Deriving the integration property of the Fourier Transform I want to derive the property of the Fourier Transform that states that if $X(j\omega) = \mathcal{F} (x(t))$ then $$\mathcal{F} \left( \int_{-\infty}^{t} x(\tau) \mathrm{d} \tau \right) = \frac{1}{j\omega} X(j\omega) + \pi X(0) \delta(\omega).$$
I started as follows, since $$x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) e^{j\omega t} \mathrm{d}\omega,$$ integrating both sides gives us $$\int_{-\infty}^{t} x(\tau) \mathrm{d} \tau = \int_{-\infty}^{t} \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) e^{j\omega\tau} \mathrm{d}\omega\mathrm{d} \tau$$ $$\int_{-\infty}^{t} x(\tau) \mathrm{d} \tau = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) \left(\int_{-\infty}^{t} e^{i\omega \tau} \mathrm{d}\tau \right) \mathrm{d}\omega.$$

How do I evaluate the inner integral inside the parantheses? My gut feeling says that it's related to the Fourier Transform of the constant function $y(t) = 1$ but I can't relate it to that since the upper limit is $t$ rather than $\infty$. How do I reach the result above?

Thank you in advance.
First, recognise the integral of $x(t)$
$$\int_{-\infty}^{t} x(\tau) d\tau \tag{1}$$
as a convolution with $u(t)$
$$ \int_{-\infty}^{t} x(\tau) d\tau =x(t) \star u(t) \tag{2}$$
where $u(t)$ is the unit-step function.
Then from the convolution property of the CTFT, one gets the following :
$$ \mathcal{F}\{ \int_{-\infty}^{t} x(\tau) d\tau  \} = \mathcal{F}\{ x(t)\star u(t) \} = X(\omega)U(\omega) \tag{3}$$ where $U(\omega)$ is the Fourier transform of the unit-step function which is $$U(\omega) = \frac{1}{j\omega} + \pi \delta(\omega) \tag{4}$$
Then the Fourier transform of the integral of $x(t)$ is
$$ \mathcal{F}\{ \int_{-\infty}^{t} x(\tau) d\tau \} = \frac{ X(\omega) }{j\omega} + \pi X(0) \delta(\omega) \tag{5}$$