I've been looking at Fourier transform, and got really confused when I looked for Fourier transform of $\sin(x)$.
$$\mathscr{F}_x[\sin(x)](\omega)=i\sqrt{\frac{\pi}{2}}(\delta(\omega-1)-\delta(\omega+1))$$
Which I understand as a zero function, with one peak heading to positive imaginary infinity at $\omega=1$ and one peak heading to negative imaginary infinity at $\omega=-1$. Even tho it is a little crazy, it makes sense.
What bothers me is the $\sqrt{\frac{\pi}{2}}$ coefficient. Why is it even neccesary, if the function has either infinite value, or zero value? How does it "not get lost", if it has no effect on the function value?
The delta function doesn't have a "value" at the spike, because it isn't a function in the conventional sense. Rather, the delta function is defined by the property that
$$\int_{-\infty}^\infty f(x) \delta(x-c) \, dx=f(c).$$
Scaling the delta function by a factor $\alpha$ then has the effect of
$$\int_{-\infty}^\infty f(x)(\alpha \delta(x-c)) \, dx=\alpha f(c).$$
Note that there are several common conventions for Fourier transform that differ in the scaling factor of $2\pi$. The scaling factor can be in front of one or the other of the Fourier transform or its inverse, or split as $\sqrt{2\pi}$ on both the transform and its inverse, or included in the complex exponential as $e^{\pm 2\pi i f t}$. You need to be careful to understand which convention you're using.