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A paper provides the following derivation: say you have to solve:

$ \displaystyle R \sim - \frac{1}{2}\int_{-\infty}^{\infty} \mathrm{d}w \frac{dp/dw~e^{2iw}}{p(w)}$

Then if we have poles at $w=w_j$, each will contribute

$\displaystyle R_j = -\frac{1}{2}e^{2iw_j} \oint dw \frac{dp/dw}{p(w)}$

Then the following case is studied: the pole $w_j$ in p'/2p arises from points $x_j$ in the complex x-plane where $p=0$ and $x_j$ is a simple zero of $p$. What I thought was that in this case we simply expand $$p (w) = p'(w_j)(w-w_j)$$ which cancels the $p'$ in the numerator of the fraction, so that $R_j \sim -\frac{1}{2} e^{2iw_j} \oint \frac{1}{w-w_j}$,trivially giving $$R_j \sim - \pi ie^{2iw_j}$$ However, the author says that $ R_j \sim - \frac{\pi i}{3}e^{2iw_j} $.

My question is: how did he get to this result?

I am not sure if it matters but $p$ is supposed to be $p=\sqrt{2m(E-V(x))}$ and $w = \int_{x_0}^x p(x') dx'$.

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up vote 1 down vote accepted

I figured it out: near one of the poles $x_j$ where $p=0$, I expanded to first order $$V(x) = E - F(x-x_j)$$ so that $$w = \int_{x_j}^x p(x') \mathrm{d}x' = \sqrt{2m} \int_0^x \sqrt{F(x-x_j)}$$ Thus, $w = \sqrt{2mF}(x-x_j)^{3/2}\frac{2}{3} = \frac{2}{3\sqrt{2mF}}p^3$ $$ p = \left(\frac{3}{2}\sqrt{2mF}w\right)^{1/3} $$ $$ p' = \frac{1}{3}\left(\frac{3}{2} \sqrt{2mF}w \right)^{-2/3} \frac{3}{2}\sqrt{2mF}$$ $$ \frac{p'}{p} = \frac{\frac{1}{3}\left(\frac{3}{2} \sqrt{2mF}w \right)^{-2/3} \frac{3}{2}\sqrt{2mF}}{\left(\frac{3}{2}\sqrt{2mF}w\right)^{1/3}} $$ $$ = \frac{1}{2}\frac{\sqrt{2mF}}{\frac{3}{2} \sqrt{2mF}w} $$ $$ \frac{p'}{p} = \frac{1}{3w}$$

Therefore, the proposed integral above becomes:

$$ R \sim -2\pi i\frac{1}{2}e^{2iw_j}\frac{1}{3} = \frac{-\pi i}{3}e^{2 iw_j} $$

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