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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 $\frac {1}{2} \delta(v)$?

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  • $\begingroup$ The first important thing about distributions is that they are not functions. They only 'live' under intervals. The definition for your $\delta$ doesn't properly reflect this. Which kind of limit are you using in the definition of your functions? What does the equality for the $\delta$ mean? Why can you exchange the limit and the integration? $\endgroup$
    – Roland
    Commented Dec 13, 2015 at 20:07
  • $\begingroup$ I've just edited the definitions. The $\delta(v)$ is equal $+\infty$ if $v=0$, and $0$, otherwise, so I think it's correct by defining as above. I didn't exchange the limit and the integration, I calculated the integral first, then plugged the limit in. $\endgroup$
    – SiXUlm
    Commented Dec 13, 2015 at 20:21
  • $\begingroup$ Ah I see what you mean by exchanging the limit and the integration. You are correct. I haven't found the reason. $\endgroup$
    – SiXUlm
    Commented Dec 13, 2015 at 20:24
  • $\begingroup$ If you regard $\delta$ as a function, it's almost everywhere zero. Thus, every integral containing it will be zero. This is not the behaviour you're expecting from the $\delta$ functional. $\endgroup$
    – Roland
    Commented Dec 13, 2015 at 23:15

1 Answer 1

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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:

$$\delta(x)=\lim_{a\rightarrow 0^+}\frac{1}{\pi}\frac{a}{a^2+x^2}$$

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