# Resideu theorem

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

https://brainmass.com/math/basic-calculus/resideu-theorem-441627

#### Solution Preview

Please see the attachment.

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:

...

#### Solution Summary

Resideu theorem is reiterated clearly in this case.