Help with proving that Improper Integral is Divergent

[student93]

Homework Statement

The problem is attached in this post.

Homework Equations

The problem is attached in this post.

The Attempt at a Solution

Lim t -> ∞ ∫ dx/xlnx from 1 to t

u-substitution:

u=lnx
du=1/x dx

Lim t -> ∞ ∫ 1/u du

Lim t -> ∞ ln u

Lim t -> ∞ ln(lnx) from 1 to t

Lim t -> ∞ ln(lnt) - ln(0)

= ∞ - ∞ = 0 (This is incorrect since the answer is that the integral is divergent).

 

[Ray Vickson]

Homework Statement

The problem is attached in this post.

Homework Equations

The problem is attached in this post.

The Attempt at a Solution

Lim t -> ∞ ∫ dx/xlnx from 1 to t

u-substitution:

u=lnx
du=1/x dx

Lim t -> ∞ ∫ 1/u du

Lim t -> ∞ ln u

Lim t -> ∞ ln(lnx) from 1 to t

Lim t -> ∞ ln(lnt) - ln(0)

= ∞ - ∞ = 0 (This is incorrect since the answer is that the integral is divergent).

$\ln(1) = 0$, not $\infty$.

Anyway, never, ever use expressions of the form $\infty - \infty$; these are meaningless and are illegal in mathematics.

 

[student93]

That's true but it's not ln(1), it's ln(ln(1)) which is ln(0), which is equal to ∞, right?

 

[Ray Vickson]

That's true but it's not ln(1), it's ln(ln(1)) which is ln(0), which is equal to ∞, right?

OK, you're right. However, $\int_a^b dx/(x \ln x) = \ln \ln b - \ln \ln a \to \infty + \infty$ as $a \to 1+$ and $b \to \infty$. As $a \to 1+$, $\ln a \to 0+$ and so $\ln \ln a \to -\infty$, or $-\ln \ln a \to +\infty$.

 

[student93]

OK, you're right. However, $\int_a^b dx/(x \ln x) = \ln \ln b - \ln \ln a \to \infty + \infty$ as $a \to 1+$ and $b \to \infty$. As $a \to 1+$, $\ln a \to 0+$ and so $\ln \ln a \to -\infty$, or $-\ln \ln a \to +\infty$.

Is there another way to prove that this integral is divergent? (For example, would the Direct Comparison Theorem work in this case?)

Also, why is ∞ + ∞ mathematically allowed, if ∞ - ∞ isn't?

 

[Ray Vickson]

Is there another way to prove that this integral is divergent? (For example, would the Direct Comparison Theorem work in this case?)

Also, why is ∞ + ∞ mathematically allowed, if ∞ - ∞ isn't?

∞ - ∞ is meaningless (along with 0/0, for example). ∞-∞ can be made equal to anything you want---anything from -∞ to +∞ and everything in between. However, ∞+∞ = +∞ is unambiguous. Think about it and you will understand why.

 

[student93]

∞ - ∞ is meaningless (along with 0/0, for example). ∞-∞ can be made equal to anything you want---anything from -∞ to +∞ and everything in between. However, ∞+∞ = +∞ is unambiguous. Think about it and you will understand why.

So ∞-∞ is an Indeterminate Form then right? So can I just L'Hopital's rule to solve for the limit instead of the other method that you showed in your previous post?

 

[student93]

So ∞-∞ is an Indeterminate Form then right? So can I just L'Hopital's rule to solve for the limit instead of the other method that you showed in your previous post?

So by using L'Hopital's rule, could I take the derivative of ln(lnx) - ln(ln1) and then try solving for the limit etc.?

 

[Ray Vickson]

So by using L'Hopital's rule, could I take the derivative of ln(lnx) - ln(ln1) and then try solving for the limit etc.?

You keep missing the point: you do not have an indeterminate form, and do not need anything like l'Hospital's rule. I suggest you look much more carefully at what you are doing, and don't rush. Anyway, I already showed you a correct argument, but for some reason you are just not "getting" it.

 

[student93]

You keep missing the point: you do not have an indeterminate form, and do not need anything like l'Hospital's rule. I suggest you look much more carefully at what you are doing, and don't rush. Anyway, I already showed you a correct argument, but for some reason you are just not "getting" it.

Ok, I think I now understand your explanation, however would it also be possible to prove this integral is divergent via the Direct Comparison Test? And if it is possible, then how would I prove that the integral is divergent via the Direct Comparison Test?