Fourier transform of $ \int_{-\infty}^{t} f(\eta )\text{d}\eta $ Suppose $f(t)$ and $F(\omega)$ are a Fourier transform pair. I want to show that $$\mathcal{F}^{-1} \left\{\frac{F(\omega)}{i\omega}\right\} = \int_{-\infty}^t f(\eta)\ \text{d}\eta$$
I start with the Fourier transform of the RHS and use integration by parts:
$$\begin{align*}
\mathcal{F}\left\{\int_{-\infty}^tf(\eta)\ \text{d}\eta \right\}\ &= \int_{-\infty}^{\infty} \int_{-\infty}^t f(\eta)\ \text{d}\eta\ e^{-i\omega t}\ \text{d}t \\&=\underbrace{\frac{-1}{i\omega}\left[\int_{-\infty}^t f(\eta)\ \text{d}\eta\  e^{-i\omega t}\right]_{t\ =-\infty}^{t\ =\ \infty}}_{=\ 0 ?}\ +\ \frac{1}{i\omega}\int_{-\infty}^{\infty}f(t)\ e^{-i\omega t}\ \text{d}t\\ &= \frac{F(\omega)}{i\omega}
\end{align*}$$
Hence $$\int_{-\infty}^t f(\eta)\ \text{d}\eta\ = \mathcal{F}^{-1} \left\{\frac{F(\omega)}{i\omega}\right\}$$
If the result is true I can see the first term on the RHS after performing the integration by parts must vanish but I'm not sure how to justify it. Any help on justifying it (or a cleaner approach to show the result) would be appreciated, thanks!
 A: One may recall that

$$
\mathcal{F}(g')(\omega)=i\omega\mathcal{F}(g)(\omega) \tag1
$$ 

applying it with
$$
g(t)=\int_{-\infty}^t f(\eta)\ \text{d}\eta, \quad g'(t)=f(t),\quad 
$$ the notation $F(\omega):=\mathcal{F}(f)(\omega)$, it gives
$$
\frac{F(\omega)}{i\omega} =\mathcal{F}\left( \int_{-\infty}^t f(\eta)\ \text{d}\eta\right)(\omega) \tag2
$$ that is 

$$
\mathcal{F}^{-1} \left\{\frac{F(\omega)}{i\omega}\right\} = \int_{-\infty}^t f(\eta)\ \text{d}\eta. \tag3
$$

Are you Ok with a proof of $(1)$ ?
A: Let $F(\omega)$ be the Fourier transform of the square integrable, continous function $f(t)$ as given by
$$F(\omega)=\int_{-\infty}^\infty f(t)e^{-i\omega t}\,dt$$
Let $I_L(\omega,)$ be defined by the integral
$$I_L(\omega)=\int_{-L}^L \int_{-\infty}^tf(t')\,dt'\,e^{-i\omega t}\,dt$$
Integrating by parts with $u=\int_{-\infty}^tf(t')\,dt'$ and $v=\frac{e^{-i\omega t}}{-i\omega}$ yields
$$I_L(\omega)=\frac{1}{i\omega}\int_{-L}^L f(t)e^{-i\omega t}\,dt+\frac{1}{i\omega}\left(e^{i\omega L}\int_{-\infty}^{-L}f(t')\,dt'-e^{-i\omega L}\int_{-\infty}^{L}f(t')\,dt'\right) \tag 1$$

Assuming that $\int_{-\infty}^t f(t')\,dt'$ is also a square integrable function, then we must have $$\lim_{L\to \infty}\int_{-\infty}^L f(t')\,dt'=0$$

Then, the boundary term on the right-hand side of $(1)$ vanishes in the limit as $L\to \infty$ and the limit of $I_L(\omega)$ becomes
$$\begin{align}
\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \left(\int_{-\infty}^t f(t')\,dt'\right)\,e^{-i\omega t}\,dt=\frac{1}{i\omega}F(\omega)}\end{align}$$
as was to be shown!
