# Contour Integral from Peskin & Schroeder Intro to QFT

1. Aug 31, 2015

### rolltider

1. I'm having some trouble with some of the contour integrals covered in Chapter 2 of Peskin & Schroeder's Intro to QTF. I'm not so much as looking for answers to the integral (in fact, the answers are given in the textbook), but I was hoping someone could point me to some resources to use to become more familiar and confident with integrals like this as so far my understanding of it is very "hand-wavy."

2. Relevant equations: In section 2.4 of P&S, the propagator for the Klein Gordon field is derived and found to have the form: $$D(x-y) = \frac{-i}{2(2\pi)^2 r} \int_{-\infty}^{\infty}{dp \frac{p e^{ipr}}{\sqrt{p^2+m^2}}}$$

Now, I can readily see how this is a contour integral along the real p-axis with (isolated?) singular points at $$p=\pm im$$

But, I'm not familiar with the contour deformation done in the book:

3. My attempt: I'm not very clear on how pushing the contour up like that doesn't change the value of the integral, or why this is easier to do in practice than to do the integral along the real line. I suppose it makes sense that you'd want to wrap around one of the poles so that you can use the residue theorem to show that the part of the contour that wraps around the pole goes to zero, since the Residue at p=+/- im is zero (or I may be completely wrong about that) leaving only the nonzero part of the contour along the branch cut, but this only makes sense to me in a "hand-wavy" sort of way.

I did take complex analysis a long time ago as an undergrad, but either my professor wasn't that great or I just have a terrible memory because I don't remember doing anything like this. From what I recall, we used the book "Complex Variables and Applications" by Churchill but that book doesn't seem to prove entirely useful when trying to review integrals like this.

Any point in the right direction would be much appreciated!

2. Aug 31, 2015

### RUber

To do this, you are essentially saying that:
*edited*
$\int_{-\infty}^\infty f(p) dp = \int_{-\infty}^\infty f(p) dp + \lim_{r\to \infty} \int_{0}^\pi f(re^{i\theta})ire^{i\theta}d\theta +\lim_{r\to \infty} \int_{\pi}^0 f(re^{i\theta}) ire^{i\theta}d\theta$
Your function appears to go to zero as r goes to infinity, so the last integral can be set to zero, leaving you with a contour around the entire upper half plane.
From there, you know that an counter-clockwise closed contour integral is equal to zero if it contains no singularities, so you break the contour into one counterclockwise contour with no singularities and one clockwise contour around the area with singularities.

3. Aug 31, 2015

### rolltider

So, please correct me if I'm wrong, but it seems to me that by writing: $$\int_{-\infty}^{\infty}{f(p)dp} = \int_{-\infty}^{\infty}{f(p)dp} + \lim_{r\rightarrow\infty}\int_{0}^{\pi}{f(re^{i\theta})ire^{i\theta}d\theta} + \lim_{r\rightarrow\infty}\int_{\pi}^{0}{f(ire^{i\theta})ire^{i\theta}d\theta}$$

what you're doing is closing the contour from -infinity to infinity on the real p-axis with a semicircle over the upper-half on the plane and then subtracting the same circle so that the value of the overall integral stays the same? And then by rewriting the last term:

$$-\lim_{r\rightarrow\infty}\int_{0}^{\pi}{f(ire^{ir\theta})ire^{ire\theta}d\theta} = \lim_{r\rightarrow\infty}\int_{\pi}^{0}{f(ire^{ir\theta})ire^{ire\theta}d\theta}$$

You claim that it goes to zero as r goes to infinity. Now, why would the last term go to zero as r goes to infinity and not the second term, too? Aren't they the same terms save for a sign? That is, unless I'm completely misunderstanding your explanation.

4. Sep 1, 2015

### rolltider

Okay, I think I get it now after thinking about it some more:

So the integral $$\int_{-\infty}^{\infty}{f(p) dp}$$ can be closed above the p-axis by adding a term (and subtracting off the same term to make sure the integral remains unchanged): $$\int_{-\infty}^{\infty}{f(p) dp} = \int_{-\infty}^{\infty}{f(p) dp} + \lim_{r\rightarrow\infty}\int_{0}^{\pi}{f(re^{i\theta})ire^{i\theta} d\theta} + \lim_{r\rightarrow\infty}\int_{\pi}^{0}{f(re^{i\theta})ire^{i\theta} d\theta}$$

But since the integrand of both the new terms go to zero as r goes to infinity, you can set the third term to zero and keep the second term at no cost. Now, since I have a closed contour, I can break down the following contour like I drew here:

And since the first contour in the image does not enclose any singularities, that term is zero leaving only the second term, which is precisely the contour deformation done in Peskin & Schroeder.

Is that correct?

5. Sep 1, 2015

### RUber

That is how I understand it.

6. Sep 1, 2015

### TSny

Equivalently, but perhaps a bit shorter, is to consider just your middle figure. The integral around the entire contour is zero. Integrals along the arcs at infinity are zero. So, the integral along the real axis must be the negative of the integral along the "dipping" portion of the contour.

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7. Sep 1, 2015

### rolltider

That makes a lot of sense, too. So basically by flipping the direction of the "dipping" portion of the contour you make the two equivalent. Makes much more sense now, thanks!