I know that $f(t) = \int_{-\infty}^\infty \frac{e^{i\omega{}t}}{1+i\omega}d\omega = 2\pi e^{-t}h(t)$ with $h(t)$ the Heaviside function.
To me it says that the inverse Laplace transform of the simple RC low-pass transfer function is the expected impulse response.
How can I proof this formula, preferably using elementary properties of integration?
Is there an intuitive argument why there is a discontinuity at 0?
EDIT:
By simple arguments I get $f(t) = \frac1{2\pi} \int_{-\infty}^\infty \frac {\cos(\omega{}t) + \omega\sin(\omega{}t)}{1+\omega^2} d\omega$.
So the question breaks down to: How to proove that $\int_{-\infty}^\infty \frac{\cos(\omega{}t)}{1+\omega^2} d\omega = \operatorname{sgn}(t)\int_{-\infty}^\infty \frac{\omega\sin(\omega{}t)}{1+\omega^2} d\omega = \pi{}e^{-|t|}$?
The easier proof that I can find for your question is just to observe that the result of the integral if transformed with Fourier transform operator is exactly what you have in the integral without the term $e^{it\omega}$. In fact what you see in that formula is the theorem of inversion of Fourier transform
$\frac1{2pi}\int^\infty_{-\infty}X(\omega)e^{it\omega}dw = \frac{f(x^+)+f(x^-)}2$
Your formula says that $\frac1{1+i\omega}$, what I called $X(\omega)$, is the Fourier antitrasform of $e^{-t}h(t)$.
The function $e^{-t}h(t)$ is not continous in the origin just for the definition of Heaviside step function. Simply it is defined as 1 if x>=0 and 0 if x<0. In a few words it's a line of height 1 from 0 to infinity, a line of height 0 from 0 to less infinity. It's the obvious that is not continous in 0 because the limit in 0 has two different values if evalueted from the right and from the left. I hope I coul helo you.