# Fourier transformation of a distribution

We have no idea how to tackle the following Fourier transformation of a distribution: $$\lim_{\epsilon\to0^+}\int_{-\infty}^{\infty}\mathrm{d} t\int_{\mathbb{R}^{d-1}} \mathrm{d}^{d-1}\vec{r} e^{-\mathrm{i}E t+\mathrm{i}\vec{p}\cdot\vec{r}}\left[-t^2+r^2-\mathrm{i}\,\epsilon t\right]^{-\alpha} \left[-t^2+r^2+\mathrm{i}\,\epsilon t\right]^{-\Delta}\ ,$$ where $$E\in\mathbb{R},\vec{p}\in\mathbb{R}^{d-1},r=\sqrt{\vec{r}\cdot\vec{r}}$$. We first tried to deform the $$t$$ integral on the complex $$t$$ plain where there are four branch cuts. Because the deformed integral is not simple, we failed to find a way to perform the remaining integrals and gain a closed-form expression.
We expect that the final form can be written as: $$|E^2-|\vec{p}||^{\Delta-d/2}[\#_1\Theta(E^2-|\vec{p}|)\Theta(E)+\#_2\Theta(E^2-|\vec{p}|)\Theta(-E)+\#_3\Theta(-E^2+|\vec{p}|)]\ ,$$ where $$\#_{1,2,3}$$ are some constants depending on $$d,\Delta,\alpha$$ and $$|\vec{p}|$$ is the magnitude of $$\vec{p}\in\mathbb{R}^{d-1}$$. We also expect that the final result is symmetric under the replacement $$(E,\alpha,\Delta)\leftrightarrow (-E,\Delta,\alpha)$$.
We would appreciate it if you show me how to perform this integral.

When $$\alpha=0$$, this integral reduces to $$\lim_{\epsilon\to0^+}\int_{-\infty}^{\infty}\mathrm{d} t\int_{\mathbb{R}^{d-1}} \mathrm{d}^{d-1}\vec{r} e^{-\mathrm{i}E t+\mathrm{i}\vec{p}\cdot\vec{r}} \left[-t^2+r^2+\mathrm{i}\,\epsilon t\right]^{-\Delta}\ ,$$ and can be performed throught the following steps. We start with the Formula: $$\frac{1}{(\tau^2+r^2)^{\Delta}}=\frac{\pi^{d/2}2^{d-2\Delta}\Gamma\left(d/2-\Delta\right)}{\Gamma(\Delta)}\int_{-\infty}^{\infty}\frac{\mathrm{d}p^d}{2\pi}\int_{\mathbb{R}^{d-1}}\frac{\mathrm{d}^{d-1}\vec{p}}{(2\pi)^{d-1}} e^{-\mathrm{i}p^d\tau-\mathrm{i}\vec{p}\cdot\vec{r}} [(p^d)^2+|\vec{p}|^2]^{\Delta-d/2}\ .$$ Putting $$\tau=\epsilon+\mathrm{i}\,t$$ and taking the limit $$\epsilon\to 0^+$$, this equation becomes: $$\lim_{\epsilon\to 0^+}\frac{1}{(-t^2+r^2+\mathrm{i}\,\epsilon t)^{\Delta}}=\lim_{\epsilon\to 0^+}\frac{\pi^{d/2}2^{d-2\Delta}\Gamma\left(d/2-\Delta\right)}{\Gamma(\Delta)}\int_{\mathbb{R}^{d-1}}\frac{\mathrm{d}^{d-1}\vec{p}}{(2\pi)^{d-1}}\int_{-\infty}^{\infty}\frac{\mathrm{d}p^d}{2\pi} e^{+tp^d-\mathrm{i}p^d\epsilon-\mathrm{i}\vec{p}\cdot\vec{r}} [(p^d)^2+|\vec{p}|^2]^{\Delta-d/2}\ .