0-1-x-1-x-6-dx-

Question Number 160194 by amin96 last updated on 25/Nov/21

\int_{0}^{1} \frac{x}{1 - x^{6}} dx = ?

Commented by mr W last updated on 25/Nov/21

= \int_{0}^{1} x (1 + x^{6} + x^{12} + x^{18} + \dots) d x

= {[\frac{x^{2}}{2} + \frac{x^{8}}{8} + \frac{x^{14}}{14} + \frac{x^{20}}{20} + \dots]}_{0}^{1}

= \frac{1}{2} + \frac{1}{8} + \frac{1}{14} + \frac{1}{20} + \dots

= \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{1}{3 n + 1}

= d i v e r g e n t!

Answered by Ar Brandon last updated on 25/Nov/21

I = \int_{0}^{1} \frac{x}{1 - x^{6}} d x = \frac{1}{2} \int_{0}^{1} \frac{d x}{1 - x^{3}} = - \frac{1}{2} \int \frac{d x}{(x - 1) (x^{2} + x + 1)}

\frac{1}{x^{3} - 1} = \frac{a}{x - 1} + \frac{b x + c}{x^{2} + x + 1} = \frac{(a + b) x^{2} + (a - b + c) x + (a - c)}{x^{3} - 1}

a = - b, a = \frac{1}{3}, c = - \frac{2}{3}

I = - \frac{1}{6} \int_{0}^{1} (\frac{1}{x - 1} - \frac{x + 2}{x^{2} + x + 1}) d x

= - \frac{\ln ∣ x - 1 ∣}{6} + \frac{1}{6} \int (\frac{1}{2} \cdot \frac{2 x + 1}{x^{2} + x + 1} + \frac{3}{2} \cdot \frac{1}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}) d x

= {[- \frac{\ln ∣ x - 1 ∣}{6} + \frac{1}{12} \ln (x^{2} + x + 1) + \frac{1}{4} \cdot \frac{2}{\sqrt{3}} \arctan (\frac{2 x + 1}{\sqrt{3}})]}_{0}^{1}

Commented by mr W last updated on 25/Nov/21

t h e i n t e g r a l d i v e r g e s .

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