Fourier transform properties (integration) proof From Signals and Systems _ Alan V. Oppenheim
There's a property of fourier transform states as below.
Fourier transform of $\int_{-\infty}^\tau x(\tau) d\tau $ equals to $\frac{ X(j\omega)}{j\omega} + \pi \delta(\omega)X(0)$
Can someone prove this?
 A: I know it is kind of late for answering the question, but it might help somebody else.
I would approach it using the convolution property and the Heaviside Step Distribution u(t).
First of all, notice that:
$$f(t)*u(t) = \int_{-\infty}^{+\infty}f(s)u(t-s)ds$$
Since, for $t-s < 0 \Longrightarrow s > t$, the integrand is zero, then:
$$f(t)*u(t) = \int_{-\infty}^{t}f(s)ds$$
Now, all that is left is to use the convolution property of the Fourier Transform:
$$\mathscr{F}\Big(\int_{-\infty}^{t} f(s)ds\Big) = \mathscr{F}(f(t)*u(t)) = F(\omega) U(\omega)$$
Since the fourier transform of the heaviside distribution is:
$$\mathscr{F}(u(t)) = \frac{1}{i\omega} + \pi \delta(\omega)$$
Then, we get:
$$\mathscr{F}\Big(\int_{-\infty}^{t} f(s)ds\Big) = F(\omega) \Big(\frac{1}{i\omega} + \pi \delta(\omega)\Big)$$
The trick here, is to see that, for all $\omega \neq 0$, the Dirac's Delta distribution is actually zero, so we take $F(0)$ instead of $F(\omega)$ for the "second product":
$$\mathscr{F}\Big(\int_{-\infty}^{t} f(s)ds\Big) = \frac{F(\omega)}{i\omega} + \pi F(0)\delta(\omega)$$
QED.
Hope that it helps!
