An exercise given by my complex analysis assistant goes as follows:
For $n \in \mathbb{N}$ and $x>0$ we define $$P_n(x) = \frac{1}{2\pi i} \oint_\Sigma \frac{\Gamma(t-n)}{\Gamma(t+1)^2}x^tdt$$ where $\Sigma$ is a closed contour in the $t$-plane that encircles the points $0,1, \dots, n$ once in the positive region.
Now, we have to prove some things like that $P_n(x)$ is a polynomial of degree $n$, which is no problem. However, the assistant claims that $P_n(x)$ "is known as a Laguerre polynomial". However, calculating $P_n(x)$ for certain $n$ and comparing to values of the Laguerre polynomials $L_n(x)$ on the internet, I find that $$(-1)^n \cdot n! \cdot P_n(x) = L_n(x)$$ Could my assistant have made a typo which explains this missing factor; or could he mean that $P_n(x)$ have similar properties as the Laguerre polynomials, perhaps?
All you need to do is to multiply the integral representation by $(-1)^n n!$, so you will have the right integral representation
$$ P_n(x) = \frac{(-1)^nn!}{2\pi i} \oint_\Sigma \frac{\Gamma(t-n)}{\Gamma(t+1)^2}x^tdt. $$
For instance,
$$ P_5(x) = 1-5\,x+5\,{x}^{2}-\frac{5}{3}\,{x}^{3}+{\frac {5}{24}}\,{x}^{4}-{\frac {1}{120 }}\,{x}^{5},$$
which agrees with the Laguerre polynomial $L_5(x)$.