Non integer derivative of $1/p(x)$

I need to find the $k$'th derivative of $1/p(x)$, where $p(x)$ is a polynomial and $k\in\mathbb{R}$

It dosen't have to be an explicit formula, an algorithm which finds a formula for some $k$ is fine.

Solution 1:

Why don't you just use the formula? For generic $f$ and for 0<$\alpha$<1 you have: $$D^\alpha(f(x))=\frac{1}{\Gamma(1-k)}\frac{\partial}{\partial x}\int_0^x\frac{f(t)}{(x-t)^\alpha}\,\mathrm{d}t$$ So after deriving [k] (the integer part) times you have the remainder $\alpha\in[0,1]$: $$D^\alpha(D^{[k]}\left(\frac{1}{p(x)}\right))=\frac{1}{\Gamma(1-k)}\frac{\partial}{\partial x}\int_0^x\frac{1}{(x-t)^k}D^{[k]}\left(\frac{1}{p(x)}\right)\,\mathrm{d}t$$

Solution 2:

Let me sketch an idea for you, based on Fourier transforms: let $f$ be a function rapidly decreasing to infinity and let $\mathcal{F}(f)(y)=\frac{1}{\sqrt{2 \pi}} \int\mathbb{e}^{\mathbb{i}yx}f(x)$ denote its Fourier transform. Notice that $\mathcal{F}(f^{(n)})(y)=(-\mathbb{i})^n y^n \mathcal{F}(f)(y)$. Then $f^{(n)}=(-\mathbb{i})^n \mathcal{F}^{-1} (y^n \mathcal{F}(f))$. This means that, for those $\alpha \in \mathbb{R}$ for which the formula meakes sense, you can define $f^{(\alpha)}=(-\mathbb{i})^\alpha \mathcal{F}^{-1} (y^\alpha \mathcal{F}(f))$, where $(-\mathbb{i})^\alpha=\mathbb{e}^{\frac{3 \pi}{2} \mathbb{i} \alpha}$. Attention must be payed at the roots of $p$! Also, if you don't like the presence of $\mathbb{i}$, you can use Laplace's transform $\mathcal{L}(f)(y)=\int\mathbb{e}^{-yx}f(x)$, but then you'll have to restrict $p$ to just $(-\infty,0)$ or $(0, \infty)$.

Note that, unlike in the answers based on the Gammma function, $\alpha$ is not restricted to be in $(0,1)$.