Doubts with non differentiable change of variable Consider the integral
$$
I=\int_{-2}^{2}f(x)dx
$$
where $f(x)$ is a differentiable function in $[-2,2]$. Let's consider the variable change
$$
y=\frac{1}{
x}
$$
$y(x)$ is non differentiable when $x=0$, and this point is included in the integration regime. Nevertheless, if we naively continue we could write
$$
I=\int_{-1/2}^{1/2}g(y)dy
$$
Nonetheless had we instead written the integral in the $x$ variable like this
$$
I=\int_{-2}^{0}f(x)dx+I=\int_{0}^{2}f(x)dx
$$
in the $y$ variable we have
$$
I=\int_{-1/2}^{-\infty}g(y)dy+\int_{\infty}^{1/2}g(y)dy
$$
and this is a different path from what we had before. Is the naive third equation right? can we proceed with such a non differentiable variable change?
 A: This is a case where the notation $\int_a^b$ is misleading. The substitution you want to does not map $[-2,2]$ to an interval. It is not even defined at $0$, so it doesn't map it to anything. When you split the integrals as you did, at zero, on each of $[-2,0)$ and $(0,2]$ the substitution works and you get 
$$
\int_{-2}^0f(x)\,dx = -\int_{-1/2}^{-\infty}f(1/y)\,\frac1{y^2}\,dy,
\ \ \ \ 
\int_{0}^2f(x)\,dx = -\int_{+\infty}^{1/2}f(1/y)\,\frac1{y^2}\,dy.
$$
To see an explicit example, take $f(x)=x^2$. Then 
$$
\int_{-2}^2f(x)\,dx=\frac{16}3.
$$
If you apply your blind substitution $y=1/x$, then $dy=-y^2\,dx$, and you would "get"
$$\tag{1}
-\int_{-1/2}^{1/2}\frac1{y^4}\,dy.
$$
This integral doesn't exist. If you were to forget about and (incorrectly) evaluate as if you could, it seems to work:
$$\tag{2}
-\int_{-1/2}^{1/2}\frac1{y^4}\,dy"="\left.\frac1{3y^3}\right|_{-1/2}^{1/2}=\frac{16}3.
$$
What's actually happening is that 
$$\tag{3}
\begin{align}
\int_{-2}^2x^2\,dx&=\int_{-2}^0x^2\,dx+\int_0^2x^2\,dx=\int_{-\infty}^{-1/2}\frac1{y^4},dy+\int_{1/2}^\infty\frac1{y^4}\,dy\\ \ \\
&=-\left.\frac1{3y^3}\right|_{-\infty}^{-1/2}-\left.\frac1{3y^3}\right|_{1/2}^\infty,
\end{align}$$
which happens to agree with the evaluation at $(2)$ because the contributions at $\pm\infty$ vanish, and so only the contributions of $\pm1/2$ matter. 
