Question

It is too long.

Answer 1

Define \(L:(0,\infty)\to\mathbb{R}\) by\[L(x)=\int_{1}^{x}\frac{dt}{t}.\]I would like to show that \(L\) is strictly increasing on its domain.

The way I approached this was by showing that\[L(x\delta)\leqslant L(x)\]for any \(\delta>1\) and \(x\in(0,\infty)\) (contradiction):\[L(x\delta)=L(x)+L(\delta)\leqslant L(x)\implies L(\delta)\leqslant0.\]However, we know that\[L(\delta)=\lim_{n\to\infty}\Delta x\sum_{i=1}^{n}\frac{1}{t_{i}^{*}}\]must be greater than zero because it is a sum of distinct positive terms of the form \(1/n\). Hence, we have a contradiction, and it must be the case that \(L(x+\epsilon)>L(x)\) for any \(\epsilon>0\) (after performing the substitution \(\delta=1+\epsilon/x\)).

What do you guys think?

Answer 2

u kidding?

Answer 3

this is impossible

Answer 4

is this university level maths?

Answer 5

You can also use the Fundamental Theorem of Calculus Part 1 and prove that the derivative is positive in the entire domain.

Answer 6

DUH!

Why didn't I think of that!? Thanks, @blockcolder!

Answer 7

No problem. :)

Answer 8

pro^

Answer 9

where are you from blackcolder?

Answer 10

Philippines. And let's talk about this in chat.

Answer 11

thanks for report btw