What Is the Fourier Transform of Propagators in QFT?

[L0r3n20]

I've been assigned the following homework:
I have to compute the spectral density of a QFT and in order to do so I have to compute Fourier tranform of the following quantity (in Minkowsky signature, mostly minus)

\rho\left(p\right) = \int \frac{1}{\left(-x^2 + i \epsilon x_0\right)^{\Delta}} e^{i p \cdot x} d^4 x

Using residual theorem I've been able to compute exactly the case for \Delta = 1 and it turns out to be \rho(p) \propto \delta(p^2) \theta(p_0) .
For the case \Delta \neq 1 it's a bit tricky but I managed to perform the integration over x_0 using the residual once again and I found (r^2 = x_i x^i)

\sum_{k=0}^{\Delta -1} {\Delta-1 \choose k} (i p_0)^k \frac{\Gamma(\Delta -1 - k)}{\Gamma(\Delta)^2} \frac{1}{(2 r)^{2 \Delta+1-k}} \left(e^{i p_0 r} + (-1)^{2 \Delta -1 - k} e^{-i p_0 r}\right)

Probably something is wrong since when I perform the remaining integrations (in spherical coordinates) I do not recover the following result:

\rho\left(p\right) = \frac{\Delta -1}{4^{\Delta} \Gamma(\Delta)^2} \theta(p_0) \delta(p^2) (p^2)^{\Delta -2}
Any help would be great.

 

[member 553083]

To compute the Fourier transform of the given expression, use the following equation:F[\rho(p)] = \int_{-\infty}^{\infty} \frac{1}{\left(-x^2 + i \epsilon x_0\right)^{\Delta}} e^{i p \cdot x} d^4 x Using integration by parts, we can rewrite the equation as:F[\rho(p)] = -\frac{i}{2\pi} \int_{-\infty}^{\infty} \frac{d^4 x}{\left(-x^2 + i \epsilon x_0\right)^{\Delta-1}} \partial_\mu e^{i p \cdot x} d^4 x Substituting in the definition of the Fourier transform, we obtain:F[\rho(p)] = -\frac{i}{2\pi} \int_{-\infty}^{\infty} \partial_\mu F[e^{i p \cdot x}] d^4 x Using integration by parts once again, we obtain:F[\rho(p)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} F[e^{i p \cdot x}] \partial_\mu \left(\frac{1}{\left(-x^2 + i \epsilon x_0\right)^{\Delta-1}}\right) d^4 x The Fourier transform of the exponential is well known and can be computed as follows:F[e^{i p \cdot x}] = (2\pi)^4 \delta(p^2) \theta(p_0)Substituting this into our expression for the Fourier transform of \rho(p), we obtain:F[\rho(p)] = \frac{1}{2\pi} (2\pi)^4 \delta(p^2) \