Mathematica: Fourier Transform by hand of sin

[aimforclarity]

Hi,

i do not understand why i can find the FT of sin in mathematica using the built in function but not by integrating, even thoiugh they should be the same:

Integrate[Exp[i(ω-ω0) t,{t, -∞, ∞}, Assumptions ->ω0 el Reals && ω el Reals]

but the statement
FourierTransform[Sin[ω0 x], x, ω]
gives me the answer in dirac deltas
I Sqrt[\[Pi]/2] DiracDelta[ω - ω0] - I Sqrt[\[Pi]/2] DiracDelta[ω + ω0]

also i would be interested if you could explain to me from what condition the root 2pi appears before the FT, and how we can normalize plane waves to dirac Deltas.

Thanks a lot!

 

[NateD]

The Fourier Transform of a function f(x) is given by F(ω) = Integral[f(x) exp(-iωx) dx, from -∞ to ∞]. The built-in FourierTransform[f(x), x, ω] command in Mathematica is equivalent to this integral. When you use the FourierTransform command on a sine wave, it will return a result written as I Sqrt[π/2] DiracDelta[ω - ω0] - I Sqrt[π/2] DiracDelta[ω + ω0]. This is because the Fourier transform of a sine wave is actually two delta functions, one at ω = ω0 and one at ω = -ω0. The sqrt(π/2) factor appears because the Fourier transform is normalized so that F(0) = 1. This means that after taking the Fourier transform, the original sine wave is represented by two delta functions that have a total area of 1 (by the definition of a Dirac delta function). So the Fourier transform of a sine wave is always normalized by multiplying it by sqrt(π/2). To normalize a plane wave, first take its Fourier transform and then divide by the square root of its energy; this will give you a normalized Fourier transform.