I would like to solve that integral and I think the easiest (only reasonable???) solution is to find the residues. The contour is large enough to contain both singularities. So to recap, I would like to find the residues at the two essential singularities of the following function
$$f(x)=\exp\Big(\frac{A}{x-1}\Big) \cdot \exp\Big(\frac{B}{x-2}\Big)$$
Since the situation is symmetric the solution for one residue should suffice for now, so let us focus on the first singularity $x=1$. One Ansatz I have in mind is the following:
$$ f(x)=\Big(1+\frac{A}{x-1} + \frac{A^2}{2! (x-1)^2} + ...\Big)\cdot \exp\Big(\frac{B}{x-2}\Big) \\ =\Big(1+\frac{A}{x-1} + \frac{A^2}{2! (x-1)^2} + ...\Big)\cdot \Big(\sum_{n=0}^{\infty} \frac{[d^n/dx^n\exp(B/(x-2))]_{x=1}}{n!} \cdot (x-1)^n\Big) \\ \rightarrow \mathrm{res}_{x=1} f(x) =\sum_{n=0}^{\infty}\frac{[d^n/dx^n\exp(B/(x-2))]_{x=1} \cdot A^{n+1}}{n! (n+1)!} $$
so the first term is expanded just following the definition of the exponential function, and the second term is Taylor-expanded around $1$ to have factors of $(x-1)$ in order to cancel the respective factors in the first factor. Then all the terms giving a $\frac{1}{x-1}$ are added, which result in the final infinite sum.
Is this result correct? I think not, but I cannot find an error in the above construction. Any help would be greatly appreciated!!!
If $\gamma$ is a contour winding once around both singularities, then by Cauchy's integral theorem we have
$$\int_{\gamma} f(z)\,dz = \int_{\lvert z\rvert = R} f(z)\,dz$$
for every $R > 2$. Choosing $R$ large, the expansion
\begin{align} f(z) &= \Biggl(1 + \frac{A}{z-1} + O\biggl(\frac{1}{(z-1)^2}\biggr)\Biggr)\Biggl(1 + \frac{B}{z-2} + O\biggl(\frac{1}{(z-2)^2}\biggr)\Biggr) \\ &= 1 + \frac{A}{z-1} + \frac{B}{(z-2)} + O(z^{-2}) \\ &= 1 + \frac{A}{z} + \frac{B}{z} + O(z^{-2}) \end{align}
and the standard estimate (ML inequality) together show
$$\int_{\gamma} f(z)\,dz = 2\pi i(A + B) + O(R^{-1}),$$
so, since the left hand side is independent of $R$,
$$\int_{\gamma} f(z)\,dz = 2\pi i(A+B)$$
follows.