Contour integral of $\frac{e^{-ux}}{Ax^2+Bx+C}$

Question

I'm having a little trouble figuring this one out. I've found the poles to be at $$\frac{-B\pm\sqrt{B^2-4AC}}{2A}$$ Am I right to assume the residues can be found by this? $$\text{Res}\bigg[ \lim_{s\to \frac{-B+ \sqrt{B^2-4AC}}{2A}}\frac{(s+\frac{-B+ \sqrt{B^2-4AC}}{2A})e^{-ux}}{s(s+\frac{-B+ \sqrt{B^2-4AC}}{2A})(s-\frac{-B- \sqrt{B^2-4AC}}{2A})} \\ +\lim_{s\to \frac{-B- \sqrt{B^2-4AC}}{2A}}\frac{(s-\frac{-B+ \sqrt{B^2-4AC}}{2A})\cdot e^{-ux}}{s(s+\frac{-B+ \sqrt{B^2-4AC}}{2A})(s-\frac{-B- \sqrt{B^2-4AC}}{2A})}\bigg]$$ My only problem is this seems a bit excessive. Is there an alternative, simpler way to find the residue here?

Answer 1

Use that

$$\operatorname{Res} \left(\frac{f(z)}{g(z)}; z_0\right) = \frac{f(z_0)}{g'(z_0)}$$

Given $z_0$ is a simple pole and $f(z_0)\neq 0$.

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Suppose that we have $$f(z) =\frac{e^{-uz}}{Az^2+Bz+C}$$ If we have $z_0$ and $z_1$ are the poles then

$$\sum_{i \in \{0,1\}}\mathrm{Res}(f,z_i)= \frac{e^{-uz_0}}{2Az_0+B}+\frac{e^{-uz_1}}{2Az_1+B}$$