Fractional derivatives of Gamma function

Question

For integer $n \geq 0$, we have $\dfrac{d^n}{ds^n} x^s = (\ln x)^n \,x^s$. From this it follows, for example, that $$\int_0^{\infty}e^{-x}\ln^n x \,dx= \Gamma^{(n)}(1)$$ Question: is there a way of defining fractional derivatives in which it makes sense to say that $$\int_0^{\infty}e^{-x}\ln^q x \, dx= \Gamma^{(q)}(1) $$ for rational $q$, and are there values of $q$ for which this can be evaluated in terms of ordinary special functions?

Answer 1

One way to check the answer to see if it holds by induction. Using $\frac{d^n}{dx^n}\frac{d^k}{dx^k}=\frac{d^{n+k}}{dx^{n+k}}$, and assuming it is continuous as a function of $n,k$, then you should be able to verify that your answer works.

Probably the easiest and neatest method, and it works great for the simple problems.

---

Second and more rigorous approach would be to use a particular definition for what $\frac{d^t}{dx^t}$ means for $t\notin\mathbb N$. One approach can be given by

$$\frac{d^t}{dx^t}f(x)=\frac1{\Gamma(1-\{t\})}\frac{d^{\lceil t\rceil}}{dx^{\lceil t\rceil}}\int_{-\infty}^x\frac{f(x)}{(x-t)^{\{t\}}}~\mathrm dt$$

In particular, this should give

$$\frac{d^t}{dx^t}e^{ax}=a^te^{ax}$$

However, as we cannot directly exchange the integrals, it appears that $\int_0^\infty t^{s-1}e^{-t}(\ln(t))^x~\mathrm dt$ is only a regularization of $\Gamma^{(x)}(s)$ in this sense.