Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 0 down vote accepted

The gap in my argument above was purely conceptual and not computational. The Direct Mapping Theorem for the Mellin transform provides regularity results for the analytic continuation of the Mellin transform of a function.

In specific, consider the formula I proved integrating by parts:

$$\int_0^1x^r\log(x)^ndx= \frac{(-1)^nn!}{(r+1)^{n+1}}.$$

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.