Let $f$ be a smooth integrable function as you commented. Then we can define the Fourier Transform as
$$\mathcal{F}[f](y) := \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)e^{-iyx}\mathrm{d}x$$
We can also derive the inverse of Fourier Transform as:
$$\mathcal{F}^c[f](y):=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)e^{+iyx}\mathrm{d}x$$
Now lets see what happen wen we apply both to a function
$$f(y) = \mathcal{F}^c[\mathcal{F}[f]](y) = \frac{1}{2\pi}\int_\mathbb{R}\left(\int_{\mathbb{R}} f(\omega)e^{-i\omega x}\mathrm{d}\omega \right)e^{+iyx}\mathrm{d}x = \int_{\mathbb{R}}f(\omega)\left(\int_{\mathbb{R}}\frac{1}{2\pi}e^{-i(\omega-y)x}\mathrm{d}x\right)\mathrm{d}\omega$$
We can see in this last equality that the function in brakets acts as a Dirac Delta. So we can relate this as de Dirac as a notation (because the Dirac Delta just have formal sense in distribution theory)
$$\delta(\omega-y) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{-i(\omega-y)x}\mathrm{d}x$$
This is an intuitive constructive way of seen that the Dirac is related to Fourier Transforms. Now for your question we can use the definition on how an Dirac Delta acts
$\hat{\delta}(\omega) = \frac{1}{(2\pi)^{1/2}}\int \delta(t-t_0)e^{-i\omega t}\mathrm{d}t = \frac{1}{(2\pi)^{1/2}}e^{-\omega t_0}$
Where we just used the above
$$f(y) = \int f (\omega)\delta(\omega-y)\mathrm{d}\omega$$
For $f(t):=e^{-i\omega t}/\sqrt{2\pi}$
Best Answer
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!