Integration by parts and infinity

[kwy]

Homework Statement

integrate (x*2e^x)/(2e^x-1)2 from x=0 to infinity

Homework Equations

The Attempt at a Solution

let t=2e^x-1 => x=ln((t+1)/2)
dt = 2e^x dx

Thus equation is now integrate (ln((t+1)/2))/t^2 dt from t=1 to infinity

Then let u = (t+1)/2 => 2du=dt
Equation now integrate 2lnu/(2u-1)^2 from u=1 to infinity

Then I started going around in circle...

 

[aim1732]

2lnu/(2u-1)^2

Integrate (2u-1)^-2 and differentiate lnu in the second integral,writing along with the first integral(the usual way you integrate by parts).Then apply partial fractions in the second in the second integral.I got ln2 as answer.

 

[kwy]

Thanks aim1732, but I'm still a little off.

diff lnu = 1/u, int (2u-1)^-2 = -1/2(2u-1)
Thus equation is now:

1) 2([-lnu/2(2u-1)] u=infinity to 1 + int(1/u)du + int(1/4u-2)du)
2) 2( 0 + int(1/u)du + int(1/4u-2)du)
3) 2([-1/u^2] + [ln(4u-2)]) where u=1 to infinity
4) 2[0-(-1)) + infinity - ln(2)]
5) 2(1 + infinity - ln(2))

Sorry for spelling out all the lines, I'm really rusty at this.
Can you please let me know where I've gone wrong?

Thanks for your assistance.

 

[aim1732]

Sorry i can not understand what you wrote.It's okay I think you are aware that putting the limits for the integral involves calculating the LHL for the upper limit and RHL for the lower one.I had to use L'Hospitals rule once.Remember all you have to do is eliminate the indeterminate forms.
If you still have problems i am going to take a hard look at your work then I guess.

 

[kwy]

Sorry, I'm going to try using LaTex to see if it makes more sense. If the lines below does not make sense, I apologise. Would you mind letting me know where you had to use L'Hopital Rule, I cannot work out where this can be applied:

My previous workings were:

2(0 + $$\int 1/u * 1/4u-2 du$$)
2($$\int 1/u du$$ + $$\int 1/(4u-2) du$$), with integration from 1 to infinity

I get:

$$\int 1/u du$$ = 1
$$\int 1/(4u-2) du$$ = infinity-ln(2)

Putting them together: 2(1 + infinity + ln2)

 

[aim1732]

Trick is in clubbing the the log terms together.

The integral as I see it is this:

ln(2u-1) - lnu -(lnu)/2u-1

This you can write as

ln[(2u-1)/u] - lnu/(2u-1)
ie. ln(2-(1/u)) - lnu/(2u-1)

Now put the limits.Notice the second term is of the form ∞/∞ for the upper limit so apply L'hospitals.

 

[kwy]

Hi aim1732

Thank you for your help. It was very rude of me to leave it so late.
The assignment has already been marked, and that question was not even included in the marking (only half of the questions are randomly chosen for marking).

Thanks again.

Cheers

kwy