Why is $\frac{d}{dx}\int^\infty_{-\infty}f(x) dx=0$? I came up to this derivative of improper integral:
$\frac{d}{dx}\int^\infty_{-\infty}f(x) dx$
and wolfram alpha tells me that it is zero. I am confused. Could someone please give a proof ot some reference of a proof?
Thank you!

An integral is essentially a summation over infinitely thin $\text{d}x$ strips for your function $f(x)$.  To illustrate that point, I can write approximately:
$$\int_{-\infty}^{\infty}f(x)\text{d}x \approx \sum_{-\infty}^{\infty}{f(x)}\Delta x = A$$
where $\Delta x$ is a small width, and where $A$ is some number that isn't  a function of $x$.  The figure below illustrates what an integral does - it takes an infinite number of those infinitely thin rectangles and sums them over the specified range (which is why the integral under a curve corresponds to the area).

Since you're summing over a range of specified $x$ (in this case the entire real domain of $x$), the output is no longer a function of $x$, and so its derivative is $0$.
If however the limits did have some $x$ dependence, then the answer would be different, though strictly, you should change the "dummy" variables that appear inside the integral, which disappear when the integral is solved, e.g.
$$\int_{-g(x)}^{g(x)}f(x')\text{d}x'$$
