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I am trying to solve the following integral in the 4D space \begin{equation} \int\exp ({-i \vec{w}^T \vec{x}}) \exp(-\frac{||\vec{x}||_2}{2}) d \vec{x} \end{equation}

I tried to follow the similar strategy shown in 3D Fourier transforms of $e^{-\beta r} $ and $re^{-\beta r} $ and solved the problem in the 4D spherical coordinate system. But the result didn't seem right.

I don't have a good idea about the polar direction in the 4D coordinate system. One thing I think I did wrong is I wrote $\vec{w}^T\vec{x}=||\vec{w}||_2 ||\vec{x}||_2 cos \theta, 0\leq \theta \leq \pi$ assuming $\vec{w}$ is parallel to the polar direction.

Wayne
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Since $e^{-\frac{||x||_2}{2}}$ is a radial function you are free to rotate your coordinate system such that $w$ is pointing to the "North Pole" in 4D. I will use the 4D spherical coordinates I derived from this post. So our integral becomes

$$4\pi \int_0^\pi \int_0^\infty e^{-(i||w||_2 \cos\theta + \frac{1}{2})r}r^3\sin^2\theta dr d\theta = 24\pi \int_0^\pi \frac{\sin^2\theta}{(i||w||_2\cos\theta + \frac{1}{2})^4}d\theta$$

where the complex coefficient exponential can be shown to have the same value as its counterpart on the real line by looking at a pie slice shaped contour in the complex plane, with the corner at the origin.

Recognize the fact that the integrand is even and can be turned into an integral over $[-\pi,\pi]$ with a factor of $\frac{1}{2}$, which lets us convert it into an integral over the unit circle in the complex plane. A long residue later:

$$\mathcal{F}(e^{-||x||_2}) = \frac{6\pi^2}{(\frac{1}{4}+||w||_2^2)^{\frac{5}{2}}}$$

Ninad Munshi
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  • Thanks for your great help. What are the singular points of $\int _{-\pi}^{\pi} \frac{\sin^2 \theta}{(i||w||_2 \cos \theta+0.5)^4} d\theta$? When I tried to follow your instructions and did it myself, I got troubles to figure out the singular points. – Wayne Sep 09 '19 at 03:14
  • @Wayne what do you mean by singular points? – Ninad Munshi Sep 09 '19 at 03:14
  • @Wayne I think I know what you mean. The trick is to turn every trig function into complex exponentials, and to multiply and divide the integrand by $ie^{i\theta}$. Then do the substitution $z=e^{i\theta}$, $\frac{1}{z} = e^{-i\theta}$, and $dz = ie^{i\theta}d\theta$. From there you can use quadratic equation to figure out the roots of the bottom polynomial, which, once when you simplify, becomes a quadratic to a fourth power. The real valued integral at the moment has no singularities. – Ninad Munshi Sep 09 '19 at 03:18
  • There are two poles $\frac{1\pm\sqrt{1+4w^2}}{2w}i$ and the pole inside the unit circle is the one $z_0$ with $-$ sign. Since $z_0$ is a forth order pole, I need to calculate residue as $\frac{1}{3!} \frac{d^3}{dz^3}f(z)\cdot (z-z_0)^4$, where $f(z)=-\frac{4z(z^2-1)^2}{(w z^2+w-i z)^4}$. The formula I get is $\frac{1}{3!} \cdot \frac{12(z_1+z)(5z_1^2z^2-z_1^2-8z_1z-z^2+5)}{(z_1-z)^7}$, where $z_1$ is the pole outside the unit circle. After I plug $z=z_0$ into the formula, I didn't get your result. Terms didn't cancel each other. I am wondering if my calculation is wrong. – Wayne Sep 09 '19 at 19:54
  • I worked it out. Thank you so much. – Wayne Sep 10 '19 at 18:53