$\alpha:(0, +\infty) \to (0, +\infty)$ continuous and increasing, $\alpha(1) = 1$. Show $\int_1^x d\alpha/\alpha = \log \alpha(x)$

Let $\alpha:(0, +\infty) \to (0, +\infty)$ continuous and increasing, with $\alpha(1) = 1$. Show that $\int_1^x d\alpha/\alpha = \log \alpha(x)$ (don't suppose $\alpha$ differentiable)

Well, the change of variable theorem for Stieltjes integration says:

Let $f:[c,d]\to\mathbb{R}$ continuous, $\alpha:[c,d]\to\mathbb{R}$ of bounded variation (rectifiable path) and $\phi:[a,b]\to[c,d]$ an homeomorphism. Then:

$$\int_c^d f(t) d\alpha = \int_a^b f\circ\phi(s)d(\alpha\circ\phi)), \mbox{ if \phi is increasing }$$

$$\int_c^d f(t) d\alpha = -\int_a^b f\circ\phi(s)d(\alpha\circ\phi)), \mbox{ if \phi is decreasing }$$

I think I'm supposed to find the inverse homeomorphism from $\alpha$ (since its continuous and injective), in a way that $\alpha\circ\phi = 1$, so the integral becomes a Riemann integral. I should then evaluate $\int_{\phi(c)}^{\phi(d)} f\circ \phi(s) ds$ where $f: t\to \frac{1}{t}$. The integral now depends on $\phi(s)$, but how to Riemann integrate it?

• I think you want $f(t)=1/\alpha$, $c=1$ and $d=x$. That way the left-hand side in the theorem is your integral $\int_{1}^{x}d\alpha/\alpha$. In that case you are left with integrating $f\circ\phi = 1/t$, which gives the logarithm that you need. Aug 22 '17 at 21:13
• The catch is that, in your statement of the change of variables theorem, $\phi$ need not be a homeomorphism; it is enough that $\phi$ is continuous and monotone. So you can replace $(f(t), \alpha(t), \phi)$ by $(t^{-1}, t, \alpha)$. Aug 23 '17 at 3:47
• But what should be $\alpha\circ\phi$ then? Aug 26 '17 at 17:08
• @SangchulLee how can you choose what $\alpha$ is going to be? The exercise gives a specirfic $\alpha$, but you don't know what it is Aug 31 '17 at 0:52
• I am not sure what are you asking. The role of $\alpha$ in your problem and the $\alpha$ in the statement are different. That is why I am suggesting to replace $(f(t), \alpha(t), \phi)$ in the theorem by $(t^{-1}, t, \alpha)$, where this $\alpha$ is now what you have in your problem. Sep 3 '17 at 21:20

Here is a simple change of variables result:

Claim. Let $f : [c, d] \to \mathbb{R}$ be Riemann integrable and $\alpha : [a, b] \to [c, d]$ be continuous and monotone increasing. Then

$$\int_{a}^{b} f(\alpha(x)) \, d\alpha(x) = \int_{\alpha(a)}^{\alpha(b)} f(y) \, dy.$$

Proof. For each $\epsilon > 0$, pick $\delta > 0$ such that $|\alpha(x) - \alpha(y)| < \epsilon$ whenever $|x-y| < \delta$. Then for any partition $\Pi = \{a = x_0 < \cdots < x_n = b\}$ with $\|\Pi\| < \delta$ and for any $i = 1, \cdots, n$, we have

$$m_i [\alpha(x_i) - \alpha(x_{i-1})] \leq \int_{x_{i-1}}^{x_i} f(\alpha(x)) \, d\alpha(x) \leq M_i [\alpha(x_i) - \alpha(x_{i-1})]$$

where

\begin{align*} m_i &= \inf \{ f(y) : y \in [\alpha(x_{i-1}),\alpha(x_i)] \} \\ M_i &= \sup \{ f(y) : y \in [\alpha(x_{i-1}),\alpha(x_i)] \}. \end{align*}

This tells that, with $\alpha(\Pi) = \{ \alpha(x_0), \cdots, \alpha(x_n)\}$ we have $\|\alpha(\Pi)\| < \epsilon$ and

$$L(f, \alpha(\Pi)) \leq \int_{a}^{b} f(\alpha(x)) \, d\alpha(x) \leq U(f, \alpha(\Pi)),$$

where $L(f, \cdot)$ and $U(f, \cdot)$ are the lower Riemann sum and the upper Riemann sum, respectively. Therefore the desired identity follows by taking $\epsilon \to 0$.