It´s a very well know result that $$e^{-x}=\frac{1}{2\pi i}\int\_{\gamma-i\infty}^{\gamma+i\infty} \Gamma(s)x^{-s}ds$$
In order to solve this integral we have to close the contour to the left and show that the integral over this path vanishes as $R\rightarrow\infty$. Unfortunately it´s not obvious to me how to proof it.
 All textbooks that I came across assume that the integral goes to zero without any proof.

What I have tried so far is the following:
I assume that we are closing this contour with a semi circle to the left, and I get the following integral: 
Let $s=Re^{i\theta}$ and $ds=iRe^{i\theta}d\theta$,
$$\int\_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \Gamma(Re^{i\theta})x^{-Re^{i\theta}}iRe^{i\theta}d\theta$$
Using Stirling´s approximation for the Gamma function we have
$$\Gamma(Re^{i\theta})\sim\sqrt{2\pi} e^{-Re^{i\theta}}(Re^{i\theta})^{Re^{i\theta}-\frac{1}{2}}$$
and $$\int\_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \Gamma(Re^{i\theta})x^{-Re^{i\theta}}iRe^{i\theta}d\theta\sim\ \sqrt{2\pi} iR \int\_{\frac{\pi}{2}}^{\frac{3\pi}{2}} e^{-Re^{i\theta}}(Re^{i\theta})^{Re^{i\theta}-\frac{1}{2}}x^{-Re^{i\theta}}e^{i\theta}d\theta$$

I took the modulus of the last integral and arrived in the following expression:

$$\sim \sqrt{2 \pi} \int\_{\frac{\pi}{2}}^{\frac{3\pi}{2}} e^{Rcos(\theta)(log(R)-log(x)-1)}e^{\frac{1}{2}log(R)}e^{-R \theta sin(\theta)}d\theta$$

But I don´t know how to evaluate the $\lim$ as $R \to \infty$

I really appreciate if someone could show this for me.
Thank you in advance.