# Inverse Fourier transform problem

Can anybody please guide me how to compute the following inverse Fourier Transform ?

$$p(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{1}{(1-j\omega\bar{x})^K}e^{-j\omega x}d\omega$$

I shall be grateful.

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Is the bar on the $x$ intentional, is $x$ complex? Is $K\ge 1$ assumed? Did you try the residuum method of complex analysis for this type of integrals? –  LutzL Feb 19 at 16:00

For any polynomials $P$ and $Q$ with $\deg Q\ge 1+\deg P$ and $\alpha>0$ one has

$$\int_{-\infty}^{\infty}\tfrac{P(x)}{Q(x)}e^{i αx}\,dx=2\pi i\sum_{z:Q(z)=0\,\land\, Im(z)α>0}Res\left(x\mapsto\tfrac{P(x)}{Q(x)}e^{i αx},z\right)$$

where $Res(f,z)$ is the residue of $f$ in $z$, i.e., developing $f(z+w)$ as Laurent series in $w$ it is the coefficient of the power $w^{-1}$.

This is mentioned, but not in detail, in the wikipedia article on contour integration.

The idea is to integrate along a box over the interval $[-R,R]$ where the part of the sides gets smaller with larger $R$ due to the degree difference and the part of the top side gets smaller for increasing height because $|e^{i αz}|=e^{-α\, Im(z)}$.

Increasing the size $R$ of the box to infinity reduces the contour integal to the integral over the real line. Going the other way by shrinking the contour over the region of holomorphy allows by the Cauchy integral theorem to reduce the contour integral to integrals along small circles around the singularities, whose values are given by the residues of the singularities, times $2\pi i$.

If $α>0$ one has to take the residues of the upper half plane, for $α<0$ the residues of the lower half plane.

In your case with the different variable names the integration variable is $ω$, $P(ω)=1$, $Q(ω)=(1-jω\bar x)^K$ with a root at $ω_0=-j\bar x^{-1}$. Inserting $ω=ω_0+\zeta$ one gets $Q(ω_0+\zeta)=(-j ζ\bar x)^K$ and

$$f(ω)=\frac1{2\pi}\frac{e^{-jxω}}{(1-jω\bar x)^K}=\frac1{2\pi}e^{-jxω_0}\frac{1+(-jxζ)+...+\tfrac1{n!}(-jxζ)^n+...}{(-j ζ\bar x)^K}$$

so that residue, the coefficient for $ζ^{-1}$, is obtained for the term $n=K-1$ of the exponential series in the numerator, that is

$$Res(f,ω_0)=\frac{j}{2\pi}e^{-\tfrac{x}{\bar x}}\frac{x^{K-1}}{(K-1)!\bar x^K}$$.

For $\bar x>0$ we get the only singularity $ω_0$ in the lower half plane. With $α=-x$ this means that the integral is only different from zero if $x>0$ so that $α<0$. And vice versa, for $\bar x<0$ non-trivial values of the integral are only possible for $x>0$.

This gives the value of the integral as

$$2\pi j\,Res(f,ω_0)=-e^{-\tfrac{x}{\bar x}}\frac{x^{K-1}}{(K-1)!\bar x^K}$$

whenever $x$ and $\bar x$ have the opposite sign.

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I am sorry this is a bit difficult to understand. The answer to the above integral is $$p(x) = \frac{1}{\Gamma(K)\bar{x}^K} x^{K-1}e^{-x/\bar{x}}$$ I donot understand how did gamma entered the equation ? I am thankful for your time –  InvizibleSoul Feb 19 at 16:26
$Γ(K)=(K-1)!$ if $K$ is an integer. If not the derivation may become more conmplicated. –  LutzL Feb 19 at 16:33
Wow @LutzL This is great! please give me a minute to digest your explanantion :) as it is hard to understand and i will get back to you if i donot understand anything. I am really grateful to you for your efforts and help –  InvizibleSoul Feb 19 at 16:33
K is an integer –  InvizibleSoul Feb 19 at 16:34
I did not understand the step where you set $\omega = \omega_0 + \zeta$ what does $\zeta$ represents ? and i didnot get <<so that the coefficient for $\zeta^{-1}$ ...>> either. –  InvizibleSoul Feb 19 at 16:40