# Time Integration property of the Fourier Transform

Given that $F(\omega)=\mathcal{F}[f(t)]$, then
$$\mathcal{F}\left[\int_{-\infty}^{t} f(t) d t\right]=\frac{F(\omega)}{j \omega}+\pi F(0) \delta(\omega) \tag{1}$$
that is, the transform of the integral of $f(t)$ is obtained by dividing the transform of $f(t)$ by $j \omega$ and adding the result to the impulse term that reflects the dc component $F(0)$. Someone might ask, "How do we know that when we take the Fourier transform for time integration, we should integrate over the interval $[-\infty, t]$ and not $[-\infty, \infty]$ ?" When we integrate over $[-\infty, \infty]$, the result does not depend on time anymore, and the Fourier transform of a constant is what we will eventually get. But when we integrate over $[-\infty, t]$, we get the integral of the function from the past to time $t$, so that the result depends on $t$ and we can take the Fourier transform of that.
If $\omega$ is replaced by 0 in Eq. (17.8), $$F(0)=\int_{-\infty}^{\infty} f(t) d t$$ indicating that the dc component is zero when the integral of $f(t)$ over all time vanishes. The proof of the time integration in Eq. (1) will be given later when we consider the convolution property.
For example, we know that $\mathcal{F}[\delta(t)]=1$ and that integrating the impulse function gives the unit step function. By applying the property in Eq. (1), we obtain the Fourier transform of the unit step function as
$$\mathcal{F}[u(t)]=\mathcal{F}\left[\int_{-\infty}^{t} \delta(t) d t\right]=\frac{1}{j \omega}+\pi \delta(\omega)$$