# Fourier transform of complex exponential

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".

• $f(t) = e^{i\omega t}\theta(t)$ where $\theta$ is the Heaviside step function. Use a Fourier transform table to find the Fourier Transforms of these functions individually, then use the product to convolution property. – Ninad Munshi Dec 19 '20 at 16:42
• I tried this: $$F[f(t)]=\int_{-\infty}^{0} f(t) e^{-i \omega^{\prime} t} d \omega+\int_{0}^{\infty} f(t) e^{-i \omega^{\prime} t} d \omega=\int_{0}^{\infty} e^{i\left(\omega-\omega^{\prime}\right) t} d \omega=\frac{i}{\omega-\omega^{\prime}}$$ Why is this wrong? – JKalle Dec 19 '20 at 18:29
• @Raffaele $PV(\frac{i}{\sqrt{2 \pi } (\alpha +\omega )})$ – reuns Dec 19 '20 at 18:29
• You meant $\int_{0}^{\infty} e^{i\left(\omega-\omega^{\prime}\right) t} d t$ and you said so it doesn't converge for $\omega-\omega'$ real – reuns Dec 19 '20 at 18:30

$$f(t)= e^{i \omega t}1_{t >0}=\lim_{a\to 0^+}f_a(t), \qquad f_a(t)=e^{(i \omega-a) t}1_{t >0}$$ 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)