Change of variable $y=\log(x)$ in a funciton $f(x)$

Question

I have a function $f(x)$ defined in $x\in[0,\infty)$ (I want $f(0)<\infty$) and want to change the variable as $y=\log(x)$. So we have a new function $u(y)$ and the followings

$f(x)dx=u(y)dy$

$ \Rightarrow f(x)=u(y)\frac{dy}{dx} =u(y)\frac{1}{x}=u(y)e^{-y}$

Because of the domain of $x$, $u(y)$ is defined in $y \in(-\infty,\infty)$. However, it gives

$f(0) =u(-\infty)\frac{1}{0}=u(-\infty)e^{\infty}$ and

$f(\infty) =u(\infty)\frac{1}{\infty}$

and they look not appropriate. Where did I make mistake?

Answer 1

Since $f(x)$ is defined on $[0,\infty]$ we are allowed to set $y=\log x \iff x=e^y$ with $y\in (-\infty,\infty)$ but note that since $e^y$, injective, has range $(0,\infty)$ the value $x=0$ is not reached.

Moreover we have

$$f(x)=u(y) \implies f'(x)dx=u'(y)dy$$

and since $u(y)=f(e^y) \implies u'(y)=e^yf'(e^y)$ by chain rule.

Answer 2

Notice: $$ f(x)dx = f(e^y)\frac{dx}{dy}dy \Rightarrow f(x) = f(e^y)\frac{dx}{dy}\frac{dy}{dx}. $$ Thus you should have no worries, since $\frac{dy}{dx}$ and its inverse will cancel out and, as expected,$$\lim_{y\to-\infty} u(y) =\lim_{x\to0} f(x) $$

Answer 3

From your definition

$$u(y)=f(x)\frac{dx}{dy}=f(e^y)e^{y}$$

and

$$f(x)=u(y)\frac{dy}{dx}=\frac{u(\log(x))}x.$$

Then

$$f(0)=\frac{u(\log(0))}0=\frac{f(0)\,e^{\log(0)}}{0}$$ and

$$f(\infty)=\frac{u(\log(\infty))}{\infty}=\frac{f(\infty)e^{\log(\infty)}}{\infty}.$$

Nothing wrong.