Question

How can I find the Fourier transform of 1/(t+2)^2?

Answer 1

`bbbF [ 1/(t+2)^2] = - omega pi e^(2 i omega) sgn(omega)`

Explanation:

There are various forms, so specifically using this definition:

• `bbbF_t[f(t)] (omega) = int _(-oo)^oo f(t) e^(-i omega t) dt = F(omega)`

With Fourier pairs written:

• `f(t) harr F(omega)`

Start by evaluating `bbbF[1/t]`, then the rest follows.

This is well-known Fourier pair, and easy to evaluate:

• `sgn(t) harr -(2 i)/omega`

The significance is the duality property of the Fourier transform:

• `f(t) harr F(omega) implies F(t) harr 2 pi f(-omega)`

So:

• `bbbF [ -(2 i)/t] = 2 pi sgn(-omega) = - 2 pi sgn(omega)`

By linearity :

• `bbbF [ 1/t] = 1/(2 i) 2 pi sgn(omega) = - i pi sgn(omega)`

From the derivative property:

`bbbF [f'(t)] = i omega F(omega)`

`implies bbbF [- 1/t^2] = i omega * ( - i pi sgn(omega) )= omega pi sgn(omega)`

By linearity :

`implies bbbF [ 1/t^2] = - omega pi sgn(omega)`

We want:

`implies bbbF [ 1/(t+2)^2]` so finish with the shift property:

• `bbbF[f(t-a)] = e^(- i a omega) F(omega)`

With `a = -2`:

`bbbF [ 1/(t+2)^2] = - omega pi e^(2 i omega) sgn(omega)`