Given $L(y) = \int_ {1}^{y}\frac{1}{t}dy$ , $y\in(0, \infty)$

Question

I am proving Theorem 37.4 , Elementary Analysis, Kenneth Ross.

Stating $$L(y) = \int_{1}^{y}\frac{1}{t}\operatorname{d}t,\quad y\in(0,\infty)$$

I have to prove that

(i) the function L is strictly increasing , continuous and differentiable in $(0,\infty)$

(ii) $L(yz)= L(y)+L(z)$, for $ y, z \in (0,\infty)$ .

For(i): Since the function $\frac{1}{t}$ is continuous in the open interval$(0,\infty)$ that contains the point 1.Then the Fundamental Theorem of Calculus ensures that $L(y) = \int_{1}^{y}\frac{1}{t}dt $ is differentiable on the interval $(0,\infty)$ and hence also continuous there.

For the part (ii) the book states that consider for any fixed z, $ g(y)= L(yz) - L(y) - L(z).$ Then using the fundamental theorem of calculus we have

$g'(y)=\frac{1}{yz}\frac{d(yz)}{dy}$ $- \frac{1}{y} - 0 = 0 $ , for any $ y\in (0,\infty).$

Then, again by the fundamental theorem again, we have

$\int_{1}^{y} g'(t) dt = g(y) - g(1)$, since $ g(1) = L(z) - L(1)- L(z)$ and $ L(1)= 0 $

This implies that $g(y)=0$.

Also, have to show (iii) $lim y\rightarrow\infty L(y)= +\infty$

**The proof in the book says : Since, $L(2) >0 $ , then $L(2^n) = n L(2) >0 $ Then as L is a strictly increasing function then $ Lim y\rightarrow \infty L(y) = Lim n \rightarrow\infty L(2^n)= \infty$

Please suggest what could be the steps.

I am thinking if we take y to be sufficiently large, then there exists a natural number n such that $ y > 2^n$ , then $ L(y) > L( 2^n)$.

How to ensure that if y approaches infinity then n also approaches infinity.**

Answer 1

To show part three you have to prove that $\int_1^{\infty}\frac{dt}{t}$ is unbounded. This should be easy if you altready know few basic resuts about convergent and divergent series.

Note that $\int_1^{\infty}\frac{dt}{t} = \sum_{n = 1}^{\infty} \int_n^{n+1}\frac{dt}{t} \ge \sum_{n = 1}^{\infty} \int_n^{n+1}\frac{dt}{n+1}$. This last inequality follows from the fact that $\frac{1}{n} \ge \frac{1}{t} \ge \frac{1}{n+1}$, $t \in [n,n+1]$.

$\sum_{n = 1}^{\infty} \int_n^{n+1}\frac{dt}{n+1} = \sum_{n = 1}^{\infty}\frac{1}{n+1}\int_{n}^{n+1}dt = \sum_{n = 1}^{\infty}\frac{1}{n+1} = \sum_{n = 2}^{\infty}\frac{1}{n} = \infty$.

Therefore $\int_1^{\infty}\frac{dt}{t} \ge \infty \Longrightarrow \int_1^{\infty}\frac{dt}{t} = \infty$.

For the divergence of the harmonic series you can give a look at http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

Answer 2

One more time: $\frac{1}{n} \ge \frac{1}{t} \ge \frac{1}{n+1}$, $t \in [n,n+1]$, therefore $L(2) = \int_1^2\frac{dt}{t} \ge \frac{1}{2}\int_1^2dt = \frac{1}{2} > 0$.

$L(2^n) = L(2 \cdot 2^{n-1}) = L(2) + L(2^{n-1})$ using the property we have already proved. Iteraring the argument we obtain $L(2^n) = L(2) + \dots + L(2) = nL(2)$.

It is also clear that $L(2^n) \le L(y)$ when $2^n \le y$, therefore $L(2^n) \le \lim_{y \to \infty} L(y)$ holds for every $n \in \mathbb{N}$. If it holds for every $n \in \mathbb{N}$ you are allowed to consider $n$ as big as you like it. This gives:

$\lim_{y \to \infty}L(y) \ge \lim_{n \to \infty}L(2^n) = \lim_{n \to \infty}nL(2) \ge \lim_{n \to \infty}n \cdot \frac{1}{2} = \infty$.

The trick is to consider the limit over $y$ before taking the limit over $n$. This is how you are allowed to take $n$ as big as you wish being sure that $y$ is still bigger.

I hope everything's clear now! :D