....advanced calculus... evaluate: i:: ∫_0 ^( 1) xH_x dx =??? ii::Σ_(n=1 ) ^∞ (H_n /(n^2 2^(n ) )) =??? iii:: ∫_0 ^( 1) ((ln(((x+1))^(1/3) ))/((x+1)(x+2)))=??? m.n. july 1970...#


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Question Number 111965 by mnjuly1970 last updated on 05/Sep/20

$. . . . a d v a n c e d c a l c u l u s . . .$ $e v a l u a t e :$ $i :: \int_{0}^{1} {x H}_{x} d x = ? ? ?$ $i i :: \underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n^{2} 2^{n}} = ? ? ?$ $i i i :: \int_{0}^{1} \frac{l n (\sqrt[3]{x + 1})}{(x + 1) (x + 2)} = ? ? ?$ $You can't use 'macro parameter character #' in math mode$
	Answered by mindispower last updated on 05/Sep/20
	$∫_0 ^1 ((ln(x+1))/((x+1)(x+2)))dx ∫_0 ^1 ((ln(x+1))/(x+1))dx_(=A) −∫_0 ^1 ((ln(x+1))/(x+2))dx_(=B) A=(1/2)[ln^2 (x+1)]_0 ^1 =((ln^2 (2))/2) B=−[ln(x+1)ln(x+2)]_0 ^1 +∫_0 ^1 ((ln(x+2))/(x+1))dx By part and x+1=u⇒ B=−ln(2)ln(3)+∫_1 ^2 ((ln(u+1))/u)du u=−y⇒B=−ln(2)ln(3)+∫_(−1) ^(−2) ((ln(1−y))/y)dy B=−ln(2)ln(3)−[−∫_(−1) ^(−2) ((ln(1−y))/y)dy] =−ln(2)ln(3)−Li_2 (−2)+Li_2 (−1) Li_2 (z)=Σ(z^k /k^2 ) Li_2 (−1)=Σ_(k≥1) (1/(4k^2 ))−Σ_(k≥0) (1/((2k+1)^2 ))=((ζ(2))/4)−{ζ(2)−((ζ(2))/4)} =−(1/2)=−(1/2)ζ(2)=−(π^2 /(12)) ∫_0 ^1 ((ln(x+1))/((x+1)(x+2)))dx=A+B=((ln^2 (2))/2)−ln(2)ln(3)−Li_2 (−2)−(π^2 /(12)) =−Li_2 (−2)−(1/2)ln(2)ln((2/9))−(π^2 /(12))$
	$\int_{0}^{1} \frac{l n (x + 1)}{(x + 1) (x + 2)} d x$ $\int_{0}^{1} \frac{l n (x + 1)}{x + 1} {d x}_{= A} - \int_{0}^{1} \frac{l n (x + 1)}{x + 2} d \underset{= B}{x}$ $A = \frac{1}{2} {[{l n}^{2} (x + 1)]}_{0}^{1} = \frac{{l n}^{2} (2)}{2}$ $B = - {[l n (x + 1) l n (x + 2)]}_{0}^{1} + \int_{0}^{1} \frac{l n (x + 2)}{x + 1} d x$ $B y p a r t a n d x + 1 = u \Rightarrow$ $B = - l n (2) l n (3) + \int_{1}^{2} \frac{l n (u + 1)}{u} d u$ $u = - y \Rightarrow B = - l n (2) l n (3) + \int_{- 1}^{- 2} \frac{l n (1 - y)}{y} d y$ $B = - l n (2) l n (3) - [- \int_{- 1}^{- 2} \frac{l n (1 - y)}{y} d y]$ $= - l n (2) l n (3) - {L i}_{2} (- 2) + {L i}_{2} (- 1)$ ${L i}_{2} (z) = Σ \frac{z^{k}}{k^{2}}$ ${L i}_{2} (- 1) = \sum_{k ⩾ 1} \frac{1}{4 k^{2}} - \sum_{k ⩾ 0} \frac{1}{{(2 k + 1)}^{2}} = \frac{ζ (2)}{4} - {ζ (2) - \frac{ζ (2)}{4}}$ $= - \frac{1}{2} = - \frac{1}{2} ζ (2) = - \frac{π^{2}}{12}$ $\int_{0}^{1} \frac{l n (x + 1)}{(x + 1) (x + 2)} d x = A + B = \frac{{l n}^{2} (2)}{2} - l n (2) l n (3) - {L i}_{2} (- 2) - \frac{π^{2}}{12}$ $= - {L i}_{2} (- 2) - \frac{1}{2} l n (2) l n (\frac{2}{9}) - \frac{π^{2}}{12}$
	Commented by mathdave last updated on 06/Sep/20

	$m i s t a k e f r o m t h e f i r s t s t e p$
	Commented by mathdave last updated on 06/Sep/20

	$i t s u p p o s e d t o b e s o m e t h i n g l i k e t h i s$ $\int_{0}^{1} \frac{\ln {(1 + x)}^{\frac{1}{3}}}{(1 + x) (2 + x)} d x = \frac{1}{3} \int_{0}^{1} \frac{\ln (1 + x)}{(1 + x)} d x - \frac{1}{3} \int_{0}^{1} \frac{\ln (1 + x)}{(2 + x)} d x$
	Commented by mindispower last updated on 06/Sep/20

	$i h a v e s t a r t e w i t h e \int_{0}^{1} \frac{l n (x + 1)}{(1 + x) (2 + x)} d x$ $j u s t b e c a u s e i d o n t w a n t w r i t e e v r e y s t e p s \frac{1}{3}$
	Commented by mathdave last updated on 06/Sep/20

	$a s i d e t h a t u d o n t w a m n a u s e d$ $e v e r y t h i n g w h e n g e t t i n g u r f i n a l$ $a n s w e r \frac{1}{3} w a s n t r e f l e c t i n g t h e r e$
	Commented by mnjuly1970 last updated on 06/Sep/20

	$t h a n k y o u s o m u c h f o r y o u$ $c a r e f u l n e s s . . .$
	Commented by mnjuly1970 last updated on 06/Sep/20

	$t h a n k y o u s o m u c h f o r y o u r$ $e f f o r t . g r a t e f u l . . . .$
	Commented by Tawa11 last updated on 06/Sep/21

	$grest sir$
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