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I'm trying to evaluate what seems to be a straightforward contour integral:

$$I=\int_{\gamma} \frac{dz}{\alpha + \beta z} $$

where $\gamma (t) = e^{-it}$, $t \in \left[ 0,\pi\right]$, $\alpha, \beta \in \mathbb{C}$, and $|\alpha| \ne |\beta|$. Explicitly,

$$I=\int_{0}^{\pi} \frac{(-ie^{-it})dt}{\alpha + \beta e^{-it}}$$ $$=-i\int_{0}^{\pi} \frac{dt}{\alpha e^{it} + \beta }$$

I want to say that

$$I=\frac{1}{\beta}\left(\log\left(\frac{\beta -\alpha}{\beta +\alpha}\right)-i \pi \right)$$

but I feel like I'm neglecting some subtle branch cut issues. Any help or insight would be greatly appreciated.

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1 Answer 1

up vote 3 down vote accepted

You do have to be careful to use a branch of the log that is analytic in a neighbourhood of the contour, which in this case is the semicircle in the lower half plane from $1$ to $-1$. Let's assume $\beta \ne 0$ (the case $\beta = 0$ being very easy). The integrand has a pole at $z = -\alpha/\beta$, which is not on the contour since $|\alpha| \ne |\beta|$. Now $\frac{\log(\alpha + \beta z)}{\beta}$ is an antiderivative, but where to put the branch cut? The branch cut must be a curve from the branch point $z = -\alpha/\beta$ to $\infty$. If you use the branch of log whose imaginary part is from $\tau$ to $\tau + 2 \pi$, the branch cut has $\alpha + \beta z = r e^{i\tau}$, i.e. $z = (-\alpha + r e^{i\tau})/\beta$, for $0 < r < \infty$. If $\beta = b e^{i\theta}$ with $b > 0$, that extends out from $-\alpha/\beta$ in the direction of argument $\tau - \theta$. You have to make sure that doesn't hit the semicircle. For example, if $-\alpha/\beta$ is in the blue region shown below, $\tau - \theta = \pi/2$ would be safe, while in the red region $\tau - \theta = -\pi/2$ would be safe.

Thus with $\alpha = 1$ and $\beta = -2 i$, $-\alpha/\beta = -i/2$ is in the blue region, $\theta = -\pi/2$, and we can take $\tau = 0$: an antiderivative is $F(z) = -\frac{i}{2} \log(1 - 2 i z)$ where the imaginary part of log is between $0$ and $2 \pi$. The integral is then $$ \eqalign{ F(-1) - F(1) &= \frac{i}{2} \left(\log(1 + 2 i) - \log(1 - 2 i)\right)\cr &= \frac{i}{2} \left( \frac{\ln 5}{2} + \arctan(2) i - \frac{\ln 5}{2} - (2 \pi - \arctan(2)) i \right)\cr & = \pi -\arctan(2)\cr}$$

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Thanks! Great explanation. –  Jackson Walters Apr 13 '12 at 15:47

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