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- Thread starter Frank Castle
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Yes. Shifting the pole by [itex]\epsilon[/itex] makes no difference to the integral in the limit as [itex]\epsilon \rightarrow 0[/itex], EXCEPT in the case where the pole is changed from being inside the contour to outside the contour. In this latter case, shifting the pole makes a discrete difference that doesn't vanish in the limit.

The most basic case is [itex]\int_{-\infty}^{+\infty} \frac{1}{z} dz[/itex]. You can argue that that integral is zero, since [itex]\frac{1}{z}[/itex] is odd, and so the integral for the region [itex]z < 0[/itex] cancels the integral for [itex]z > 0[/itex]. But if you shift the pole by [itex]\pm i \epsilon[/itex], you get [itex]\pm \pi i[/itex]. Shifting the pole by [itex]+i \epsilon[/itex] gives the same result as making a little counterclockwise semicircular detour around the pole, and shifting the pole by [itex]-i \epsilon[/itex] gives the same result as a clockwise detour.

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So is it essentially analogous to deforming the contour, in the sense that one can deforming the contour as long as one doesn't pass through the poles. In this case, one can shift the poles as long as they don't pass through the contour?!Yes. Shifting the pole by ϵϵ\epsilon makes no difference to the integral in the limit as ϵ→0ϵ→0\epsilon \rightarrow 0, EXCEPT in the case where the pole is changed from being inside the contour to outside the contour. In this latter case, shifting the pole makes a discrete difference that doesn't vanish in the limit.

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