Proof that the definite integral of 1/ln(x) doesn't exist

[JulienB]

Homework Statement

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.

Homework Equations

Integrals, limits

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?

Thx a lot in advance for your answers.Julien.

 

[Samy_A]

Homework Statement

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.

Homework Equations

Integrals, limits

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?

Thx a lot in advance for your answers.Julien.

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 $\displaystyle \int_1^2 \frac{1}{x-1} dx$.

 

[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.

 

[Samy_A]

@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.

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 ...

 

[JulienB]

@Samy_A Aha alright I get it. Thanks a lot for your answer, that was very helpful!Julien.

 

[Blanci1]

the divergence of said integral 1/ln x arises only very close to x=1. At that point log x is basically a simple straight line passing through a zero y=0 with gradient 1 or some finite value or whatever. So the reciprocal 1/ln x shoots off to infinity at x=1 in just the same way that y=1/x shoots off to infinity at x=0. So you just need to study how this latter área shoots off to infinity, which it indeed does. ( whereas other diverging functions manage to give a finite área such as y=1/ sqrt(x) from 0 to 1 say, corresponding to how área under 1/x^2 gives finite area from 1 to infinity. Hope that helps.

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

Integrals, limits

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?

Thx a lot in advance for your answers.Julien.

By definition, the integral you seek is
$$\lim_{a \to 0+} \int_{1+a}^2 \frac{1}{\ln x} \, dx \hspace{5ex}(1)$$
By a change of variables, the integral in (1) can be written as
$$I(a) = \int_a^1 \; \frac{1}{\ln(1+u)} \, du $$ Now $1/\ln(1+u) \sim 1/u$ for small $u > 0$, so look at
$$f(u) = \frac{1}{\ln(1+u)} - \frac{1}{u},$$
which can be extended from $(0,1]$ to $[0,1]$ as a (bounded) continuous function. We have
$$I(a) = \int_{a}^1 f(u) \,du + \int_a^1 \frac{1}{u} \, du.$$

 

[epenguin]

To show the divergence of the integral I would argue more natural and less clever would be to make the substitution $y=ln x$ .Then

$$\int ^{2}_{x=1}\dfrac {dx}{\ln x}=\int ^{\ln 2}_{y=0}\dfrac {e^{y}}{y}dy$$

Write out the numerator as exponential series and we have$$\int ^{\ln 2}_{0}\left( \dfrac {1+y+y^{2}/2!+y3/3!+\ldots }{y}\right) _{i}dy$$

$$=\int ^{\ln 2}_{0}\dfrac {dy}{y}+\int ^{\ln 2}_{0}\left( 1+\dfrac {y}{2!}+\dfrac {y^{2}}{3!}+\ldots \right) dy$$The second of these terms is the integral of a function which is everywhere finite in the interval, and is therefore finite.Whether the whole is finite therefore depends on whether the first term is, and you have probably already seen this if you have seen any example of a divergent integral. Anyway we can integrate it explicitly giving$$\left[ \ln x\right] ^{\textit{ln 2}}_{0}=\ln(ln 2)-\ln 0= \ln(\ln 2) --\infty =\infty $$PS The reason I claim the above approach is most natural is that it follows immediately from the definition of $ln$ as inverse of exponential, whereas the series expansion of $ln$ involve some extra knowledge or considerations. However there is not a huge difference between the different approaches given above.I think there is a risk of the techie formula grinding in all cases being taken for the proof and obscuring the real essence of the argument. Which is that the definite integral is the difference between a lower limit and an upper limit, and that the lower one is infinite but the upper one is finite. The latter is finite because it is sum of an infinite series, which however can be shown to be finite*. This was implicit in all arguments made - I think we miss the point if we do not make it explicit. Unless... if we went into how we came by these infinite series in the first place… we find that we don't need to invoke this principle after all?

*(If both were infinite the matter would have to be settled by looking at the differences which might or might not be finite; examples not to mind but I daresay there are examples of both cases.)