Prove that probability density function is smooth There is a probability density function that depends on non-deterministic ($v$) and random ($x$) parameters:
$Pr(v)=\int_{G(v)} dP$,
where $G (v)$ is the "goal" region, the probability of getting into it must be determined and that, in turn, depends on $v$ and $x$.
It is known also that the function does not have jumps.
For example, we consider the normal distribution function:
$Pr(v)=\frac{1}{\sqrt{2\pi \sigma^2}}\int_{G(v)} e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$
So that:
$Pr(v)=\frac{1}{\sqrt{2\pi \sigma^2}}\int_{G(v)} e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx=\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty} I_{G(x)} (x) e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx=\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-2v}^{2v} e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$,
where $I (x)$ is the identity function that returns 1 or 0 depending on whether it hits or does not hit $G (v)$, $G (v) = [- 2v, 2v]$, $v = [1,2]$.
MY QUESTION IS: How can I show that this probability function is smooth? I know that to prove it I need to show that it is infinitely differentiable, but I do not see an elegant way to do it.
I will really appreciate your help! 
 A: From the hints above I conclude that I should use The Fundamental Theorem of Calculus (https://education.ti.com/html/t3_free_courses/calculus89_online/mod16/mod16_lesson3.html) here, which says that If $f$ is continuous on $[a, b]$, then the function $F(x)=\int_a^x f(t)dt$ has a derivative at every point in $[a, b]$, and the derivative is:
$\frac{d}{dx}F(x)=\frac{d}{dx}\int_a^xf(t)dt=f(x)$ (1)
That is, the derivative of the definite integral of the upper limit is the variable. This is true regardless of the value of the lower limit a. The function named F is the same as the area function that was previously explored.
So in my case:
I can take $-2v<=v_0<=2v$ such that:
$Pr(v)=\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-2v}^{2v} e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx=\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-2v}^{v_0} e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$
so using (1) we have that:
$Pr(v)=\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-2v}^{v_0} e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx=\frac{1}{\sqrt{2\pi \sigma^2}} e^{-\frac{(v_0-\mu)^2}{2\sigma^2}}$,
which is smooth, derivatives of all order exists and we can conclude that $Pr(v)$ is also smooth.
