Does the integral of ln(x)/(1+exp(x)) from 2 to ∞ converge or diverge?

  • Thread starter hoho2666
  • Start date
In summary, the limit of In(x)/(1+exp(x)) as x approaches infinity is equal to 0. The function is defined for all values of x from 2 to infinity and its behavior as x approaches 2 from the right is undefined, approaching negative infinity. There are no vertical asymptotes for this function. The graph will approach the x-axis as x approaches infinity and will have a vertical asymptote at x=2.
  • #1
hoho2666
1
0
determine whether the integral In(x)/(1+exp(x)) from 2 to ∞? converges or diverges?
 
Physics news on Phys.org
  • #2
What is this function In(x)? Do you mean ln(x)? Assuming so, I think you can do a simple comparison test:

0 < ln(x) / (1+exp(x)) < x / (1+exp(x)) < x / exp(x)

And it's fairly easy to show the integral of x/exp(x) converges.
 

FAQ: Does the integral of ln(x)/(1+exp(x)) from 2 to ∞ converge or diverge?

What is the limit of In(x)/(1+exp(x)) as x approaches infinity?

The limit of In(x)/(1+exp(x)) as x approaches infinity is equal to 0.

Is the function In(x)/(1+exp(x)) defined for all values of x from 2 to infinity?

Yes, the function In(x)/(1+exp(x)) is defined for all values of x from 2 to infinity.

What is the behavior of In(x)/(1+exp(x)) as x approaches 2 from the right?

The behavior of In(x)/(1+exp(x)) as x approaches 2 from the right is undefined, as the function approaches negative infinity.

Does the function In(x)/(1+exp(x)) have any vertical asymptotes?

No, the function In(x)/(1+exp(x)) does not have any vertical asymptotes.

How does the graph of In(x)/(1+exp(x)) look like for values of x from 2 to infinity?

The graph of In(x)/(1+exp(x)) will approach the x-axis as x approaches infinity, and it will have a vertical asymptote at x=2.

Similar threads

Replies
5
Views
447
Replies
17
Views
3K
Replies
3
Views
698
Replies
3
Views
1K
Replies
37
Views
3K
Replies
3
Views
1K
Back
Top