definite integral

**Random Variable** · December 22nd 2010, 11:00 AM

Math Challenge Problem:

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

**FernandoRevilla** · December 22nd 2010, 11:49 AM

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

**Random Variable** · December 22nd 2010, 12:30 PM

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

**chisigma** · December 22nd 2010, 12:42 PM

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)

Merry Christmas from Italy

$\chi$ $\sigma$

**FernandoRevilla** · December 22nd 2010, 12:54 PM

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

**Ackbeet** · December 22nd 2010, 01:08 PM

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.

**Random Variable** · December 22nd 2010, 01:19 PM

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.

**chisigma** · December 22nd 2010, 01:24 PM

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!...

Merry Christmas from Italy

$\chi$ $\sigma$

**Random Variable** · December 22nd 2010, 01:32 PM

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 $0<t<\infty$ )

now integrate by parts

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

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

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

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

$= \sum^{\infty}_{k=1} (-1)^{k} \ \frac{1}{k^{2}} = -\frac{\pi^{2}}{12}$

**Bruno J.** · December 22nd 2010, 03:55 PM

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?

**FernandoRevilla** · December 22nd 2010, 10:26 PM

Originally Posted by Bruno J.

I'm not saying you're wrong,

You can say it.

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

**TheCoffeeMachine** · December 29th 2010, 08:20 PM

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!

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