It is obvious for Mahler, not for me!

Question

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.

Answer 1

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.