Problem in residue calculation in case of essential singularity The function I'm dealing with is expressed as follows 
$$\psi\left(v\right)=\frac{1}{\left(v-\beta\right)\left(v+\beta\right)\left(v-\gamma_1\right)\left(v+\gamma_2\right)}\mbox{exp}\left[-\frac{A_1}{v-\beta}+\frac{A2}{v+\beta}\right]$$
where 
$$A_1=\frac{\beta^{2}}{2}\left[\alpha_1\beta+\alpha_2\right]$$
$$A_2=\frac{\beta^{2}}{2}\left[\alpha_1\beta-\alpha_2\right]$$
$\alpha_1$, $\beta$, $\gamma_1$ and $\gamma_2$ are all real and positive parameters while $\alpha_2$ is real but could be positive or negative or zero
I want to find the solution to the following integral 
$$I=\int_{-i\infty}^{i\infty}\dfrac{\psi\left(v\right)}{v}\;dv$$
In calculating this integral I need to calculate the residues of the poles on either the left or the right half plan of the complex variable $v$. I found difficulty getting the residues of the pole at either $v=\beta$ or $v=-\beta$ in the following cases:


*

*$\alpha_2>\alpha_1 \beta$, which in this case we have $A_1>0$ and $A_2<0$ and hence we have zeros in the exponential term at $v=\beta$ and $v=-\beta$ which results in that the residues at $v=\beta$ and $v=-\beta$ have zero values. I believe this might not be correct and I need help to verify this 

*$\alpha_2<-\alpha_1 \beta$, which in this case we have $A_1<0$ and $A_2>0$ and hence we have extra poles in the exponential term at $v=\beta$ and $v=-\beta$. In such case I'm not sure how could we calculate the residues at $v=\beta$ and $v=-\beta$
I appreciate the help in calculating the residues for the two cases I mentioned, or, if someone can think of other methods, other than the residue theorem, to calculate the integral $I$ that would be also a great help
 A: What removable singularity?
$v = \pm \beta$ are essential singularities, not poles.  However, we can still try to find the residues.   For convenience, let $s = v - \beta$.
$$\dfrac{1}{v(v-\beta)(v+\beta)(v-\gamma_1)(v+\gamma_2)} = \sum_{k=-1}^\infty a_k s^k $$
where 
$$
a_{-1} = {\frac {1}{ 2\left( -\gamma_{{1}}+\beta \right) {\beta}^{2}
 \left( \gamma_{{2}}+\beta \right) }}$$ 
and for $k \ge 0$,
$$ a_k = \dfrac{(-1)^{k+1}}{\gamma_1 (\beta + \gamma_1)(\gamma_2 + \gamma_1)(\beta - \gamma_1)^{2+k}} 
+ \dfrac{(-1)^{k+1}}{\gamma_2 (\beta - \gamma_2)(\gamma_2 + \gamma_1)(\beta + \gamma_2)^{2+k}}
+ \dfrac{(-1)^k}{\gamma_1 \gamma_2 \beta^{3+k}} + \frac{(-1)^k}{2^{2+k} \beta^{3+k} (\beta + \gamma_1)(\beta - \gamma_2)} $$
$$ \exp(A_2/(v+\beta)) = \sum_{m=0}^\infty \dfrac{A_2^m}{m! (v+\beta)^m} = \sum_{k=0}^\infty \sum_{m=0}^\infty \dfrac{A_2^m (m)_k}{(2\beta)^m m! k!} s^k$$
(where $(m)_k = m (m+1) \ldots (m+k-1)$ is a Pochhammer symbol)
$$ \exp(-A_1/(v - \beta)) = \sum_{k=0}^\infty \dfrac{(-A_1)^k}{k! s^k}$$
Unfortunately, we have a mixture of both positive and negative powers of $s$ here, so to get the residue you'll need a complicated sum of infinitely many terms.
