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Question Number 33351 by caravan msup abdo. last updated on 14/Apr/18

p r o v e t h a t

\int_{0}^{1} \frac{{(- l n x)}^{p}}{1 + x^{2}} = p! \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{p + 1}}

p i n t e g r .

Commented by math khazana by abdo last updated on 19/Apr/18

l e t p u t A_{p} = \int_{0}^{1} \frac{{(- l n x)}^{p}}{1 + x^{2}} d x

A_{p} = {(- 1)}^{p} \int_{0}^{1} \frac{{(l n x)}^{p}}{1 + x^{2}} d x

= {(- 1)}^{p} \int_{0}^{1} (\sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n}) {(l n x)}^{p} d x

= \sum_{n = 0}^{\infty} {(- 1)}^{n + p} \int_{0}^{1} x^{2 n} {(l n x)}^{p} d x l e t f i n d

I_{n, p} = \int_{0}^{1} x^{n} {(l n x)}^{p} d x b y p a r t s

I_{n, p} = {[\frac{1}{n + 1} x^{n + 1} {(l n x)}^{p}]}_{0}^{1} - \int_{0}^{1} \frac{1}{n + 1} x^{n + 1} \frac{p}{x} {(l n x)}^{p - 1}

= - \frac{p}{n + 1} \int_{0}^{1} x^{n} {(l n x)}^{p - 1} d x = \frac{- p}{n + 1} I_{n, p - 1} \Rightarrow

I_{n, p} = \frac{(- p) (- (p - 1))}{{(n + 1)}^{2}} I_{n, p - 2} = \dots . = \frac{{(- 1)}^{p} p!}{{(n + 1)}^{p}} I_{n, 0}

I_{n, 0} = \int_{0}^{1} x^{n} d x = \frac{1}{n + 1} \Rightarrow I_{n, p} = \frac{{(- 1)}^{p} p!}{{(n + 1)}^{p + 1}} \Rightarrow

\int_{0}^{1} x^{2 n} {(l n x)}^{p} d x = \frac{{(- 1)}^{p} p!}{{(2 n + 1)}^{p + 1}} a n d

A_{p} = \sum_{n = 0}^{\infty} {(- 1)}^{n + p} \frac{{(- 1)}^{p} p!}{{(2 n + 1)}^{p + 1}}

= p! \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{p + 1}} .