0-1-log-1-x-7-1-x-7-dx-

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} = \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}}))