I have this question here, which I am unsure of my final answer:
Firstly, I times the numerator and denominator by the conjugate 1-jt – resulting me in the alternate form of the function
Next, I've found the Fourier Transform of the first term using duality property
And lastly, I've found the Fourier Transform of the next term
As a result, the final Fourier Transform of x(t) would be 0. Have I gone wrong somewhere in my reasoning?
Best Answer
There is an important trick to solve such problems. If you have a standard transform pair and you just interchange the variables \$t\$ and \$\omega\$, then you can just use your table if you know the following:
$$\mathcal{F}\{f(t)\}=F(\omega)\Longrightarrow\mathcal{F}\{F(t)\}=2\pi f(-\omega)\tag{1}$$
This is a consequence of the fact that the Fourier transform and the inverse transform are essentially identical, apart from the factor \$2\pi\$ and a minus sign in the exponent. This is exactly what you see in (1): you get a factor of \$2\pi\$ and you have to invert the frequency variable \$\omega\$.
So in your case the standard transform pair is
$$\mathcal{F}\{e^{-t}u(t)\}=\frac{1}{1+j\omega}$$
from which you get using (1)
$$\mathcal{F}\left\{\frac{1}{1+jt}\right\}=2\pi e^{\omega}u(-\omega)\tag{2}$$
PS: The mistake in your calculations is in the last derivative. You compute the derivative as if you considered the function \$e^{-\omega}\$ instead of \$e^{-|\omega|}\$. The correct derivative is
$$\pi\text{sign}(\omega)e^{-|\omega|}$$
which gives the answer
$$X(\omega)=\pi e^{-|\omega|}(1-\text{sign}(\omega))=2\pi e^{-|\omega|}u(-\omega)=2\pi e^{\omega}u(-\omega)$$
which is of course identical to (2).