Find $\displaystyle \int_0^1\frac{x^2}{(1-x)\ln^2(1-x)}\,dx.$
You can rewrite it as $\displaystyle \int_0^1{\sum_{n=0}^{\infty}x^{n+2}}\cdot\bigg(-\sum_{n=0}^{\infty}\frac{x^{n}}{n}\bigg)^{-2}$...which then is equal to$\displaystyle \int_0^1{\sum_{n=0}^{\infty}x^{n+2}\cdot{2\sum_{n= 0}^{\infty}\sum_{n=0}^{\infty}\frac{n^2}{x^{2n}}}}$...and then $\displaystyle 2\int_0^1{\sum_{n=0}^{\infty}\sum_{n=0}^{\infty}\s um_{n=0}^{\infty}\frac{x^{2-n}}{n}}$...so wed get $\displaystyle 2\sum_{n=0}^{\infty}\sum_{n=0}^{\infty}\sum_{n=0}^ {\infty}\frac{x^{3-n}}{n(3-n)}$ evaluated from 0 to 1...I know I know I did it completely wrong...but will someone point out where? I just did this for novelty
Hello,
Let $\displaystyle u=\ln(1-x)$
--------------
$\displaystyle du=-\frac{dx}{1-x}$
--------------
$\displaystyle e^u=1-x$
$\displaystyle x=1-e^u$
$\displaystyle x^2=(1-e^u)^2$
--------------
If x=0 then $\displaystyle u=0$
If x=1 then $\displaystyle u=-\infty$
--------------
Inverting the extremities of the integral with the - sign coming from du
$\displaystyle \int_0^1\frac{x^2}{(1-x)\ln^2(1-x)} dx = \int_{-\infty}^0 \frac{(1-e^u)^2}{u^2} du$
Where did i go wrong ? :'(
is that then equal to $\displaystyle \int{\bigg(-2\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n-1}}{n!}+\sum_{n=0}^{\infty}\frac{(-2)^{n}x^{n-2}}{n!}+1\bigg)dx}$$\displaystyle =-2\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n}}{n\cdot{n!}}+\sum_{n=0}^{\infty}\frac{ (-2)^{n}x^{n-1}}{(n-1)n!}+x$ evaluated from 0 to ∞?