Cauchy P.V. of an Improper Integral

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SUMMARY

The discussion centers on the legitimacy of choosing a semicircular contour in the lower half-plane versus the upper half-plane for evaluating a Fourier Transform Integral. It is established that the choice of contour depends on the behavior of the integrand, specifically whether the term \( e^{i \omega t} \) diverges as \( \omega \) approaches \( i \infty \) or \( -i \infty \). The key takeaway is that the contribution along the infinite semicircle must vanish, which is determined by the sign of the exponent in \( \omega t \). For instance, using \( e^{-i \omega t} \ necessitates closing the contour below the real axis to avoid divergence.

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Go37Pi
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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?
 
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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.
 
Go37Pi said:
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?

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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