I'm solving problems on the Gamma random variables and there is this question where it wants me to calculate the Mellin transform of sum of two independent Gamma variables from their moment generating function (basically the Laplace transform where kernel is $e^{sx}$ instead of $e^{-sx}$, i.e. $\mathbb E\left[ e^{sx}\right]$ where $x$ is your random variable) using the following formula

$$M(\alpha) = \frac{1}{2\pi i}\Gamma(\alpha)\int_{c-i\infty}^{c+i\infty} L(s)(-s)^{-\alpha} \, ds, \qquad \operatorname{Re}\{\alpha\}>0$$

where $c\in \mathbb{R}^+$ is some positive real valued constant. We know that the moment generating function for Gamma variable $x\sim \Gamma\left(k,\theta\right)$ is $$\mathbb E\left[ e^{sx}\right] = \frac{1}{(1-\theta s)^k}, \quad \left(1-\theta s\right)> 0$$

where $\theta , k \in \mathbb{R^+}$ are positive real valued parameters. Therefore for sum of two independent Gamma variables $x_1\sim \Gamma(k_1,\theta_1),x_2\sim \Gamma(k_2,\theta_2)$ we have

$$\mathbb E\left[ e^{s(x_1+x_2)}\right] = E\left[ e^{s(x_1)}\right] E\left[ e^{s(x_2)}\right] = \frac{1}{(1-\theta_1 s)^{k_1} (1-\theta_2 s)^{k_2}}, \quad (1-\theta_1 s)> 0, (1-\theta_2 s)>0$$

Therefore to get the Laplace transform first we have to apply $s\rightarrow-s$ to get

$$L(s) = \frac{1}{(1+\theta_1 s)^{k_1} (1+\theta_2 s)^{k_2}}, \quad (1+\theta_1 s)> 0, (1+\theta_2 s)>0$$

and then I have to calculate the following integral

$$M(\alpha) = \frac{1}{2\pi i}\Gamma(\alpha)\int_{c-i\infty}^{c+i\infty} \frac{(-s)^{-\alpha}}{(1+\theta_1 s)^{k_1} (1+\theta_2 s)^{k_2}} \, ds, \qquad \operatorname{Re}\{\alpha\}>0.$$

My question: I don't know how to calculate the last integral! If it was a simple pole or poles with integer power then I simply used partial fraction decomposition and used the Residue theorem, however the power values of $k_1,k_2$ are real numbers and the singularity is essential! I would appreciate if you can show me how to deal with this beast!

Thanks in advance!