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.

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 ...

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

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

Re: About a improper integral

Quote:

Originally Posted by

**skeeter** note for x __>__ 2 ...

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

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

huh...how stupid I am

thank you for helping