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Resideu theorem

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Show that integral from 0 to infinity of 1/(x^3+1)dx=2pi(sqrt3)/9
by integrating 1/(z^3+1) around circular sector 0<thetha<2/3pi and 0<r<p and let p go to infinity

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Solution Summary

Resideu theorem is reiterated clearly in this case.

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Problem:

Show that integral from 0 to infinity of 1/(x^3+1)dx=2pi(sqrt3)/9 by integrating 1/(z^3+1) around circular sector 0<2/3pi and 0.

Solution:

We have to compute the real integral
( 1)
using the residues theorem..
Let's consider the complex function
( 2)
and the contour (C) bounding the circular sector of angle 2/3 and a radius R which extends to infinity, as in the figure below:

The closed contour (C) includes 3 pieces:
...

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