Can someone please help me explain this fallacy $$\Omega = \int_{-1}^1 \frac{dx}{1+x²}\ =\ -\int_{-1}^1\frac{dy}{1+y²} $$ $$ \text{ by using the transformation }\;  x= \frac{1}{y}$$ $$ \Omega = - \Omega ,\ 0\,\text{hence } \Omega =0 $$
but the integral is obviously $$\frac{π}{2} $$
 A: This is just a slightly different take on aschepler's answer, showing a few more steps:
$$\begin{align}
\int_{-1}^1{1\over1+x^2}\,dx
&=\int_{-1}^0{1\over1+x^2}\,dx+\int_0^1{1\over1+x^2}\,dx\\
&=\int_{-1}^{-\infty}{1\over1+(1/y)^2}\,{-dy\over y^2}+\int_\infty^1{1\over1+(1/y)^2}\,{-dy\over y^2}\\
&=\int_{-\infty}^{-1}{1\over y^2+1}\,dy+\int_1^\infty{1\over y^2+1}\,dy
\end{align}$$
where the negative sign from $dx=-dy/y^2$ goes into reversing the limits of integration, i.e., $\int_b^a=-\int_a^b$.
A: Although $x=-1$ implies $y=-1$ and $x=1$ implies $y=1$, this does not mean the integral on $y$ can be written "from $-1$ to $1$". We need to consider the actual regions of integration, not just the boundary points.
$-1 \leq x \leq 1$ implies $y \leq -1$ or $y \geq 1$. (We can ignore the single point $x=0$.) So the correct transformation is:
$$ \int_{-1}^1 \frac{dx}{1+x^2} = \int_{-\infty}^{-1} \frac{dy}{1+y^2} + \int_1^{\infty} \frac{dy}{1+y^2} $$
The minus sign from $dy = -\frac{dx}{x^2}$ is removed, because when dealing with sets rather than just a monotone increasing or decreasing function on single intervals, we must use the absolute value of the derivative.
