# Homework Help: Proof that definite integral of 1/lnx doesn't exist

1. Apr 10, 2016

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! I'm having a hard time to find a way to cleanly prove that ∫(1/ln(x)) dx between 1 and 2 doesn't exist. At first I thought it was because it's not bounded (Riemann criterion I believe), but then I looked at another unbounded definite integral such as ∫lnx dx between 0 and 1 and it does exists! I've seen some proofs with li(x) but I haven't done that in class, so it'd be strange to use that in an exam.

2. Relevant equations

Integrals, limits

3. The attempt at a solution

Absolutely no idea. I've been trying to compare it with ∫lnx dx between 0 and 1 but it didn't really help:

o1 ln(x) dx = lim a→0+a1 ln(x) dx
= lim a→0+ (1(ln(1) - 1) - a(ln(a) - 1)) = -1

Is there a similar method to show that the integral of 1/lnx diverges?

Julien.

2. Apr 10, 2016

### Samy_A

Try to determine $\displaystyle \lim_{x \rightarrow 1+} \frac{x-1}{\log x}$.
That will allow you to determine the divergence using the easier integral $\displaystyle \int_1^2 \frac{1}{x-1} dx$.

3. Apr 11, 2016

### JulienB

@Samy_A Hi and thanks for your answer. I am not sure to understand what you are suggesting me. Using L'Hôpital, I find that the limit you ask for is 1. As for the integral, I would say:

12 1/(x-1) = lim a→1+ (ln(2-1) - ln(1-1)) = -∞

Is that correct? Not sure what's the link between the limit and the integral.. Moreover, it is indeed easy to find the antiderivative of 1/(x-1) but I have no clue on how to determine the antiderivative of lnx :/

Julien.

4. Apr 11, 2016

### Samy_A

You have a sign error, $\displaystyle \int_1^2 \frac{1}{x-1} dx = + \infty$.

$\displaystyle \lim_{x \rightarrow 1+} \frac{x-1}{\log x}=1$ is correct.

That means that for $x$ close to 1 (say for $1<x <a \leq2$), you have (for example), $\frac{x-1}{\log x} \geq \frac{1}{2}$. (We don't care about the exact value of $a$, just that it exists.)

That means that for $x \in ]1,a[$, $\frac{1}{\log x} \geq \frac{1}{2}\frac{1}{x-1}$.

You know that the integral of the second function diverges ...

5. Apr 11, 2016