I'm stuck with this problem: Evaluate $\int_1^t \frac{1}{x} dx, t>1$ as a regulated integral.
Earlier a regulated function $f:[a,b] \to \mathbb{R}$ was introduced as one for which for all $\epsilon > 0$ there exists a step function $φ_\epsilon$ such that $|f(x)-φ_\epsilon(x)| \leq \epsilon$ for all $x\in [a,b]$, or equivalently if it is the uniform limit of a sequence $(φ_k)_{k\in \mathbb{N}}$ of step functions on $[a,b]$. The regulated integral was then defined as the limit of the sequence $(\int_a^b φ_k)_{k\in \mathbb{N}}$ of integrals.
So I would need to
- construct a sequence $(φ_n)_{n\in \mathbb{N}}$ of step functions on $[1,t]$,
- then show that for all $\epsilon > 0$ there exists a $k\in\mathbb{N}$ such that $|f(x)-φ_k(x)| \leq \epsilon$ for all $x\in [1,t]$ (I'm using this definition because uniform convergence hasn't been given a systematic treatment).
- And lastly take the limit of the sequence of integrals $(\int_1^t φ_n)_{n\in \mathbb{N}}$.
In 1) you can supposedly use the partition $$Z_n: 1=z_0 < z_1 = t^{1/n} < z_2 = t^{2/n} < ... < z_n = t$$ for $φ_n$ and I guess you can define $φ_n(x)=\frac 1{z_i}$ for all $x\in(z_{i-1},z_i],i=1,...,n$ with $φ_n(1)=1$. Is this reasonable and how would you go about 2) and 3)? Thank you for your help.
Your definition of $\phi _n$ is fine. For $z_i \leq x \leq z_{i+1}$ we have $|\phi_n (x)- \frac 1 x | =| \frac 1 {z_i} - \frac 1x |$ = $\frac {|x- z_i|} {x z_i}$. Note that the denominator is greater than 1. Hence $|\phi_n (x)- \frac 1 x | <|x-z_i| \leq (z_{i+1}-z_i) =t^{i/n} (t^{1/n} -1) \leq t (t^{1/n} -1)$. Noting that the last quantity $\to 0$ as $n \to \infty$ we get part 2). For 3) I leave it to you to get the relation $\int_ 1 ^{t} \phi_n = n(t^{1/n} -1)$. It is a standard fact that the last quantity $\to \log t$ as $n \to \infty$.