nice-calculus-prove-that-0-1-ln-1-x-2-x-dx-pi-2-12-m-n-1970-

Question Number 122525 by mnjuly1970 last updated on 17/Nov/20

\dots n i c e c a l c u l u s \dots

p r o v e t h a t :::::≫ Ψ = \int_{0}^{1} \frac{l n (1 - x)}{2 - x} d x = - \frac{π^{2}}{12} ✓

\dots m . n . 1970 \dots

Answered by Dwaipayan Shikari last updated on 17/Nov/20

\int_{0}^{1} \frac{l o g (1 - x)}{2 - x} d x = \int_{0}^{1} \frac{l o g (x)}{1 + x} d x

= \int_{0}^{1} l o g (x) \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} x^{n}

= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{1} x^{n} l o g x d x

= \underset{n ⩾ 0}{\sum^{\infty}} {(- 1)}^{n} (- \int_{0}^{1} \frac{x^{n}}{(n + 1)} d x)

= - \underset{n ⩾ 0}{\sum^{\infty}} {(- 1)}^{n} \frac{1}{{(n + 1)}^{2}} = - \frac{π^{2}}{12}

Commented by mnjuly1970 last updated on 17/Nov/20

p e r f e c t \dots