I have been assigned to make the Fourier transform of the following equation:
$$\frac{d g(t)}{d t}+A g(t)=B f(t)$$
where:
$$f(t)=\left\{\begin{aligned}
0, & \quad t<0 \\
e^{i \omega t}, & \quad t \geq 0
\end{aligned}\right.$$
Making the Fourier transform of the left-hand side gives:
$$\begin{array}{l}
g(t)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \hat{g}(\omega) e^{i \omega t} d \omega \Rightarrow \frac{d g}{d t}=\frac{1}{2 \pi} \int_{-\infty}^{\infty} i \omega \hat{g}(\omega) e^{i \omega t} d \omega \\
\Rightarrow F[g(t)]=\hat{g}(\omega), \quad F\left[\frac{d g}{d t}\right]=i \omega \hat{g}(\omega)
\end{array}$$
However, the Fourier transform of $f(t)$ does not seem to converge. $f(t)$ is supposed to represent an oscillating field and I am doing the Fourier transform to find an expression for $g(t)$ "after the transient term".
Best Answer
$$f(t)= e^{i \omega t}1_{t >0}=\lim_{a\to 0^+}f_a(t), \qquad f_a(t)=e^{(i \omega-a) t}1_{t >0}$$ The Fourier transform of $f_a$ converges without trouble $$F[f_a](w) = \frac1{a-i(w-\omega)}$$ But $f$ is not $L^1$ nor $L^2$ and its Fourier transform integral diverges.
The problem is to understand $\lim_{a\to 0^+} F[f_a]$ which converges in the sense of distributions but not in $L^2$ nor in $L^1$.
For a Schwartz function $\phi$ then $$\lim_{a\to 0^+} \int_{-\infty}^\infty \frac1{a-i(w-\omega)}\phi(w)dw=\lim_{h\to 0^+} \int_{-\infty}^{-h}+\int_h^\infty \frac{\phi(w)}{-i(w-\omega)}dw+\pi\phi(\omega)$$ Thus $$F[f](w) = PV(\frac1{-i(w-\omega)})+\pi \delta(w-\omega)$$ ($PV$ for principal value)