Let $0\le a \le 1$ and $0\le a_1 \le 1$ and $L \in {\mathbb Z}$. Consider a following contour integral: \begin{equation} S_L(a,a_1) := \oint z^{L+1} \cdot \frac{1}{1-(1-a) z}\cdot \frac{1}{1-(1-a_1) z}\cdot \frac{1}{z-(1-a)}\cdot \frac{1}{z-(1-a_1)} \frac{d z}{2 \pi \imath} \end{equation} where the contour is a circle of radius one centered at the origin. Now, by expanding each of the inverse linear functions in an infinite series and then integrating the resulting series term by term and then by carefully carrying out the resulting sum using elementary methods we have found that: \begin{equation} S_L(a,a_1) :=\frac{(a_1-2) a_1 (1-a)^{\left| L\right| +1}+(2-a) a (1-a_1)^{\left| L\right| +1}}{(a-2) a (a_1-2) a_1 (a-a_1) (-a a_1+a+a_1)} \end{equation} Now, the question is is it possible to show the identity in question in some other simpler way, for example using the Cauchy' residue theorem?
Closed form expression for a contour integral
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contour-integration
1 Answers
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In this case the answer is actually trivial. There are two poles each one of first order inside the contour of integration one at $z=(1-a)$ and another one at $z=(1-a_1)$. Denote the integrand of the contour integral as $f(z)$. Then from the Cauchy' residue theorem we have: \begin{equation} S_L(a,a_1) = \left.(z-(1-a)) f(z)\right|_{z\rightarrow 1-a} + \left.(z-(1-a_1)) f(z)\right|_{z\rightarrow 1-a_1} \end{equation} which, after a straightforward calculation, gives the expression on right hand side .