Computing Fourier transform of power law I'm trying to compute the Fourier transform of 
$$f(\mathbf{r}) = \frac{1}{r^\alpha}$$
where $\mathbf{r} \in \mathbb{R}^n$. For sufficiently large $\alpha$, the Fourier transform exists. One well-known example in physics is the case $\alpha = n-2$, which is the Coulomb potential; its Fourier transform is $1/k^2$.
For a general $\alpha$, we have
$$\widetilde{f}(\mathbf{k}) \propto \frac{1}{k^{n-\alpha}}$$
by ‘physical arguments'. Specifically, since $f$ is rotationally invariant, $\widetilde{f}$ must be as well. Since $f$ is scale invariant, so must $\widetilde{f}$, so it must be a power law. The exponent of the power law can then be found by dimensional analysis.
I tried to do the calculation explicitly to get the proportionality constant, but it got very messy. How is that calculation done?
 A: Provided that $\alpha$ is in the right range, it is quite easy to prove that the Fourier transform is radial and has the scaling you calculate; it then reduces to the calculation of an integral to find the constant.
For $f(r)$ integrable and radial, the Fourier transform is also radial, because we can write
$$ \int_{\mathbb{R}^n} f(\lvert x \rvert) e^{-2\pi i k \cdot x} \, dx = \int_0^{\infty} f(r) r^{n-1} \left( \int_{S^{n-1}} e^{-2\pi i \lvert k \rvert r \cos{\theta}} \, dn \right) \, dr, $$
and the inside integral is a function of $r$. Which function? It turns out that it is close to a Bessel function; in fact, we have
$$ \int_{S^{n-1}} e^{-2\pi i a \cos{\theta}} \, dn = 2\pi a^{1-n/2} J_{n/2-1}(2\pi a), $$
which we can show by expanding the exponential in a power series and integrating term-by-term. Thus the Fourier transform of $r^{-\alpha}$ is
$$ \int_{0}^{\infty} r^{n-\alpha-1} 2\pi (\lvert k \rvert r)^{1-n/2} J_{n/2-1}(2\pi \lvert k \rvert r) \, dr, $$
and setting $u= \lvert k \rvert r$ gives the correct scaling. One would then evaluate $ 2\pi \int_{0}^{\infty} u^{n/2-\alpha} J_{n/2-1}(2\pi u) \, du $, but there is an easier way.

We have
$$ \frac{1}{r^{\alpha}} = \frac{2\pi^{\alpha/2}}{\Gamma(\alpha/2)} \int_{0}^{\infty} \lambda^{\alpha-1} e^{-\pi \lambda^2 r^2} \, d\lambda, $$
and the latter is easy to Fourier transform: interchanging the order of integration, we have
$$ \int_{\mathbb{R}^n} e^{-\pi \lambda^2 \lvert x \rvert^2} e^{-2\pi i k \cdot x} \, dx = \lambda^{-n}e^{-\pi \lvert k \rvert^2/\lambda^2} $$
Now,
$$ \frac{2\pi^{\alpha/2}}{\Gamma(\alpha/2)} \int_{0}^{\infty} \lambda^{\alpha-n-1} e^{-\pi \lvert k \rvert^2/ \lambda^2 } \, d\lambda = \frac{2\pi^{\alpha/2}}{\Gamma(\alpha/2)} \int_{0}^{\infty} \mu^{(n-\alpha)-1} e^{-\pi \lvert k \rvert^2 \mu^2 } \, d\mu \\
= \frac{2\pi^{\alpha/2}}{\Gamma(\alpha/2)} \frac{\Gamma((n-\alpha)/2)}{2\pi^{n/2-\alpha/2}} \frac{1}{\lvert k \rvert^{n-\alpha}} \\
= \frac{\pi^{\alpha-n/2}\Gamma((n-\alpha)/2)}{\Gamma(\alpha/2)} \frac{1}{\lvert k \rvert^{n-\alpha}}, $$
setting $\mu = 1/\lambda$, and the result holds if $0<\alpha<n$.
