Let $k,m$ and $\rho$ be positive integers. In "Zur Approximation der Exponentialfunktion und des Logarithmus. Teil II"
Mahler considered the complex integrals $A\_k(x)=\frac1{2i\pi}\int\_{\mathcal C\_0}\frac{e^{zx}\mathrm dz}{\prod\_{h=0}^m(z+k-h)^\rho}$ where $\mathcal C\_0$ is a circle centered in $0$ with radius lesser than $1$. He asserted that $A\_k=\sum\_{l=0}^{\rho-1}\frac{c\_l}{l!}x^l$ with $c\_l=\frac1{2i\pi}\int\_{\mathcal C\_0}\frac{z^{l-\rho}\mathrm dz}{\prod\_{\substack{0\le h\le m\\ h\ne k}}(z+k-h)^\rho}$. I do not manage to prove this. Any answer would be welcome.



It is obvious for Mahler, not for me!
These are just the details of prets's comment. Note that
\begin{align\*}
\frac{e^{zx}}{z^\rho}&=\sum\_{l\ge0}\frac{(zx)^l}{z^\rho l!}=
\sum\_{l\ge0}\frac{z^{l-\rho}x^l}{l!}
=\sum\_{l=0}^{\rho-1}\frac{z^{l-\rho}x^l}{l!}+\sum\_{l\ge\rho}\frac{z^{l-\rho}x^l}{l!}\\
&=\sum\_{l=0}^{\rho-1}\frac{z^{l-\rho}x^l}{l!}+\sum\_{n\ge0}\frac{z^{n}x^{\rho+n}
}{(\rho+n)!}.
\end{align\*}


The left sum is what you want and the second is an entire function. Then you may apply Cauchy's integral theorem.


