While playing aroud with integrals I stumbled across following identity:
$\newcommand{\sgn}{\text{sgn}}$ $\sgn(g(x))f(x) = \frac{d}{dx}\left(\int\sgn(g(x))f(x)dx\right) = \frac{d}{dx}\left(\sgn(g(x))\int f(x)dx\right)$
Which can be easily verified using the chainrule. The way I interpret this is that when trying to find an antiderivative of a function multiplied by the sign of another function, one can pull the sign-function out of the integral to get a different yet still valid antiderivative.
Example:
$I(x) = \int\frac{|x^3\sin(x)|}{x^3} dx = \int\sgn(x^3\sin(x))\frac{x^3\sin(x)}{x^3} dx =\int\sgn(x\sin(x))\cdot \sin(x) dx = \sgn(x\sin(x))\int \sin(x) dx = \sgn(x\sin(x))(c-\cos(x))$
The annoying part about this is that the sign-function is also multiplied to the constant, but the derivative still yields the original function.
$I'(x) = 0\cdot (c-\cos(x))+ \sgn(x\sin(x))\sin(x) = \sgn(x^3\sin(x))\frac{x^3\sin(x)}{x^3} = \frac{|x^3\sin(x)|}{x^3}$
Is this legitimate or are there some flaws with this?
It's been a few months and I realized the whole thing is trivial.
$\newcommand{\sgn}{\text{sgn}}$
$\displaystyle \int\sgn(g(x))f(x)dx = \begin{cases} \displaystyle \int f(x)dx,& g(x)>0\\ \displaystyle \int 0 \, dx=0 = 0\cdot\int f(x) dx, & g(x) = 0\\ \displaystyle \int -f(x)dx = -\int f(x)dx, & g(x)<0 \end{cases} = \sgn(g(x))\int f(x)dx$