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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 $0<2\Delta<d$. (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: $$ \lim_{\epsilon\to0^+}\int_{-\infty}^{\infty}\mathrm{d} t\int_{\mathbb{R}^{d-1}} \mathrm{d}^{d-1}\vec{r} e^{-\mathrm{i}E t+\mathrm{i}\vec{p}\cdot\vec{r}} \left[-t^2+r^2+\mathrm{i}\,\epsilon t\right]^{-\Delta}\\ =\frac{\pi^{d/2+1}2^{d-2\Delta+1}}{\Gamma(\Delta)\Gamma\left(\Delta-\frac{d-2}{2}\right)}[E^2-|\vec{p}|^2]^{\Delta-d/2} \Theta(E)\Theta[E^2-|\vec{p}|^2]\ . $$

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