# Proof of the Direct mapping Theorem for Mellin transform.

I cannot understand an integration in the proof of the Direct mapping Theorem for the Mellin transform. A statement of the Theorem, together with an outline of the standard proof, can be found at page 15 of this book.

After decomposing the Mellin transform into a sum, the following integral, which the author declares to be "easily computable", has to be examined:

$$\int_0^1 \sum_{k,b} c_{k,b} x^{s+b-1} log(x)^k dx, \qquad b\in\mathbb{R}, k\in\mathbb{N},$$

And $s\in\mathbb{C}$ is bounded from below, but in principle it could be $\Re(s)<-b$. (And $\Re(s)<-b$ indeed happens, for example applying this Theorem to obtain a meromorphic extension for the Gamma function.)

Integrating by parts I proved:

$$\int_0^1 x^rlog(x)^n = \frac{(-1)^n n!}{(r+1)^{n+1}}, \qquad \forall \:r\in\mathbb{R}_{\geq -1},\:n\in\mathbb{N}.$$

But it seems to me that he claims this result to hold for any $r\in\mathbb{R}$.

Question Do you see a gap in my arguments? Or do you know where to find a detailed proof of the Direct mapping Theorem for Mellin transform?

Thank you very much!

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$$\int_0^1x^r\log(x)^ndx= \frac{(-1)^nn!}{(r+1)^{n+1}}.$$
The left hand side makes sense only for $r\geq-1$, but the right hand side is well defined for any $r\in\mathbb{R}$ (properly for any $r\in\mathbb{C}$). This actually defines the analytic continuation of $\int_0^1 x^r\log(x)^n dx$ to the whole complex plane. And we can use the equality for any $r\in\mathbb{R}$ as the Author of the book claims.