Fractional integration

Remarks

If $f \in L^p(\R^n)$, an element of the Lp space, with $1 \le p < n/\alpha$, the integral converges absolutely for almost every $x$. With the Fourier transform $\hat f(\xi) = \int f(x) e^{-2\pi i x \cdot \xi} \, dx$, the constant $\gamma(\alpha)$ is chosen so that $\widehat{I_\alpha f}(\xi) = (2\pi\abs{\xi})^{-\alpha} \hat f(\xi)$, for example for Schwartz functions $f$. In this sense $I_\alpha = (-\Delta)^{-\alpha/2}$, and $I_\alpha I_\beta = I_{\alpha + \beta}$ when $\alpha + \beta < n$.

Example: $n = 3$ and $\alpha = 2$. Then $\gamma(2) = \pi^{3/2} \cdot 4 \cdot \Gamma(1)/\Gamma(1/2) = 4\pi$, since $\Gamma(1/2) = \sqrt{\pi}$. So $I_2 f(x) = \frac{1}{4\pi}\int_{\R^3} \frac{f(y)}{\abs{x - y}} \, dy$ is the Newtonian potential, and $-\Delta I_2 f = f$ for smooth $f$ with compact support.

Non-example: the order $\alpha = n$ fails the condition $0 < \alpha < n$. Then the kernel $\abs{x - y}^{0}$ is the constant $1$, so the formula gives $\int_{\R^n} f(y) \, dy$, which does not depend on $x$ and smooths nothing. Also $\Gamma((n - \alpha)/2) = \Gamma(0)$ is undefined, so the normalization breaks down.