I read some proofs of Fourier transform of Heaviside function on this site but I don't really understand (because I haven't learned about distribution yet). I'm trying to derive it myself based on what I learned so far.
I use the following definitions of Heaviside and Dirac delta function:
$$H(t)= \lim_{a\to 0^{+}} H_a(t) \text {, where } H_a(t) = \begin{cases} e^{-at}, & t \geq 0 \\ 0, & \text{otherwise} \end{cases}$$ $$\delta(v)= \lim_{a\to 0^{+}} \delta_a(v) \text {, where } \delta_a(v)= \begin{cases} \frac {1}{a}, & |v| \leq \frac{a}{2} \\ 0, & \text{otherwise} \end{cases}$$ Now, we have: $$\mathcal {F}\{H(t)\}=\lim_{a\to 0^{+}}\mathcal {F}\{H_a(t)\}=\lim_{a\to 0^{+}} \int_{0}^{\infty} e^{-at}e^{-2i \pi vt}dt = \lim_{a\to 0^{+}} \frac {1}{a+2i \pi v}$$ $$=\lim_{a\to 0^{+}} \frac {a-2i \pi v}{a^2+4 \pi^2 v^2} =\lim_{a\to 0^{+}} \frac {a}{a^2+4 \pi^2 v^2}+\lim_{a\to 0^{+}} \frac {-2i \pi v}{a^2+4 \pi^2 v^2}$$
The second term corresponds to $\frac {1}{2i \pi v}$. But why does the first term correspond to $\frac {1}{2} \delta(v)$?
The second term is one form of a nascent delta function. It is in fact the Poisson kernel in the upper half plane, which converges to the Dirac delta impulse:
$$\delta(x)=\lim_{a\rightarrow 0^+}\frac{1}{\pi}\frac{a}{a^2+x^2}$$