nice-calculus-please-calculate-I-0-1-1-x-1-6-dx-notice-x-is-the-fractionl-of-x-

Question Number 129805 by mnjuly1970 last updated on 19/Jan/21

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

p l e a s e c a l c u l a t e :::

I := \int_{0}^{1} {\frac{1}{\sqrt[6]{x}}} d x

n o t i c e : {x} i s t h e f r a c t i o n l o f^{″} x^{″} .

\dots \dots \dots \dots \dots .

Answered by mindispower last updated on 19/Jan/21

l e t u = \frac{1}{^{6} \sqrt{x}} \Rightarrow x = \frac{1}{u^{6}} \Rightarrow d x = \frac{- 6}{u^{7}} d u

\Leftrightarrow 6 \int_{1}^{\infty} {u} . \frac{d u}{u^{7}}

= 6 \sum_{k ⩾ 1} \int_{k}^{k + 1} [u - k] . \frac{d u}{u^{7}}

= 6 \int_{1}^{\infty} \frac{d u}{u^{6}} - 6 \sum_{k ⩾ 1} \int_{k}^{k + 1} \frac{k}{u^{7}} d u

= \frac{6}{5} + \sum_{k ⩾ 1} {[\frac{k}{u^{6}}]}_{k}^{k + 1} = \frac{6}{5} + \sum_{k ⩾ 1} \frac{k}{{(k + 1)}^{6}} - \frac{1}{k^{5}}

= \frac{6}{5} + \sum_{k ⩾ 1} (\frac{1}{{(k + 1)}^{5}} - \frac{1}{k^{5}}) - \frac{1}{{(k + 1)}^{6}})

= \frac{6}{5} - 1 - ζ (6) + 1 = \frac{6}{5} - ζ (6) = \frac{6}{5} - \frac{π^{6}}{945}