I was having a look into the signum function on Wikipedia, and it gives the definition of it as: $$sgn(x)= \begin{cases} 1, & x>0 \\ 0, & x=0 \\ -1, & x<0 \end{cases} $$

However, if I integrate this function for all $x$, I wind up with $|x|+c$ (assuming that the constant of integration is the same for each condition). $$\int sgn(x)\,dx= \begin{cases} \int 1\,dx, & x>0 \\ \int 0\,dx, & x=0 \\ \int -1\,dx, & x<0 \end{cases} \\= \begin{cases} x, & x>0 \\ 0, & x=0 \\ -x, & x<0 \end{cases} +c \\= \begin{cases} x, & x\ge0 \\ -x, & x<0 \end{cases} +c \\= |x|+c $$

If I differentiate $|x|+c$, however, I get the result of $\frac{|x|}{x}$ which assumes x is not zero, but is different from the definition. Wikipedia mentions that this definition is true for $x\ne0$; but how would I return to the original definition from the derivative - assuming the integration was correct?


Well, $|x|$ isn't differentiable at $x=0$, so you can't hope to retrieve anything there. If $f$ is a general (not necessarily continuous) $L^1_{loc}$-function (i.e. integrable over compact intervals), then defining $F(x)=\int_0^x f(t)\textrm{d}t,$ it's only true that $F'(x)=f(x)$ holds almost everywhere. In your case, you get the conclusion for every $x\neq 0$.

If you look at the usual proof of the fundamental theorem of calculus, it relies heavily on the fact that it deals with the integration of continuous functions.


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