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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