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To evaluate an integral of the form $\int_{-\infty}^{\infty}f(x) dx$, where the rational function $f(x) =p(x)/q(x)$ is continuous on $(−∞, ∞)$, by residue theory we replace $x$ by the complex variable $z$ and integrate the complex function $f$ over a closed contour $C$ that consists of the interval $[−R, R]$ on the real axis and a semicircle $C_R$ of radius large enough to enclose all the poles of $ f(z) = p(z)/q(z)$ in the upper half-plane $Im(z) > 0$.

Doubt: I am confused why we consider semicircle $C_R$ in the upper half-plane $Im(z) > 0$ only. What would happen if instead of taking semicircle $C_R$ in the upper half plane we take in lower half plane $Im(z) < 0$? Similarly, why we can't consider full circle instead of considering semicircles.

Here is the image

I would be really grateful to you for your help.

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  • $\begingroup$ As $R$ increases, you will find that the contribution of the semi-circular arc in the contour above decreases. What remains behind is the integral over the real line of the function that you want. This is why we consider a semicircle. $\endgroup$ – астон вілла олоф мэллбэрг Nov 29 '16 at 3:35
  • $\begingroup$ For rational functions $\frac{p(x)}{q(x)}$ with $deg(q) > deg(p)$ you can take the lower or upper semi-circle the result will be the same, but for $\frac{p(x)}{q(x)}e^{i \omega x}$ if $Re(\omega) > 0$ you'll have no choice and only the upper semi-circle works $\endgroup$ – reuns Nov 29 '16 at 3:41
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The real line from $[-R,R]$ plus the semicircle $|z| = R$ creates a closed contour.

The evaluation of the integral over the closed contour can be evaluated at the enclosed poles.

As R gets to be big enough, every pole on the plane above the real line is enclosed inside the contour.

You hope (and must demonstrate) that as $R$ gets to be big enough the path integral over the semi-circle goes to $0.$

That would then mean that the integral along the real line equals the integral of the closed contour equals the evaluation of the residues.

If you wanted to, could you enclose the poles below the real line? Yes, you could. However, you are now traversing the contour in a clockwise direction. And that is going to flip the sign of the result.

Doing the full circle... in the example given, how does that create a closed contour with the real line?

You will get there. When you integrate from $0$ to $\infty$, but I don't want to spoil the surprise. (or explain in detail how to define the contours.)

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  • $\begingroup$ Thank you very much for the answer. It seems to me that if we are doing with the full circle then we will be having four curves, say $C_i$,$ i = 1,2 ,3 ,4$. Integration over $C_1$ and $C_2$ will be cancel over the real axis and other two integrations will be canceled over semi circle. As they are equal and in a opposite direction. Am I right? $\endgroup$ – srijan Nov 29 '16 at 5:17
  • $\begingroup$ That is the idea, the contour is a small circle or radius r, a large circle of radius R, and positive real axis from r to R, and a line just below the real axis from R back to r. As it turns out, the linear sections don't necessarily cancel out. $\endgroup$ – Doug M Nov 29 '16 at 17:03
  • $\begingroup$ Thank you very much for your clear and detailed reply. I am blessed. Thanks :) $\endgroup$ – srijan Nov 30 '16 at 8:09

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