$$ Let us assume $$0<2\Delta. (We will extend the final form to any $$\Delta$$ by analytic continuation) Then the integrand goes to zero in the limit $$|p^d|\to \infty$$. Therefore from Jordan's lemma, we can deform $$p^d$$ integral so that the deformed contour picks up the discontinuity along the negative imaginary axis on the complex $$p^d$$ plain: $$\lim_{\epsilon\to 0^+} \int_{-\infty}^{\infty}\frac{\mathrm{d}p^d}{2\pi} e^{+tp^d-\mathrm{i}p^d\epsilon} [(p^d)^2+|\vec{p}|^2]^{\Delta-d/2} \\ =\lim_{\epsilon,o\to 0^+}(-\mathrm{i})\int_{0}^{\infty}\frac{\mathrm{d}p^0}{2\pi} e^{+\mathrm{i}t p^0 -p^0\epsilon} \left\{[(-\mathrm{i}p^0+o)^2+|\vec{p}|^2]^{\Delta-d/2}-[(-\mathrm{i}p^0-o)^2+|\vec{p}|^2]^{\Delta-d/2}\right\}\\ =\lim_{\epsilon,o\to 0^+}(-\mathrm{i})\int_{0}^{\infty}\frac{\mathrm{d}p^0}{2\pi} e^{+\mathrm{i}t p^0 -p^0\epsilon} \left\{[-(p^0)^2+|\vec{p}|^2-\mathrm{i}\,o]^{\Delta-d/2}-[-(p^0)^2+|\vec{p}|^2+\mathrm{i}\,o]^{\Delta-d/2}\right\}\\$$ We introduced infinitesimal real positive number $$o$$ to detour the branch cut. By using the identity $$\lim_{o\to 0^+}\left[(x+\mathrm{i}\,o)^\lambda-(x-\mathrm{i}\,o)^\lambda\right]=2\mathrm{i}(-x)^\lambda \Theta(-x)\sin\pi\lambda$$, the $$p^d$$ integral results in: $$\lim_{\epsilon\to 0^+} \int_{-\infty}^{\infty}\frac{\mathrm{d}p^d}{2\pi} e^{+tp^d-\mathrm{i}p^d\epsilon} [(p^d)^2+|\vec{p}|^2]^{\Delta-d/2} \\ = \int_{0}^{\infty}\frac{\mathrm{d}p^0}{2\pi} e^{+\mathrm{i}t p^0 }[(p^0)^2-|\vec{p}|^2]^{\Delta-d/2} \Theta[(p^0)^2-|\vec{p}|^2]2\sin\pi\left(d/2-\Delta\right)\\ =\int_{-\infty}^{\infty}\frac{\mathrm{d}p^0}{2\pi} e^{+\mathrm{i}t p^0 }[(p^0)^2-|\vec{p}|^2]^{\Delta-d/2} \Theta(p^0)\Theta[(p^0)^2-|\vec{p}|^2]2\sin\pi\left(d/2-\Delta\right)$$ Putting altogether and use the Euler's reflection formula $$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}$$ we obtain: $$\lim_{\epsilon\to 0^+}\frac{1}{(-t^2+r^2+\mathrm{i}\,\epsilon t)^{\Delta}}\\ =\frac{\pi^{d/2+1}2^{d-2\Delta+1}}{\Gamma(\Delta)\Gamma\left(\Delta-\frac{d-2}{2}\right)}\int_{-\infty}^{\infty}\frac{\mathrm{d}p^0}{2\pi} \int_{\mathbb{R}^{d-1}}\frac{\mathrm{d}^{d-1}\vec{p}}{(2\pi)^{d-1}}e^{+\mathrm{i}p^0 t-\mathrm{i}\vec{p}\cdot\vec{r}} \\ \quad \times [(p^0)^2-|\vec{p}|^2]^{\Delta-d/2} \Theta(p^0)\Theta[(p^0)^2-|\vec{p}|^2]\ .$$ Inverting this Fourier integral, we end up with: $$\lim_{\epsilon\to0^+}\int_{-\infty}^{\infty}\mathrm{d} t\int_{\mathbb{R}^{d-1}} \mathrm{d}^{d-1}\vec{r} e^{-\mathrm{i}E t+\mathrm{i}\vec{p}\cdot\vec{r}} \left[-t^2+r^2+\mathrm{i}\,\epsilon t\right]^{-\Delta}\\ =\frac{\pi^{d/2+1}2^{d-2\Delta+1}}{\Gamma(\Delta)\Gamma\left(\Delta-\frac{d-2}{2}\right)}[E^2-|\vec{p}|^2]^{\Delta-d/2} \Theta(E)\Theta[E^2-|\vec{p}|^2]\ .$$