Integral involving power of logarithm

Question

I was wondering if we can compute the following integral:

$$ I = \int_{0}^{1}{e^{\alpha y} y^{\beta} (\ln y)^m {\rm d}y} $$

where $m \in \mathbb{N}$, $\alpha > 0$, $\beta>0$.

Answer 1

Using the series expansion for $e^x$ we obtain

$$\begin{align} I=&\int_0^1y^\beta \ln^m y \sum_{k=0}^\infty \frac{(\alpha y)^k}{k!}\,dy\\ &=\int_0^1\sum_{k=0}^\infty \frac{y^\beta \ln^m y (\alpha y)^k}{k!}\,dy\\ &=\sum_{k=0}^\infty\int_0^1 \frac{y^\beta \ln^m y (\alpha y)^k}{k!}\,dy\\ &=\sum_{k=0}^\infty \frac{e^{i \pi m} \alpha^k m! (\beta+k+1)^{-m-1}}{k!}\\ &= (-1)^m m!\sum_{k=0}^\infty\frac{\alpha^k}{k!}(\beta+k+1)^{-m-1}\\ &=\frac{m!(-1)^m}{(\beta+1)^{m+1}} \, _{m+1}F_{m+1}(\beta+1,\beta+1,\dots, \beta+1;\beta+2,\beta+2,\dots, \beta+2;\alpha)\\ \end{align}$$

I'm not very familiar with hypergeometrics and you may be able to simplify it further.

Answer 2

The classical reference "Table of Integrals, Series and Products" by Gradshtein and Ryzhnik (Academic Press 2nd printing, 1981) contains this formula :

(4.358) $\int_1^{\infty}x^{\nu-1}e^{-\mu x}(ln(x))^m dx=\dfrac{d^m}{d\nu^m}[\mu^{-\nu}\Gamma(\mu,\nu)] \ , m=0,1..., Re \mu >0, Re \nu >0.$

A further change of variable X=1/x is needed to get the appropriate bounds of integration.

(Such integrals are usually computed by complex integration and/or derivation with respect to a parameter under the integration symbol).