Explore BrainMass
Share

Explore BrainMass

    Resideu theorem

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    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

    © BrainMass Inc. brainmass.com October 10, 2019, 3:55 am ad1c9bdddf
    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.

    $2.19