Math Challenge Problem:
$\displaystyle \int^{2}_{1} \frac{\ln(x-1)}{x} \ dx $
Let's denote:
$\displaystyle I=\displaystyle\int^{2}_{1} \dfrac{\ln(x-1)}{x} \ dx $
with the substitution:
$\displaystyle t=\ln (x-1)$ we obtain:
$\displaystyle I=\displaystyle\int^{0}_{-\infty} \dfrac{te^t\;dt}{e^t+1} $
Now, apply twice integration by parts.
Fernando Revilla
Setting $\displaystyle x-1= u$ the integral becomes...
$\displaystyle \displaystyle \int_{0}^{1} \frac{\ln u}{1+u}\ du$ (1)
Now if You consider that is...
$\displaystyle \displaystyle \frac{1}{1+u} = \sum_{n=0}^{\infty} (-1)^{n}\ u^{n}$ (2)
... and...
$\displaystyle \displaystyle \int_{0}^{1} u^{n}\ \ln u = - \frac{1}{(n+1)^{2}}$ (3)
... You obtain...
$\displaystyle \displaystyle \int_{0}^{1} \frac{\ln u}{1+u}\ du = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} = - \frac{\pi ^{2}}{12}$ (4)
Merry Christmas from Italy
$\displaystyle \chi$ $\displaystyle \sigma$
So what?. You can find $\displaystyle I$ (definite integral) by means of an equation.
Fernando Revilla
Some years ago I 'found' the following formula...
$\displaystyle \displaystyle \int_{0}^{1} u^{n}\ \ln^{m} u\ du = (-1)^{m}\ \frac{m!}{(n+1)^{m+1}}$ (1)
In that case is $\displaystyle m=1$ so that...
$\displaystyle \displaystyle \int_{0}^{1} u^{n}\ \ln u\ du = - \frac{1}{(n+1)^{2}}$ (2)
Thank a lot to Ackbeet!...
Merry Christmas from Italy
$\displaystyle \chi$ $\displaystyle \sigma$
My solution:
Let $\displaystyle x = 1+e^{-t} $
$\displaystyle = -\int^{0}_{\infty} \frac{\ln(e^{-t})}{1+e^{-t}} \ e^{-t} \ dt $
$\displaystyle = - \int^{\infty}_{0} \frac{te^{-t}}{1+e^{-t}} \ dt $
$\displaystyle = \int^{\infty}_{0} t \sum^{\infty}_{k=1} (-1)^{k} e^{-kt} \ dt $ (since $\displaystyle e^{-t} < 1 $ for $\displaystyle 0<t<\infty $ )
now integrate by parts
$\displaystyle 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 $
$\displaystyle 0 + \int^{\infty}_{0} \sum^{\infty}_{k=1} (-1)^{k} \ \frac{e^{-kt}}{k} $
$\displaystyle = \sum^{\infty}_{k=1} \int^{\infty}_{0} (-1)^{k} \ \frac{e^{-kt}}{k}
$ (assuming$\displaystyle f_{k}(t) = (-1)^{k} \ \frac{e^{-kt}}{k} $ converges uniformly on $\displaystyle 0 \le t<\infty $)
$\displaystyle = -\sum^{\infty}_{k=1} (-1)^{k+1} \ \frac{1}{k^{2}} $
$\displaystyle = \sum^{\infty}_{k=1} (-1)^{k} \ \frac{1}{k^{2}} = -\frac{\pi^{2}}{12} $
You can say it.
I can't. I had an stupid computational mistake. Sorry.but I really don't see how you can find $\displaystyle I$ by integration by parts or by "means of an equation" - perhaps you can be more precise and provide your solution?
Fernando Revilla
I'm able to show that:
$\displaystyle \begin{aligned} \int_{0}^{1}\frac{\ln\left(x+1\right)}{x}\;{dt} &=\int_{0}^{1}\bigg(\sum_{k = 0}^{\infty}\frac{(-x)^k}{k+1}\bigg) \;{dx} = \displaystyle \sum_{k=0}^{\infty}\bigg(\int_{0}^{1}\frac{(-x)^k}{k+1} \;{dx}\bigg)\\ &= \sum_{k=0}^{\infty}\bigg[\frac{(-1)^{k}x^{k+1}}{(k+1)^2}\;{dx}\bigg]_{x = 0}^{x = 1} = \sum_{k=0}^{\infty}\frac{(-1)^{k}}{(k+1)^2} = \frac{\pi^2}{12}.\end{aligned}$
But I'm lacking the change of variable to show this:
$\displaystyle \displaystyle \int^{2}_{1} \frac{\ln(x-1)}{x} \ dx = -\int_{0}^{1}\frac{\ln\left(x+1\right)}{x}$
What 'substitution' could do the trick? Mental block!