find-U-n-0-1-x-n-arctan-x-dx-with-n-integr-nstural-

Question Number 124919 by mathmax by abdo last updated on 07/Dec/20

find U_{n} = \int_{0}^{1} x^{n} \arctan (x) dx with n integr nstural

Commented by mindispower last updated on 07/Dec/20

b y p a r t = {[\frac{x^{n + 1}}{n + 1} \tan^{- 1} (x)]}_{0}^{1} - \frac{1}{n + 1} \int_{0}^{1} \frac{x^{n + 1}}{1 + x^{2}} d x

= \frac{π}{4 (n + 1)} - \frac{1}{n + 1} \int_{0}^{1} \sum_{k ⩾ 0} {(- 1)}^{k} x^{n + 1 + 2 k} d x

= \frac{π}{4 (n + 1)} - \frac{1}{(n + 1)} \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{n + 2 + 2 k}

= \frac{π}{4 (n + 1)} - \frac{1}{n + 1} \sum_{k ⩾ 0} \frac{(n + 2 + 2 (2 k + 1)) - (n + 2 + 4 k)}{(n + 2 + 2.2 k) (n + 2 + 2 (2 k + 1)}

= \frac{π}{4 (n + 1)} - \frac{1}{n + 1} \sum_{k ⩾ 1} \frac{2}{(n - 2 + 4 k) (n + 4 k)}

= \frac{π}{4 (n + 1)} - \frac{1}{8 (n + 1)} \sum_{k ⩾ 1} \frac{1}{(\frac{n - 1}{4} + k) (\frac{n}{4} + k)}

= \frac{π}{4 (n + 1)} - \frac{1}{8 (n + 1)} . \frac{Ψ (\frac{n}{4}) - Ψ (\frac{n - 1}{4})}{\frac{n}{4} - \frac{n - 1}{4}}

= \frac{π}{4 (n + 1)} - \frac{1}{2 (n + 1)} (Ψ (\frac{n}{4}) - Ψ (\frac{n - 1}{4})), n ⩾ 1

n = 0

w e g e t \frac{π}{4} - \int_{0}^{1} \frac{x}{1 + x^{2}} = \frac{π}{4} - l n (\sqrt{2})

Commented by Bird last updated on 07/Dec/20

t h a n k s s i r

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