Question

fourier transform of e^-2x
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Answer 1

Step-by-step explanation:

Caveat: I'm using the normalization f^(ω)=∫∞−∞f(t)e−itωdt.

A cute way to to derive the Fourier transform of f(t)=e−t2 is the following trick: Since

f′(t)=−2te−t2=−2tf(t),

taking the Fourier transfom of both sides will give us

iωf^(ω)=−2if^′(ω).

Solving this differential equation for f^ yields

f^(ω)=Ce−ω2/4

and plugging in ω=0 finally gives

C=f^(0)=∫∞−∞e−t2dt=π−−√.

I.e.

f^(ω)=π−−√e−ω2/4.

Thanks yes that seems familiar could you explain how you get to the step ................ taking the Fourier transfom of both sides will give us

iωf^(ω)=−2if^′(ω).

– S F

Those should be familiar "rules" for Fourier transforms: The Fourier transform of f′(t) is iωf^(ω) and the FT of tf(t) is −if^′(ω). If they are not familiar, they follow fairly easily from the definition of the Fourier transform. – mrf

Furthermore why does e^-infinity - e^ infinity = square root(pi)

Hope this helps you to gain Knowledge in Fourier transform

Thank you