About a improper integral

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November 27th 2011, 06:45 AM
wong999

About a improper integral

I would like to ask about this integral:
integral from 2 to infinity [ ln x /(exp(x) +1) ]

I think it diverges, but I don't know how to show it diveges.

first, I thought [ ln x /(exp(x) +1) ]can't be integrate,
so I considered if it can be proved by comparing it with simpler integral.
I had try to find a smaller integral and then prove it diveges,
hence it could conclude that integral from 2 to infinity [ ln x /(exp(x) +1) ] also diveeges.
However, I could't think out such these smaller integral.

So how can I proved it, or it actually converges?
Please help me.
November 27th 2011, 07:18 AM
skeeter

Re: About a improper integral

Quote:

Originally Posted by wong999

I would like to ask about this integral:
integral from 2 to infinity [ ln x /(exp(x) +1) ]

I think it diverges, but I don't know how to show it diveges.

first, I thought [ ln x /(exp(x) +1) ]can't be integrate,
so I considered if it can be proved by comparing it with simpler integral.
I had try to find a smaller integral and then prove it diveges,
hence it could conclude that integral from 2 to infinity [ ln x /(exp(x) +1) ] also diveeges.
However, I could't think out such these smaller integral.

So how can I proved it, or it actually converges?
Please help me.

note for x > 2 ...

$\frac{\ln{x}}{e^x + 1} < \frac{\ln{x}}{e^x} < \frac{x}{e^x}$

and $\int_2^\infty xe^{-x} \, dx = \frac{3}{e^2}$ converges
November 27th 2011, 05:20 PM
wong999

Re: About a improper integral

Quote:

Originally Posted by skeeter

note for x > 2 ...

$\frac{\ln{x}}{e^x + 1} < \frac{\ln{x}}{e^x} < \frac{x}{e^x}$

and $\int_2^\infty xe^{-x} \, dx = \frac{3}{e^2}$ converges

huh...how stupid I am
thank you for helping