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    Contour integration to solve improper integrals

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    Use the contour shown (Fig. 7.11, attached) with R--> infinity to prove that

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    © BrainMass Inc. brainmass.com December 24, 2021, 11:46 pm ad1c9bdddf
    https://brainmass.com/math/complex-analysis/contour-integration-solve-improper-integrals-600736

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    SOLUTION This solution is FREE courtesy of BrainMass!

    We need to calculate the improper integral
    (1.1)
    We can rewrite the integrand:
    (1.2)
    And now we replace it by the complex number z:
    (1.3)
    And we are going to integrate over the contour:

    The problem is that we have a removable singularity at the origin, which is part of the path that we want to integrate over.
    Instead we divide the closed path into four parts.
    (1): The path from to
    (2): The path along the semi-circle centered about the origin of radius e
    (3): The path from to R
    (4): The enclosing path that closes the contour via and
    So we get:
    (1.4)
    If we convert this to complex integration:
    (1.5)
    And we can close the contour with path (4):
    (1.6)
    Where is path number (2) - along the semicircle of radius
    Now, since there is no pole inside this closed contour, the sum of these integrals is zero (remember that the contour goes around the singularity), hence:
    (1.7)
    Along the contour the integrand disappears since as we have
    (1.8)
    So by the estimation principle the path integral along is zero.

    Around the semicircle we can write:
    (1.9)
    And:
    (1.10)
    Since is arbitrarily small we can simply say that
    Then:
    (1.11)
    So:
    (1.12)
    And therefore:

    (1.13)

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

    © BrainMass Inc. brainmass.com December 24, 2021, 11:46 pm ad1c9bdddf>
    https://brainmass.com/math/complex-analysis/contour-integration-solve-improper-integrals-600736

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