Apparent paradox in non-null derivative but null primitive

Question

Suppose you have the function $\frac{\partial \eta}{\partial y}=\frac{1}{|y|^3}-\frac{1}{y^3}$, now this function is $0$ if $y>0$, and different from $0$ if $y<0$, but integrating this we get: $$ \eta=\int^y\frac{1}{|y|^3}-\frac{1}{y^3}=\frac{1}{-2|y|^2}-\frac{1}{-2y^2} +C=\frac{1}{-2y^2}-\frac{1}{-2y^2}+C=0+C,\space \forall x \in \mathbb{R} $$ And this implies that $\frac{\partial \eta}{\partial y}=0,\space \forall x \in \mathbb{R}$, which is not correct as seen above. Where is the loophole in my reasoning?

Answer 1

Confirming part of your calculation: $$\int y^{-3}\, dy = \frac{y^{-2}}{-2} + C = -\frac1{2y^2} + C.$$

With $f(y) = \lvert y\rvert^{-3},$ however, we have $$ f(y) = \begin{cases} y^{-3} & y > 0, \\ -y^{-3} & y < 0, \end{cases} $$ with the function undefined at $0.$

We then have the antiderivative on $(0,\infty),$ $$\int \lvert y\rvert^{-3}\, dy = \int y^{-3}\, dy = -\frac1{2y^2} + C_1,$$ and the antiderivative on $(-\infty,0),$ $$\int \lvert y\rvert^{-3}\, dy = \int -y^{-3}\, dy = \frac1{2y^2} + C_2.$$

There is not a good way to integrate across zero.

Alternatively, let $g(y) = \dfrac1{\lvert y\rvert^3} - \dfrac1{y^3}.$ Then $$ g(y) = \begin{cases} 0 & y > 0, \\ -2y^{-3} & y < 0. \end{cases} $$ If we define $g(0) = 0$ then we have the antiderivative on $[0,\infty),$ $$\int g(y)\,dy = \int 0\, dy = C_3,$$ and the antiderivative on $(-\infty,0),$ $$\int g(y)\,dy = \int -2y^{-3}\, dy = \frac1{y^2} + C_4.$$

There is still no way to make this into a single function that is continuous at zero. So the best we can do is to say that $$\int\left(\frac1{\lvert y\rvert^3} - \frac1{y^3}\right)\,dy = \begin{cases} C_3 & y \geq 0, \\ \dfrac1{y^2} + C_4 & y < 0. \end{cases} $$