# Complex analysis integration. Strange result

1. Nov 10, 2012

### McLaren Rulez

Hi,

Consider the real variable $x$ and some real constant $x_{0}$. I want to integrate

$$\int_{-\infty}^{\infty}\frac{x_{0}}{x_{0}-x}$$

This blows up when the denominator is zero but we can still take the principal value of the integral. That is, we notice that the integral is an odd function around $x_{0}-x = \epsilon$ and $x-x_{0} = \epsilon$ so we ignore the integral from $-\epsilon$ to $\epsilon$.

The integral can now be done by contour integration. We take the upper semicircle as shown in the attached image with a small semicircle around our singularity. So here I have my first question:

1) Is it true that the integral along the curve $\Gamma$ is zero? How can I prove it? The ML inequality didn't work for me.

Anyway, assuming it is zero, we see that our real integral is just the negative of the integral around the small semicircle near $x_{o}$. And we can work that integral out to be

$$\int dz\frac{x_{0}}{x_{0}-z}$$

Expressing $z=x_{0}+\epsilon e^{i\theta}$ we get $dz = i\epsilon e^{i\theta}$

This makes the integral

$$\int_{\pi}^{0} d\theta i \epsilon e^{i\theta}\frac{x_{0}}{-\epsilon e^{i\theta}} = i\pi x_{0}$$

That is $$\int_{-\infty}^{\infty}\frac{x_{0}}{x_{0}-x} = -i\pi x_{0}$$

And that raises another question

2) Why is the integral of a real function giving me a complex number as the result? How did the i get in there?

Thank you very much for your help :)

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2. Nov 10, 2012

### lurflurf

This is a common thing. Once complex numbers are involved they tend to show up. The integral does not exist in the usual sense, so a new definition is made. This new integral might as well be complex as any thing else. Similar things happen when a real number is approximated by a complex one. We might have 2~2.001+.001i, in some sense this is a bad approximation, while in another sense it is good.

3. Nov 11, 2012

### haruspex

There are two semicircles in the path. According to my calculation their integrals cancel, leaving zero for the portion on the real line. That makes sense to me since the function is antisymmetric about the centre of arc of the semicircles.
Also, it's not clear to me that this procedure makes the integral $$\int_{-\infty}^{\infty}\frac{x_{0}}{x_{0}-x}$$ meaningful. You may be able to evaluate $$lim_{ε→0, R→∞} (\int_{-R}^{-ε}\frac{x_{0}}{x_{0}-x}+\int_{ε}^{R}\frac{x_{0}}{x_{0}-x})$$, but it's a stretch to claim that's an evaluation of $$\int_{-\infty}^{\infty}\frac{x_{0}}{x_{0}-x}$$.

4. Nov 11, 2012

### McLaren Rulez

Thank you both for the replies.

I was assuming that the larger semicircle would somehow integrate to zero (as they usually do for contour integration). Is that not true then? That was my first question.

Also, let's say I just wanted
$$lim_{ε→0, R→∞} (\int_{-R}^{-ε}\frac{x_{0}}{x_{0}-x}+\int_{ε}^{R}\frac{x_{0}}{x_{0}-x})$$

Are you saying this is zero?

5. Nov 11, 2012

### jackmell

It's equivalent to looking at:

$$\lim_{\epsilon\to 0}\left(\int_{-1}^{\epsilon} \frac{1}{x} dx+\int_{\epsilon}^1 \frac{1}{x}dx\right)$$
$$=\lim_{\epsilon\to 0} \left(\log(x)\biggr|_{-1}^{-\epsilon}+\log(x)\biggr|_{\epsilon}^{1}\right)$$
$$=\lim_{\epsilon\to 0}\left(\ln|\epsilon|+\pi i-(\pi i)-\ln|\epsilon|)\right)$$
$$=0$$

6. Nov 12, 2012

### McLaren Rulez

I see.

This is strange because this result is part of the integration to calculate a Lamb shift. Maybe I am making a mistake somewhere because the answer should not be zero.

Thank you for the help.

7. Nov 12, 2012

### Mute

You didn't include the contribution from the large semi-circle of radius R. It does not tend to zero as the radius is taken to infinity. Its contribution will cancel out the contribution from the small semi-circle. The principle value of the integral is thus zero.

It is a standard way of calculating the principal value of an integral. The limit version you wrote down is essentially the definition (though what you wrote has a typo - the limits should be $x_0 - \epsilon$ and $x_0 + \epsilon$, rather than $\mp \epsilon$), and is how such divergent integrals are often treated in calculations which require a sensible answer.