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Consider the real variable [itex]x[/itex] and some real constant [itex]x_{0}[/itex]. I want to integrate

[tex]\int_{-\infty}^{\infty}\frac{x_{0}}{x_{0}-x}[/tex]

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 [itex]x_{0}-x = \epsilon[/itex] and [itex]x-x_{0} = \epsilon[/itex] so we ignore the integral from [itex]-\epsilon[/itex] to [itex]\epsilon[/itex].

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 [itex]\Gamma[/itex] 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 [itex]x_{o}[/itex]. And we can work that integral out to be

[tex]\int dz\frac{x_{0}}{x_{0}-z}[/tex]

Expressing [itex]z=x_{0}+\epsilon e^{i\theta}[/itex] we get [itex]dz = i\epsilon e^{i\theta}[/itex]

This makes the integral

[tex]\int_{\pi}^{0} d\theta i \epsilon e^{i\theta}\frac{x_{0}}{-\epsilon e^{i\theta}} = i\pi x_{0}[/tex]

That is [tex]\int_{-\infty}^{\infty}\frac{x_{0}}{x_{0}-x} = -i\pi x_{0}[/tex]

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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# Complex analysis integration. Strange result

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