computations problem with reverse Fourier transform

Question

The reverse Fourier transform of a function is defined as $u(x):=\frac{1}{(2\pi )^{\frac{3}{2}}}\int_{\mathbb{R}^{3}}^{ }e^{-ikx}u(k)dv(k)$. I have to prove that when $u(k)=-\frac{1}{(2\pi )^{\frac{3}{2}}}\cdot \frac{e^{ikx_{0}}}{k^{2}}$ then $u(x)=\frac{1}{4\pi \left | x-x_{0} \right |}$, where $x,x_{0},k\epsilon \mathbb{R}^{3}$. Our professor also gave us the hint $\int_{0}^{+\infty }\frac{sint}{t}dt=\frac{\pi }{2}$. Can someone please help? I really have no clue how to accomplish this .

Answer 1

Dropping the constant multiplier for the convenience $$I=\int_{\mathbb{R}^{3}}^{ }e^{-i\vec k\vec x}\frac{e^{i\vec k\vec x_0}}{|\vec k|^2}d^3\vec k=\int_{\mathbb{R}^{3}}^{ }e^{-i\vec k(\vec x-\vec x_0)}\frac{d^3\vec k}{|\vec k|^2}$$ Setting the $Z$ axis along the vector ($\vec x-\vec x_0$) and going to polar system of coordinates $$I=2\pi\int_0^{\infty}\frac{k^2}{k^2} dk\int_0^\pi e^{-ik|\vec x-\vec x_0|\cos\theta}\sin\theta \,d\theta=2\pi\int_0^{\infty}dk\int_{-1}^1 e^{-ik|\vec x-\vec x_0|\,t}dt$$ $$=2\pi\int_0^{\infty}\frac{e^{-ik|\vec x-\vec x_0|}-e^{ik|\vec x-\vec x_0|}}{-ik|\vec x-\vec x_0|}dk=\frac{4\pi}{|\vec x-\vec x_0|}\int_0^{\infty}\frac{\sin(k|\vec x-\vec x_0|)}{k}dk$$ $$=\frac{4\pi}{|\vec x-\vec x_0|}\int_0^{\infty}\frac{\sin(t)}{t}dt=\frac{2\pi^2}{|\vec x-\vec x_0|}$$

Another way is to use Schwinger trick.

Denoting $(\vec x-\vec x_0)=\vec a =(a_1, a_2,... a_n) \text{ in } R^n$ $$I_n=\int_{\mathbb{R}^{n}}^{ }e^{-i\vec k\vec a}\frac{d^n\vec k}{|\vec k|^2}=\int_{-\infty}^{\infty}..\int_{-\infty}^{\infty}dk_1...dk_n\int_{0}^{\infty}ds\,e^{-s(k_1^2+... k_n^2)}e^{-i(k_1a_1+..k_na_n)}$$ $$=\int_{0}^{\infty}ds\int_{-\infty}^{\infty}dk_1\exp{\Bigl(-s\bigl(k_1^2+\frac{ia_1}{s}-\frac{a^2_1}{4s^2}\bigr)-\frac{a^2_1}{4s}}\Bigr)\int_{-\infty}^{\infty}dk_2e^{(...)}...\int_{-\infty}^{\infty}dk_ne^{(...)}$$ $$=\int_{0}^{\infty}ds\exp\Bigl(-\frac{a^2_1}{4s}-..-\frac{a^2_n}{4s}\Bigr)\biggl(\int_{-\infty}^{\infty}dte^{-s(t-b)^2}\biggr)^n\, \,(t=k_1, k_2...)$$ $$=\int_{0}^{\infty}ds\exp\Bigl(-\frac{a^2_1+a^2_2+..a^2_n}{4s}\Bigr)\frac{\pi^{\frac{n}{2}}}{s^\frac{n}{2}}=(s=1/4t)\,\,\,2^{n-2}\pi^{\frac{n}{2}}\int_{0}^{\infty}e^{-|\vec a|^2t}t^{\frac{n}{2}-2}dt$$ $$=\frac{2^{n-2}\pi^{\frac{n}{2}}}{|\vec a|^{n-2}}\Gamma\Bigl(\frac{n}{2}-1\Bigr)$$ $$I_n=\frac{2^{n-1}\pi^{\frac{n}{2}}}{(n-2)|\vec x-\vec x_0|^{n-2}}\Gamma\Bigl(\frac{n}{2}\Bigr)$$ For $n=3$ $$I_3=\frac{2\pi^{2}}{|\vec x-\vec x_0|}$$