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    Integral from 0 to infinity of x^3 sin(2x)/(x^2+1)^2

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    I need the solution of:
    integral from 0 to infinity of x^3 sin(2x)/(x^2+1)^2 dx

    © BrainMass Inc. brainmass.com March 4, 2021, 11:39 pm ad1c9bdddf
    https://brainmass.com/math/integrals/residue-theorem-442849

    Solution Preview

    Put

    I = integral from 0 to infinity of x^3 sin(2x)/[(x^2+1)^2] dx

    Then because the integer is an even function, the integral from minus infinity to infinity is twice as large:

    2 I = integral from -infinity to infinity of x^3 sin(2x)/[(x^2+1)^2] dx

    We can replace sin(2x) by the imaginary part of exp(2 i x) in here:

    2 I = integral from -infinity to infinity of x^3 Im[exp(2 i x)]/[(x^2+1)^2] dx

    = Im [ integral from -infinity to infinity of x^3 exp(2 i x)/[(x^2+1)^2] dx ]

    With the replacement of sin(2x) by Im[exp(2 i x)], we can now complete the contour by adding a half circle in the upper half plane, the integral along this circle will vanish in the limit of the radius going to infinity. Note that in some cases you can have a factor x in ...

    Solution Summary

    We explain how to compute this integral using contour integration methods.

    $2.49

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