Author Topic: Section 0201 quiz6-question 2B  (Read 211 times)

Houze Xu

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Section 0201 quiz6-question 2B
« on: March 18, 2021, 01:09:00 PM »
f(x) = e^−α|x|sin(βx)
Answer: f1(x) = e^−α|x|
f1^(k) = (2/pi)^1/2(α/α^2 + k^2)
f^(k) = F(f(x)) = -i/2(F(e^−α|x|e^iβx)-F(e^−α|x|e^-iβx))
F(e^−α|x|e^iβx) = (2/pi)^1/2(α/α^2 +(k^2- β^2)^2)
F(e^−α|x|e^-iβx))= (2/pi)^1/2(α/α^2 +(k^2+β^2)^2)
f^(k) = -i/(2pi)^1/2(α/α^2 +(k^2- β^2)^2 - α/α^2 +(k^2+β^2)^2 )