Cfrgtkky

I want to compute the Fourier transform of $f(x)=e^{-2pi {mid x mid}}$ 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^{-2pi {mid x mid}}} e^{-2pi x cdot y } dx
end{equation}

Could anyone please help me?

From Bracewell, the n-dimensional Fourier Transform

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

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

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

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

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

For $n=1$

$$begin{align*}F(q) &= 2pisqrt{q} int_{0}^{infty} e^{-2pi r} sqrt{dfrac{2}{pi(2pi qr)}}cos(2pi qr)space sqrt{r} space dr\
\
&= 2 int_{0}^{infty} e^{-2pi r} cos(2pi qr)space dr \
\
&=int_{0}^{infty} e^{-2pi r} left(e^{i2pi qr} + e^{-i2pi qr}right)space dr \
\
&= left[dfrac{e^{-2pi r}e^{i2pi qr}}{-2pi (1-iq)} + dfrac{e^{-2pi r}e^{-i2pi qr}}{-2pi (1+iq)}right]Biggr|_0^infty\
\
&= dfrac{1}{2pi}left[dfrac{1}{1-iq}+dfrac{1}{1+iq}right]\
\
&= dfrac{1}{2pi} dfrac{2}{1+q^2}
end{align*}$$

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

For $n=2$

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

For $n =3$

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

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