... nice calculus ... please calculate ::: I:= ∫_0 ^( 1) {(1/( (x)^(1/6) )) }dx notice: {x} is the fractionl of ′′ x ′′. ................


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Question Number 129805 by mnjuly1970 last updated on 19/Jan/21
$... nice calculus ... please calculate ::: I:= ∫_0 ^( 1) {(1/( (x)^(1/6) )) }dx notice: {x} is the fractionl of ′′ x ′′. ................$
$. . . n i c e c a l c u l u s . . .$ $p l e a s e c a l c u l a t e :::$ $I := \int_{0}^{1} {\frac{1}{\sqrt[6]{x}}} d x$ $n o t i c e : {x} i s t h e f r a c t i o n l o f^{″} x^{″} .$ $. . . . . . . . . . . . . . . .$
	Answered by mindispower last updated on 19/Jan/21
	$let u=(1/(^6 (√x)))⇒x=(1/u^6 )⇒dx=((−6)/u^7 )du ⇔6∫_1 ^∞ {u}.(du/u^7 ) =6Σ_(k≥1) ∫_k ^(k+1) [u−k].(du/u^7 ) =6∫_1 ^∞ (du/u^6 )−6Σ_(k≥1) ∫_k ^(k+1) (k/u^7 )du =(6/5)+Σ_(k≥1) [(k/u^6 )]_k ^(k+1) =(6/5)+Σ_(k≥1) (k/((k+1)^6 ))−(1/k^5 ) =(6/5)+Σ_(k≥1) ((1/((k+1)^5 ))−(1/k^5 ))−(1/((k+1)^6 ))) =(6/5)−1−ζ(6)+1=(6/5)−ζ(6)=(6/5)−(π^6 /(945))$
	$l e t u = \frac{1}{^{6} \sqrt{x}} \Rightarrow x = \frac{1}{u^{6}} \Rightarrow d x = \frac{- 6}{u^{7}} d u$ $\Leftrightarrow 6 \int_{1}^{\infty} {u} . \frac{d u}{u^{7}}$ $= 6 \sum_{k ⩾ 1} \int_{k}^{k + 1} [u - k] . \frac{d u}{u^{7}}$ $= 6 \int_{1}^{\infty} \frac{d u}{u^{6}} - 6 \sum_{k ⩾ 1} \int_{k}^{k + 1} \frac{k}{u^{7}} d u$ $= \frac{6}{5} + \sum_{k ⩾ 1} {[\frac{k}{u^{6}}]}_{k}^{k + 1} = \frac{6}{5} + \sum_{k ⩾ 1} \frac{k}{{(k + 1)}^{6}} - \frac{1}{k^{5}}$ $= \frac{6}{5} + \sum_{k ⩾ 1} (\frac{1}{{(k + 1)}^{5}} - \frac{1}{k^{5}}) - \frac{1}{{(k + 1)}^{6}})$ $= \frac{6}{5} - 1 - ζ (6) + 1 = \frac{6}{5} - ζ (6) = \frac{6}{5} - \frac{π^{6}}{945}$
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