Suppose I have to evaluate $$\int_0^\infty f(x) \text{d}x$$ And I would like to use the substitution $x^2=\frac{t}{1-t}$.
Now I must find the new integration interval in the variable $t$, I suppose I need to do a thing like this: if $x=0$ we have $0=\frac{t}{1-t}$ and so $t=0$, if $x\to\infty$ then $\frac{t}{1-t} \to \infty$ and this is true when $t \to 1^-$.
My questions are:
(1) is this correct? Is there a rigorous way to to this?
(2) I know that when I make substitutions in integrals they must be bijective, but I'm confused if I must verify that $x^2$ is bijective or $\frac{t}{1-t}$; in this case they are both increasing, so maybe it can help. Thanks.
$x^{2}=\frac t {1-t}$ iff $t=\frac {x^{2}} {1+x^{2}}$. Using this it is easy to check that $x \to t$ is indeed bijective from $(0,\infty)$ on,to $(0,1)$. So the substitution is legitimate the integral transforms to an integral from $0$ to $1$.