Integral of $z^n e^{az}$ along the negative real axis Let $n$ be a positive integer (or zero) and $a$ a real, positive number. Consider the contour show below and the integral
$$
\begin{equation}
I(a)=\int_{-\infty}^{(0^+)}\mathrm{d}z\:z^n \mathrm{e}^{az}.
\end{equation}
$$

Since there's no pole at the origin and along the axis the integral is well behaved, I'm inclined to say that the contribution is zero, but it still remains to add the contribution at infinity which I'm not sure how to obtain.
 A: Since $n$ is a positive integer and $a$ is a positive real number, I think you can just rewrite this integral in terms of the Gamma Function. I came to this just from familiarity of the Gamma Function.
The Gamma Function is defined as
$$
\Gamma(n+1) = \int_0^{\infty}dx\;x^ne^{-x}.
$$
When $n$ is a positive integer or $0$, one has $\Gamma(n+1) = n!$.
Now let's massage the Gamma Function to take on the form more familiar to what you have in the problem.
Let $x = -ay$ for some positive real number $a$, and $n$ a positive integer or $0$. Then
$$
\Gamma(n+1) = \int_0^{\infty}dx\;x^ne^{-x} = \int_0^{-\infty}dy\;(-a)(-ay)^ne^{ay} \\= (-1)a^{n+1}\int_0^{-\infty}dy\;y^ne^{ay}
= a^{n+1}\int_{-\infty}^0dy\;y^n e^{ay}.
$$
So
$$
\int_{-\infty}^0 dy\;y^ne^{ay} = \frac{\Gamma(n+1)}{a^{n+1}} = \frac{1}{a}\frac{n!}{a^{n}}.
$$
I am not sure if this answers your question or if you really want to evaluate Integral about the (open) Contour. If the Contour is really at the heart of what you want, then I suppose you can divide it up to three sections

*

*From $\infty \rightarrow 0$ with $y \rightarrow y-i\varepsilon$

*The semicircle from $-i\varepsilon \rightarrow i\varepsilon$ in the counter-clockwise direction

*From $0 \rightarrow \infty$ with $y \rightarrow y+i\varepsilon$
then take $\lim_{\varepsilon \to 0}$.

I can provide a full expression for the sum of the first and third bullet by expanding the product $(y\pm i\varepsilon)^n$, but suffice it to say it is strictly imaginary and proportional to $\varepsilon$
$$
(1) + (3) = \int_{-\infty}^0 dy\;e^{ay}\big[(y-i\varepsilon)^ne^{-ia\varepsilon} - (y+i\varepsilon)^ne^{+ia\varepsilon}\big] \\= -2i\times \text{Im}\int_{-\infty}^0 dy\;e^{ay}(y+i\varepsilon)^ne^{+ia\varepsilon}.
$$
Notice that $\lim_{\varepsilon \to 0}(1) + (3) = 0$.
The second section can be written in polar coordinates ($y\to \varepsilon e^{i\theta}$) as
$$
\int_{\mathcal{C}}dy\;y^ne^{ay} = 
\varepsilon^{n+2}\int_{\pi}^0d\theta \;e^{in\theta} \exp\{ae^{i\theta}\}.
$$
This also finite which can be computed by doing a Taylor expansion on $\exp\{ae^{i\theta}\}$.
Notice also that $\lim_{\varepsilon \to 0}(2) = 0$.
Therefore $\lim_{\varepsilon \to 0} (1) + (2) + (3) = 0$. Without taking this limit however, the expression is finite. Nor does it blow up.
