distributional derivative of $x^{-1/2}$.

Question

Consider the distribution defined for $\phi \in C_{c}(\mathbb{R})$ by $$T(\phi) = \int_{-\infty}^{\infty} |x|^{-1/2} \phi(x) dx.$$ Compute its derivative $T^{\prime}(\phi)$.

Attempt: I use $(T^{\prime}, \phi) = -(T, \phi^{\prime})$. How to remove its dominant singularity? I think the answer should be $-\frac{1}{2} |x|^{-3/2} + $ some terms...

Answer 1

Hint: Break the integral into two parts - from $0\to\infty$ and $-\infty\to 0$.

Answer 2

Note that $\dfrac{d}{dx} |x|^{-1/2} = - \text{sgn}(x) |x|^{-3/2}/2$ (where sgn is the signum function), not $- |x|^{-3/2}/2$.

$T'(\phi)$ is the Cauchy principal value of $- \displaystyle\frac{1}{2} \int_{-\infty}^\infty \text{sgn}(x) |x|^{-3/2} \phi(x)\; dx$