How to compute the Fourier transform of $e^{-2\pi {\mid x \mid}}$?

Question

I want to compute the Fourier transform of $f(x)=e^{-2\pi {\mid x \mid}}$ where $x$ is the vector in $\mathbb{R}^n$. I know that the Fourier transform of a radial function is radial but cannot see how to use the fact to calculate the following integrate \begin{equation} \hat{f}(y)=\int_{\mathbb{R}^n} {e^{-2\pi {\mid x \mid}}} e^{-2\pi x \cdot y } dx \end{equation}

Could anyone please help me?

Answer 1

From Bracewell, the n-dimensional Fourier Transform

$$F(s_1, \dots, s_n) = \int_{-\infty}^{\infty} \dots \int_{-\infty}^{\infty} f(x_1,\dots, x_n) e^{-2\pi i (x_1 s_1 + \dots + x_n s_n)} \space dx_1 \dots dx_n $$

when there is n-dimensional symmetry, can be manipulated into

$$F(q) = \dfrac{2\pi}{q^{\frac{1}{2}n-1}} \int_{0}^{\infty} f(r) J_{\frac{1}{2}n-1}(2\pi qr)\space r^{\frac{1}{2}n} \space dr$$

Where $r$ and $q$ are the radii from the origin in the original domain and transform domain respectively.

So that the cases of $n=1$ and $n=3$ don't seem so foreign, it is probably worth noting that

$$J_{-\frac{1}{2}}(x) = \sqrt{\dfrac{2}{\pi x}}\cos(x)$$ $$J_{\frac{1}{2}}(x) = \sqrt{\dfrac{2}{\pi x}}\sin(x)$$

I'm not quite sure I know what you meant by $|x|$ in your original problem statement, but if you just meant the magnitude of $x$, then the specific transform you wanted to compute looks like

$$F(q) = \dfrac{2\pi}{q^{\frac{1}{2}n-1}} \int_{0}^{\infty} e^{-2\pi r} J_{\frac{1}{2}n-1}(2\pi qr)\space r^{\frac{1}{2}n} \space dr$$

For $n=1$

$$\begin{align*}F(q) &= 2\pi\sqrt{q} \int_{0}^{\infty} e^{-2\pi r} \sqrt{\dfrac{2}{\pi(2\pi qr)}}\cos(2\pi qr)\space \sqrt{r} \space dr\\ \\ &= 2 \int_{0}^{\infty} e^{-2\pi r} \cos(2\pi qr)\space dr \\ \\ &=\int_{0}^{\infty} e^{-2\pi r} \left(e^{i2\pi qr} + e^{-i2\pi qr}\right)\space dr \\ \\ &= \left[\dfrac{e^{-2\pi r}e^{i2\pi qr}}{-2\pi (1-iq)} + \dfrac{e^{-2\pi r}e^{-i2\pi qr}}{-2\pi (1+iq)}\right]\Biggr|_0^\infty\\ \\ &= \dfrac{1}{2\pi}\left[\dfrac{1}{1-iq}+\dfrac{1}{1+iq}\right]\\ \\ &= \dfrac{1}{2\pi} \dfrac{2}{1+q^2} \end{align*}$$

which is exactly what one would expect for the 1 dimensional Fourier Transform.

For $n=2$

$$\begin{align*}F(q) &= 2\pi \int_{0}^{\infty} e^{-2\pi r} J_{0}(2\pi qr)\space r \space dr\\ \\ &\mbox{(Hankel Transform table lookup)}\\ \\ &= \dfrac{2\pi (2\pi)}{\left[(2\pi q)^2 + (2\pi)^2\right]^{\frac{3}{2}}}\\ \\ &= \dfrac{1}{2\pi} \dfrac{1}{\left(1+q^2\right)^{\frac{3}{2}}} \end{align*}$$

For $n =3$

$$\begin{align*}F(q) &= \dfrac{2\pi}{\sqrt{q}} \int_{0}^{\infty} e^{-2\pi r} \sqrt{\dfrac{2}{\pi(2\pi qr)}}\sin(2\pi qr)\space r^{\frac{3}{2}} \space dr \\ \\ &= \dfrac{2}{q} \int_{0}^{\infty} e^{-2\pi r} \sin(2\pi qr)\space r \space dr \\ \\ &\mbox{(Wolfram Alpha invocation in lieu of integration by parts)} \\ &= \dfrac{1}{(2\pi)^2}\dfrac{4}{(1+q^2)^2}\\ \end{align*}$$

And I suppose I'll let you keep on going until you see a pattern in terms of $n$ emerge.