# I Deformation of contour of integration or shifting poles

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1. Feb 11, 2017

### Frank Castle

As I understand it, in order to compute a contour integral one can deform the contour of integration, such that it doesn't pass through any poles of the integrand, and the result is identical to that found using the original contour of integration considered. However, I have seen applications (particularly in QFT) where one considers an integral on the real axis, and instead of deforming the contour of integration, one shifts the poles of the integrand by a small positive/negative imaginary term in such a way that they don't pass through the contour. Is the justification for this that it is simply a convient trick that is equivalent to deforming the integration contour in the limit as the shift tends to zero and hence leads to the same result for the integral?

2. Feb 11, 2017

### stevendaryl

Staff Emeritus
Yes. Shifting the pole by $\epsilon$ makes no difference to the integral in the limit as $\epsilon \rightarrow 0$, 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.

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

3. Feb 12, 2017

### Frank Castle

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?!