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Cauchy P.V. of an Improper Integral

  1. Jun 21, 2006 #1
    I was doing a Fourier Transform Integral, and was wondering if it would be legitimate for me to choose a semicircle CR on the lower half-plane below the real axis rather than choosing a semicircle CR on the upper half-plane above the real axis. I would expect it to be valid because the contour integral should have equal areas regardless of whether it is oriented above or below the Real axis. Is this right to expect?
  2. jcsd
  3. Jun 22, 2006 #2


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    It's not clear whether you're asking about the Fourier transform or its inverse. I presume your interest is in the inverse transform.

    You can decide which contour to used based on whether [itex]e^{i \omega t}[/itex] diverges as [itex]\omega[/itex] goes to [itex]i \infty[/itex] or [itex]-i \infty[/itex] and that depends on the sign of t.
  4. Jun 22, 2006 #3


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    You should not think in terms or *area*. You really are doing a *line* integral. So it's not a question of whether the area is the same, it's the question of whether the contribution *along* the upper and lower contours is the same. What you must ensure is that the contribution along the infinite semi-circle vanishes. As Tide pointed out, this is determined by the sign of the exponent in omega t. For example, if you have [itex] e^{-i \omega t} [/itex], if you close with a contour above the real axis, as t goes to [itex] + i \infty [/itex], the exponent will go to [itex] e^{ + \infty} [/itex] which means that the contribution along that contour will blow up! So you *must* close below in that case.
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