Derivation of integration property of Fourier transform For $f \in \mathcal{S}(\mathbb R)$, where $\mathcal{S}$ is a Schwartz space, denote its Fourier transform as
$$
\hat{f}(\omega) := \mathcal{F}f(t) := \int_{-\infty}^{\infty} f(t) e^{-2 \pi i \omega t} dt
$$
and corresponding inverse Fourier transform as
$$
\mathcal{F}^{-1} \hat{f}(\omega) := \int_{-\infty}^{\infty} e^{2 \pi i t \omega} \hat{f}(\omega) d\omega
$$
I would like to know what is the formula for:
$$
\mathcal{F} \left[ \int_{-\infty}^t f(\tau) d\tau \right]
$$
What I have so far is:
\begin{alignat}{2}
 f(\tau) &= \mathcal{F}^{-1}\hat{f}(\omega) \qquad \implies \\
 \int_{-\infty}^t f(\tau) d\tau &= \int_{-\infty}^t \mathcal{F}^{-1}\hat{f}(\omega) d\tau \\
  &= \int_{-\infty}^t \left[ \int_{-\infty}^\infty e^{2 \pi i \tau \omega} \hat{f}(\omega) d\omega \right] d\tau \\
  &= \int_{-\infty}^\infty \hat{f}(\omega) \int_{-\infty}^t e^{2 \pi i \tau \omega} d\tau\, d\omega \\
  &= \int_{-\infty}^\infty \hat{f}(\omega) (2 \pi i \omega)^{-1}\left[ e^{2 \pi i \tau \omega} \right]_{-\infty}^t d\omega \\
\end{alignat}
and then I have a problem that the limit
$$
\lim_{\tau \to -\infty}e^{2 \pi i \tau \omega}
$$
is not well defined.
I wonder how did they get the corresponding result on this page:
 A: Simple way is to calculate the Fourier transform of the distribution $\theta(x)$ and apply convolution theorem. To get the first part, it is convenient to replace $\theta (x)$ by $\theta(x) e^{-\epsilon x}$ and take the limit $\epsilon \to 0$ through positive values only after calculating the Fourier transform. This is correct because Fourier transformation is continuous in the space of distributions with weak-* topology.
A: *

*When $f \in S(\mathbb{R})$ and $\int_{-\infty}^\infty f(t) dt = 0$ then $\int_{-\infty}^t f(\tau)d\tau \in S(\mathbb{R})$. 
Integrating by parts $$\mathcal{F}[\int_{-\infty}^t f(\tau)d\tau](\xi) = \int_{-\infty}^\infty\int_{-\infty}^t f(\tau) d\tau e^{-2i \pi \xi t} dt = \int_{-\infty}^\infty f(t) \frac{e^{-2i \pi \xi t}}{2i \pi \xi} dt = \frac{1}{2i \pi \xi} \mathcal{F}[ f(t)](\xi)$$

*Otherwise $\mathcal{F}[\int_{-\infty}^t f(\tau)d\tau]$ exists only as the Fourier transform of a tempered distribution, and the result becomes $$\mathcal{F}[\int_{-\infty}^t f(\tau)d\tau](\xi) = \frac{1}{2i \pi \xi} \mathcal{F}[ f(t)](\xi)+  \frac{\int_{-\infty}^\infty f(t) dt}{2}\delta(\xi)$$ 
