Is there a general method for calculating a contour integral when you have a pole on the contour?
For example, how do I integrate, $\frac{1}{z-1}$ over the unit circle centred at the origin?
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1If the order of the singularity is $\geq 1$ then the integral won't exist in the standard sense, though you could try to obtain its principal value. – Antonio Vargas Mar 21 '14 at 19:51
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Your contour would look something like this except you would have an indent at $(1,0)$. Since we are excluding the pole, by Cauchy's integral formula, the integral is equal to zero. Then we can write $$ \int_0^{2\pi}\frac{dz}{z-1} $$ as $$ \int_0^{2\pi}\frac{dz}{z-1} = \lim_{\epsilon\to 0}\biggl[\int_{\epsilon}^{2\pi-\epsilon}\frac{dz}{z-1}+\int_{\pi}^0\frac{i\epsilon e^{i\theta}}{(1+\epsilon e^{i\theta})-1}d\theta\biggr] = 0 $$ where I let $z=1+\epsilon e^{i\theta}$ in the second integral so $dz = i\epsilon e^{i\theta}d\theta$. Then we have \begin{align} \int_0^{2\pi}\frac{dz}{z-1}&= -i\lim_{\epsilon\to 0}\int_{\pi}^0\frac{\epsilon e^{i\theta}}{(1+\epsilon e^{i\theta})-1}d\theta\\ &= -i\lim_{\epsilon\to 0}\int_{\pi}^0d\theta\\ &= i\pi \end{align}
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I might be wrong but shouldn't the second integral inside you limit technically not be from 0 to $\pi$ but slightly off of those values because in both directions we have advanced slightly along the main circle. I guess the argument for the validity of the above is then it only gives order $\epsilon$ corrections which vanish with the limit. – Löwe Simon Sep 13 '16 at 18:17
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@LöweSimon it has been awhile since I looked at this but I believe it should have bounds $[\pi-\epsilon, \epsilon]$ which I suppose I neglected. Feel free to make any corrections if they are needed. – dustin Sep 13 '16 at 18:56
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Sir, In second equation you wrote the integral equal to $0$ and in last you wrote same integral equal to $iπ$? How? Second thing i would like to ask how $-i$ appears? – Akash Patalwanshi May 07 '20 at 13:27