We learnt about the following statement:
Corollary: Assume that $(X_1, \dots, X_n)$ has density $p(x_1, \dots, x_n) = \prod_{i=1}^n q_i(x_i)$ where $q_i$ is positive and measurable on $\mathbb{R}$. Then the random variables $X_1, \dots, X_n$ have density $p_i = c_i q_i$ for some $c_i > 0$.
Now I'm looking at the following exercise where this statement was used.
Exercise (I'm mentioning only the relevant part):
$(U,V)$ has density $\frac{1}{\sqrt{4\pi}}e^{-\frac{u^2}{4}}\frac{1}{\sqrt{4\pi}}e^{-\frac{v^2}{4}}$. From the mentioned Corollary it follows that $U$ and $V$ are independnt since the density of $(U,V)$ can be written as the product positive and measurable functions on $\mathbb{R}$. I'm fine with that.
Now the part that confuses me, is that one says the density of $U = \frac{1}{\sqrt{4\pi}}e^{-\frac{u^2}{4}}$ and $V = \frac{1}{\sqrt{4\pi}}e^{-\frac{v^2}{4}}$. But the Corollary states that the density of $U = c_1\frac{1}{\sqrt{4\pi}}e^{-\frac{u^2}{4}}$ for some constant $c_1$ and $V= c_2 \frac{1}{\sqrt{4\pi}}e^{-\frac{v^2}{4}}$ for some constant $c_2$. How do I know that $c_1 = c_2 = 1?$
Answer (thanks to Kavi Rama Murthy)
Density of $U = c_1\frac{1}{\sqrt{4\pi}}e^{-\frac{u^2}{4}}$. But since we know that any density integrates to 1 we have:
$\int c_1\frac{1}{\sqrt{4\pi}}e^{-\frac{u^2}{4}} = c_1 \int \frac{1}{\sqrt{4\pi}}e^{-\frac{u^2}{4}}$. Now substituting $v = \frac{u}{\sqrt{2}} \implies \frac{dv}{du} = \frac{1}{\sqrt{2}} \iff du = \sqrt{2}dv$ and $u = \sqrt{2}v$ we get
$\int \frac{1}{\sqrt{4\pi}} e^{-\frac{(\sqrt{2}v)^2}{4}} \sqrt{2}dv = \int \frac{1}{2 \sqrt{\pi}}e^{-\frac{v^2}{2}}dv = \frac{1}{\sqrt{2\pi}} \int e^{-\frac{v^2}{2}}dv = 1.$
Hence, $c_1 = 1$. The same applies for $c_2$.
Any density function integrates to $1$. So $c_1\int\frac 1 {\sqrt {4\pi}} e^{-u^{2}/4}du=1$. Making the substitution $v=\frac u {\sqrt 2}$ and using the fact that the standard normal density integrates to $1$ we see that $c_1$ must be $1$.