\frac{1}{2} \left(\log \left(\frac{s^2}{4 t^2}+\pi ^2\right)-2 \log (\pi )\right)$$ Which completes the evaluation of this case. Note that this can be trivially generalized to cases that $m=2, 3$, namely elementary closed forms of IFT that $I_2=\sum_{i,j} a_i a_j K_2(r_1 b_i, r_2 b_j)$ and $I_3=\sum_{i,j,k} a_i a_j a_k K_3(r_1 b_i, r_2 b_j, r_3 b_k)$, where

• $\small K_2(s,y)=\int_{\mathbb R^+}\frac{1-e^{-s w}}{w}\frac{1-e^{-y w}}{w}\cos(2\pi t w)dw=-\frac{1}{2} s \log \left(s^2+4 \pi ^2 t^2\right)+\frac{1}{2} s \log \left((s+y)^2+4 \pi ^2 t^2\right)+\frac{1}{2} y \log \left((s+y)^2+4 \pi ^2 t^2\right)-2 \pi t \tan ^{-1}\left(\frac{2 \pi t}{s+y}\right)+2 \pi t \tan ^{-1}\left(\frac{2 \pi t}{s}\right)-\frac{1}{2} y \log \left(4 \pi ^2 t^2+y^2\right)+2 \pi t \tan ^{-1}\left(\frac{2 \pi t}{y}\right)-\pi ^2 t$

• $\scriptsize K_3(s,y,z)=\int_{\mathbb R^+}\frac{1-e^{-s w}}{w}\frac{1-e^{-y w}}{w}\frac{1-e^{-z w}}{w}\cos(2\pi t w)dw=\frac{1}{4} \left(2 \log (s) s^2+\log \left(\frac{4 \pi ^2 t^2}{s^2}+1\right) s^2-2 \log (s+y) s^2-\log \left(\frac{4 \pi ^2 t^2}{(s+y)^2}+1\right) s^2-2 \log (s+z) s^2+2 \log (s+y+z) s^2-\log \left(\frac{4 \pi ^2 t^2}{(s+z)^2}+1\right) s^2-8 \pi t \tan ^{-1}\left(\frac{2 \pi t}{s}\right) s+8 \pi t \tan ^{-1}\left(\frac{2 \pi t}{s+y}\right) s+8 \pi t \tan ^{-1}\left(\frac{2 \pi t}{s+z}\right) s-8 \pi t \tan ^{-1}\left(\frac{2 \pi t}{s+y+z}\right) s-4 y \log (s+y) s-2 y \log \left(\frac{4 \pi ^2 t^2}{(s+y)^2}+1\right) s-4 z \log (s+z) s+4 y \log (s+y+z) s+4 z \log (s+y+z) s-2 z \log \left(\frac{4 \pi ^2 t^2}{(s+z)^2}+1\right) s-8 \pi t y \tan ^{-1}\left(\frac{2 \pi t}{y}\right)+8 \pi t y \tan ^{-1}\left(\frac{2 \pi t}{s+y}\right)-8 \pi t z \tan ^{-1}\left(\frac{2 \pi t}{z}\right)+8 \pi t z \tan ^{-1}\left(\frac{2 \pi t}{s+z}\right)+8 \pi t y \tan ^{-1}\left(\frac{2 \pi t}{y+z}\right)+8 \pi t z \tan ^{-1}\left(\frac{2 \pi t}{y+z}\right)-8 \pi t y \tan ^{-1}\left(\frac{2 \pi t}{s+y+z}\right)-8 \pi t z \tan ^{-1}\left(\frac{2 \pi t}{s+y+z}\right)-8 \pi ^2 t^2 \log (s)-4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{s^2}+1\right)+8 \pi ^2 t^2 \log (2 \pi t)-8 \pi ^2 t^2 \log (y)+2 y^2 \log (y)+8 \pi ^2 t^2 \log (s+y)-2 y^2 \log (s+y)-4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{y^2}+1\right)+y^2 \log \left(\frac{4 \pi ^2 t^2}{y^2}+1\right)+4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{(s+y)^2}+1\right)-y^2 \log \left(\frac{4 \pi ^2 t^2}{(s+y)^2}+1\right)-8 \pi ^2 t^2 \log (z)+2 z^2 \log (z)+8 \pi ^2 t^2 \log (s+z)-2 z^2 \log (s+z)+8 \pi ^2 t^2 \log (y+z)-2 y^2 \log (y+z)-2 z^2 \log (y+z)-4 y z \log (y+z)-8 \pi ^2 t^2 \log (s+y+z)+2 y^2 \log (s+y+z)+2 z^2 \log (s+y+z)+4 y z \log (s+y+z)-4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{z^2}+1\right)+z^2 \log \left(\frac{4 \pi ^2 t^2}{z^2}+1\right)+4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{(s+z)^2}+1\right)-z^2 \log \left(\frac{4 \pi ^2 t^2}{(s+z)^2}+1\right)+4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{(y+z)^2}+1\right)-y^2 \log \left(\frac{4 \pi ^2 t^2}{(y+z)^2}+1\right)-z^2 \log \left(\frac{4 \pi ^2 t^2}{(y+z)^2}+1\right)-2 y z \log \left(\frac{4 \pi ^2 t^2}{(y+z)^2}+1\right)+(s+2 \pi t+y+z) (s-2 \pi t+y+z) \log \left(\frac{4 \pi ^2 t^2}{(s+y+z)^2}+1\right)\right)$

There is no essential difficulty of calculating $K_2$ and $K_3$ in spite of their complicated forms (due to which I used Mathematica to generate the result). Take $K_2$ for example, just apply Feynman's trick again on $s$ and use the lemma again. For $K_3$, just differentiate $s$ twice, use the lemma, and integrate back twice (with good care of boundary values taken). The method described here should work for all $m>3$, since essentially we are integrating nothing but the forms $\int s^k \log(s^2+a^2) ds$ and $\int s^k \tan^{-1}(s)ds$, which will be expressible by $\log, \tan^{-1}$ again (no matter how harge $m$ is) without using polylog terms (like in, for instance, this question).

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For OP's convenience, this command should generate the expression of $K_3$.

Integrate[(1 - Exp[-s w]) (1 - Exp[-y w]) (1 - Exp[-z w])/w^3*
Cos[2 Pi t w], {w, 0, Infinity}, 
Assumptions -> s > 0 && y > 0 && z > 0 && t > 0] // FullSimplify

One may use them to write down explicit forms of IFT for certain $m$ and parameters. End of story.