As stated in the title, let $m\geq k$ be natural numbers ($k$ possibly zero) and let $p(z)=\prod\_{j=0}^m(z-\lambda\_j)$ where $\lambda\_j\in\mathbb R$ for all $j$. We have by the residue theorem that
\begin{align}
&\frac{1}{2\pi i}\lim\_{R\rightarrow\infty}\int\_{\lvert z\rvert=R}\frac{z^k}{p(z)}\mathrm d z=\sum\_{j=0}^m\mathrm{Res}\left(\frac{z^k}{p(z)};\lambda\_j\right)\\&=\sum\_{j=0}^m \lambda\_j^k\prod\_{l\neq j}(\lambda\_j-\lambda\_l)^{-1}.
\end{align}

Now the thing claimed (which I am not able to prove) is that 

\begin{align}
\frac{1}{2\pi i}\lim\_{R\rightarrow\infty}\int\_{\lvert z\rvert=R}\frac{z^k}{p(z)}\mathrm d z=
\begin{cases}
1&\text{ if }k=m,\\
0&\text{ if }k<m.
\end{cases}
\end{align}

This would then imply that $a\_j=\prod\_{l\neq j}(\lambda\_j-\lambda\_l)^{-1}$ is the solution to the system of equations

\begin{align}
\sum\_{j=0}^m \lambda\_j^k a\_j=
\begin{cases}
1&\text{ if }k=m,\\
0&\text{ if }k<m.
\end{cases}
\end{align}
which is needed in order to establish the theorem cited below.

*Wagner, Peter*, [**A new constructive proof of the Malgrange-Ehrenpreis theorem**](http://dx.doi.org/10.4169/193009709X470362), Am. Math. Mon. 116, No. 5, 457-462 (2009). [ZBL1229.35007](https://zbmath.org/?q=an:1229.35007).