Ω=∫_0 ^1 ((log(1+x^7 ))/(1+x^7 ))dx=?


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Question Number 162219 by mathlove last updated on 27/Dec/21

$Ω = \underset{0}{\int^{1}} \frac{l o g (1 + x^{7})}{1 + x^{7}} d x = ?$
	Answered by amin96 last updated on 27/Dec/21

	$x^{7} = - t \frac{d t}{d x} = - 7 x^{6} = - 7 t^{\frac{6}{7}}$ $Ω = \frac{1}{7} \int_{- 1}^{0} \frac{l n (1 - t)}{(1 - t) t^{\frac{6}{7}}} d t = - \frac{1}{7} \underset{n = 1}{\sum^{\infty}} H_{n} \int_{- 1}^{0} t^{n - \frac{6}{7}} d t =$ $= - \frac{1}{7} \underset{n = 1}{\sum^{\infty}} H_{n} {[\frac{t^{n + \frac{1}{7}}}{n + \frac{1}{7}}]}_{- 1}^{0} = - \frac{1}{7} \underset{n = 1}{\sum^{\infty}} \frac{H_{n} {(- 1)}^{n + \frac{1}{7}}}{(n + \frac{1}{7})} =$ $= \frac{1}{7} \underset{n = 1}{\sum^{\infty}} \frac{H_{n} {(- 1)}^{n}}{(n + \frac{1}{7})} = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n} H_{n}}{7 n + 1}$
	Answered by mindispower last updated on 30/Dec/21
	$1+x^7 =Π_(k=0) ^6 (x−a_k ),a_k =e^(i(1+2k)(π/7)) ,k∈{0,6} (1/(1+x^7 ))=(1/7)Σ_(k=0) ^6 ((−a_k )/(x−a_k )). log(1+x^7 )=Σ_(j=0) ^6 ln(x−a_j ) ⇔(1/7)Σ_(j=0) ^6 Σ_(k=0) ^6 −a_k ∫_0 ^1 .((ln(x−a_j ))/(x−a_k ))dx ∫_0 ^1 ((ln(x−a_j ))/(x−a_k ))dx,y=x−a_k =∫_(−a_k ) ^(1−a_k ) ((ln(y+a_k −a_j ))/y)dy y=(a_j −a_k )z⇔ ∫_(−((ak)/(a_j −a_k ))) ^((1−a_k )/(a_j −a_k )) ((ln((a_k −a_j )(1−z)))/z)dz =ln(a_k −a_j )ln(1−(1/a_k ))+∫_(−(a_k /(a_j −a_k ))) ^((1−a_k )/(a_j −a_k )) ((ln(1−x))/x)dx Li_2 (z)=−∫_0 ^z ((ln(1−t))/t)dt We Get ln(a_k −a_j )ln(1−(1/a_k ))+Li_2 ((a_k /(a_k −a_j )))−Li_2 (((a_k −1)/(a_k −a_j ))) We Get ∫^1 _0 ((ln(1+x^7 ))/(1+x^7 ))dx=Σ_(j=0) ^6 Σ_(k=0) ^6 ((−a_k )/7)[ln(a_k −a_j )ln(1−(1/a_k ))+Li_2 ((a_k /(a_k −a_j )))−Li_2 (((a_k −1)/(a_k −a_j ))))$
	$1 + x^{7} = \underset{k = 0}{\prod^{6}} (x - a_{k}), a_{k} = e^{i (1 + 2 k) \frac{π}{7}}, k \in {0, 6}$ $\frac{1}{1 + x^{7}} = \frac{1}{7} \underset{k = 0}{\sum^{6}} \frac{- a_{k}}{x - a_{k}} .$ $l o g (1 + x^{7}) = \underset{j = 0}{\sum^{6}} l n (x - a_{j})$ $\Leftrightarrow \frac{1}{7} \underset{j = 0}{\sum^{6}} \underset{k = 0}{\sum^{6}} - a_{k} \int_{0}^{1} . \frac{l n (x - a_{j})}{x - a_{k}} d x$ $\int_{0}^{1} \frac{l n (x - a_{j})}{x - a_{k}} d x, y = x - a_{k}$ $= \int_{- a_{k}}^{1 - a_{k}} \frac{l n (y + a_{k} - a_{j})}{y} d y$ $y = (a_{j} - a_{k}) z \Leftrightarrow$ $\int_{- \frac{a k}{a_{j} - a_{k}}}^{\frac{1 - a_{k}}{a_{j} - a_{k}}} \frac{l n ((a_{k} - a_{j}) (1 - z))}{z} d z$ $= l n (a_{k} - a_{j}) l n (1 - \frac{1}{a_{k}}) + \int_{- \frac{a_{k}}{a_{j} - a_{k}}}^{\frac{1 - a_{k}}{a_{j} - a_{k}}} \frac{l n (1 - x)}{x} d x$ ${L i}_{2} (z) = - \int_{0}^{z} \frac{l n (1 - t)}{t} d t$ $W e G e t$ $l n (a_{k} - a_{j}) l n (1 - \frac{1}{a_{k}}) + {L i}_{2} (\frac{a_{k}}{a_{k} - a_{j}}) - {L i}_{2} (\frac{a_{k} - 1}{a_{k} - a_{j}})$ $W e G e t$ ${\int_{0}}^{1} \frac{l n (1 + x^{7})}{1 + x^{7}} d x = \underset{j = 0}{\sum^{6}} \underset{k = 0}{\sum^{6}} \frac{- a_{k}}{7} [l n (a_{k} - a_{j}) l n (1 - \frac{1}{a_{k}}) + {L i}_{2} (\frac{a_{k}}{a_{k} - a_{j}}) - {L i}_{2} (\frac{a_{k} - 1}{a_{k} - a_{j}}))$
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