# definite integral

• Dec 22nd 2010, 11:00 AM
Random Variable
definite integral
Math Challenge Problem:

$\int^{2}_{1} \frac{\ln(x-1)}{x} \ dx$
• Dec 22nd 2010, 11:49 AM
FernandoRevilla
Quote:

Originally Posted by Random Variable
Math Challenge Problem:

$\int^{2}_{1} \frac{\ln(x-1)}{x} \ dx$

Let's denote:

$I=\displaystyle\int^{2}_{1} \dfrac{\ln(x-1)}{x} \ dx$

with the substitution:

$t=\ln (x-1)$ we obtain:

$I=\displaystyle\int^{0}_{-\infty} \dfrac{te^t\;dt}{e^t+1}$

Now, apply twice integration by parts.

Fernando Revilla
• Dec 22nd 2010, 12:30 PM
Random Variable
$\int \frac{t e^{t}}{e^{t}+1} \ dt$ does not have an antiderivative that can be expressed as a finite number of elementary functions. So it's not quite that straightforward.
• Dec 22nd 2010, 12:42 PM
chisigma
Quote:

Originally Posted by Random Variable
Math Challenge Problem:

$\int^{2}_{1} \frac{\ln(x-1)}{x} \ dx$

Setting $x-1= u$ the integral becomes...

$\displaystyle \int_{0}^{1} \frac{\ln u}{1+u}\ du$ (1)

Now if You consider that is...

$\displaystyle \frac{1}{1+u} = \sum_{n=0}^{\infty} (-1)^{n}\ u^{n}$ (2)

... and...

$\displaystyle \int_{0}^{1} u^{n}\ \ln u = - \frac{1}{(n+1)^{2}}$ (3)

... You obtain...

$\displaystyle \int_{0}^{1} \frac{\ln u}{1+u}\ du = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} = - \frac{\pi ^{2}}{12}$ (4)

http://digilander.libero.it/luposaba...ato&#91;1].jpg

Merry Christmas from Italy

$\chi$ $\sigma$
• Dec 22nd 2010, 12:54 PM
FernandoRevilla
Quote:

Originally Posted by Random Variable
$\int \frac{t e^{t}}{e^{t}+1} \ dt$ does not have an antiderivative that can be expressed as a finite number of elementary functions. So it's not quite that straightforward.

So what?. You can find $I$ (definite integral) by means of an equation.

Fernando Revilla
• Dec 22nd 2010, 01:08 PM
Ackbeet
Quote:

Originally Posted by chisigma
Setting $x-1= u$ the integral becomes...
$\displaystyle \int_{0}^{1} \frac{\ln u}{1+u}\ du = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{2}} = \frac{\pi ^{2}}{12}$ (4)

Mathematica and WolframAlpha both give the negative of your answer, so I'm guessing you have a sign error in there somewhere. Not sure where, though.
• Dec 22nd 2010, 01:19 PM
Random Variable
Quote:

Originally Posted by Ackbeet
Mathematica and WolframAlpha both give the negative of your answer, so I'm guessing you have a sign error in there somewhere. Not sure where, though.

I think he just left out the negative by mistake.
• Dec 22nd 2010, 01:24 PM
chisigma
Some years ago I 'found' the following formula...

$\displaystyle \int_{0}^{1} u^{n}\ \ln^{m} u\ du = (-1)^{m}\ \frac{m!}{(n+1)^{m+1}}$ (1)

In that case is $m=1$ so that...

$\displaystyle \int_{0}^{1} u^{n}\ \ln u\ du = - \frac{1}{(n+1)^{2}}$ (2)

Thank a lot to Ackbeet!...

http://digilander.libero.it/luposaba...ato&#91;1].jpg

Merry Christmas from Italy

$\chi$ $\sigma$
• Dec 22nd 2010, 01:32 PM
Random Variable
My solution:

Let $x = 1+e^{-t}$

$= -\int^{0}_{\infty} \frac{\ln(e^{-t})}{1+e^{-t}} \ e^{-t} \ dt$

$= - \int^{\infty}_{0} \frac{te^{-t}}{1+e^{-t}} \ dt$

$= \int^{\infty}_{0} t \sum^{\infty}_{k=1} (-1)^{k} e^{-kt} \ dt$ (since $e^{-t} < 1$ for $0 )

now integrate by parts

$t \sum_{k=1}^{\infty} (-1)^{k+1} \frac{e^{-kt}}{k} \Big|^{\infty}_{0} - \int^{\infty}_{0} \sum^{\infty}_{k=1} (-1)^{k+1} \ \frac{e^{-kt}}{k} \ dt$

$0 + \int^{\infty}_{0} \sum^{\infty}_{k=1} (-1)^{k} \ \frac{e^{-kt}}{k}$

$= \sum^{\infty}_{k=1} \int^{\infty}_{0} (-1)^{k} \ \frac{e^{-kt}}{k}
$
(assuming $f_{k}(t) = (-1)^{k} \ \frac{e^{-kt}}{k}$ converges uniformly on $0 \le t<\infty$)

$= -\sum^{\infty}_{k=1} (-1)^{k+1} \ \frac{1}{k^{2}}$

$= \sum^{\infty}_{k=1} (-1)^{k} \ \frac{1}{k^{2}} = -\frac{\pi^{2}}{12}$
• Dec 22nd 2010, 03:55 PM
Bruno J.
Quote:

Originally Posted by FernandoRevilla
So what?. You can find $I$ (definite integral) by means of an equation.

Fernando Revilla

How do you expect to get $\pi^2$ in there? I'm not saying you're wrong, but I really don't see how you can find $I$ by integration by parts or by "means of an equation" - perhaps you can be more precise and provide your solution?
• Dec 22nd 2010, 10:26 PM
FernandoRevilla
Quote:

Originally Posted by Bruno J.
I'm not saying you're wrong,

You can say it. :)

Quote:

but I really don't see how you can find $I$ by integration by parts or by "means of an equation" - perhaps you can be more precise and provide your solution?
I can't. I had an stupid computational mistake. Sorry. :)

Fernando Revilla
• Dec 29th 2010, 08:20 PM
TheCoffeeMachine
I'm able to show that:

\begin{aligned} \int_{0}^{1}\frac{\ln\left(x+1\right)}{x}\;{dt} &=\int_{0}^{1}\bigg(\sum_{k = 0}^{\infty}\frac{(-x)^k}{k+1}\bigg) \;{dx} = \displaystyle \sum_{k=0}^{\infty}\bigg(\int_{0}^{1}\frac{(-x)^k}{k+1} \;{dx}\bigg)\\ &= \sum_{k=0}^{\infty}\bigg[\frac{(-1)^{k}x^{k+1}}{(k+1)^2}\;{dx}\bigg]_{x = 0}^{x = 1} = \sum_{k=0}^{\infty}\frac{(-1)^{k}}{(k+1)^2} = \frac{\pi^2}{12}.\end{aligned}

But I'm lacking the change of variable to show this:

$\displaystyle \int^{2}_{1} \frac{\ln(x-1)}{x} \ dx = -\int_{0}^{1}\frac{\ln\left(x+1\right)}{x}$

What 'substitution' could do the trick? Mental block!