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In summary, when computing a contour integral, one can deform the contour of integration or shift the poles of the integrand by a small imaginary term in order to avoid passing through the poles. This is justified by the fact that in the limit as the shift tends to zero, the result is equivalent to deforming the contour. However, if the pole is shifted from inside to outside the contour, there will be a discrete difference in the result. Therefore, shifting the poles is analogous to deforming the contour as long as they do not pass through each other.

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Frank Castle said:

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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stevendaryl said: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.

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?!

Deformation of contour of integration involves changing the path of integration in a complex plane. This is typically done to simplify the calculation of a complex integral by avoiding singularities or branch points.

Sometimes, the original path of integration may be difficult or impossible to solve using traditional methods. Deforming the contour allows us to avoid these difficulties and ultimately obtain a solution.

Deforming the contour does not change the value of the integral as long as the new contour does not enclose any additional poles or singularities. This is known as the Cauchy-Goursat theorem.

Shifting poles refer to the process of moving a pole (singularity) of a complex function to a different location in the complex plane. This can be done to simplify the calculation of a complex integral by avoiding the pole or choosing a better contour for integration.

Yes, when deforming the contour, it is important to ensure that the new contour does not cross any branch cuts or enclose any additional poles or singularities. Additionally, the new contour should still be closed and the original integrand should remain analytic within the new contour.

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