confusion about 1 over t
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A short working note by Phil (dated 4.5.13) from the Spectral Theory Book folder of incorporated material. It tries three ways to Fourier transform x(t)=1/t: contour closing for a generalized FT giving 2πi θ(-ω), principal value plus or minus iπδ(t), and integrating straight through the pole to get -iπ sgn(ω). It says the material was later moved into Appendix C and the summary.
AI-written summary; may contain errors.
Extracted text (machine-read; may contain errors)
Paradox on computing FT of 1/t PhL 4.5.13
This stuff now appears in Appendix C (c) and summary (g). I did get a verification of the sgn formula as shown below. The "paradox" is that it depends whether you use regular or generalized FT.
I am unable to compute the Fourier Transform of x(t) = 1/t
Method 1
Start off with
X(ω) = !Syntax Error, Idt x(t) e-iωt projection = transform (1.1)
Try this for x(t) = 1/t. In the generalized FT, I would run the contour below the pole.
For ω > 0, we can close down because -iωt = -iω(-i∞) = -ω∞ . There is no pole, so get 0.
For ω < 0 we can close up, we get the pole CCW so get 2πi. The result is then
X(ω) = 2πi θ(-ω)
This is like the FT of the Heaviside step situation.
Method 2
Use the actual FT, so we have
X(ω) = !Syntax Error, Idt (1/t) e-iωt projection = transform (1.1)
This is undefined, so we try if using
= pf(1/t) ± iπδ(t)
Then we find
X(ω) = !Syntax Error, Idt e-iωt = dt (1/t) e-iωt ± !Syntax Error, Idt iπδ(t) e-iωt
Then the integral is easy to do
dt (1/t) [ cos(ωt)- isin(ωt) ] = -i dt (1/t) sin(ωt)
= -2i !Syntax Error, Idt (1/t)sin(ωt) = -2i (π/2) signum(ω)
= -iπ signum(ω) ω is real
Then we get
X(ω) = -iπ signum(ω) ±iπ
This result makes no sense at all to me, I don't even know what it means.
Method 3
If I just assume the integral goes right through the pole I get
X(ω) = dt (1/t) e-iωt = -iπ signum(ω)
Then the Fourier Transform of 1/(πt) is given by
FT [1/(πt)] = -iπ signum(ω)
This seems to agree with my PDF source