Advanced-Calculus-0-1-ln-1-1-x-x-3-dx-3-2-3-solution-I-B-P-1-2x-2-ln-3-

Question Number 160734 by mnjuly1970 last updated on 05/Dec/21

You can't use 'macro parameter character #' in math mode

Φ = \int_{0}^{1} {(\frac{{l n}^{} (\frac{1}{1 - x})}{x})}^{3} d x \overset{?}{=} 3 (ζ (2) + ζ (3))

- - - - s o l u t i o n - - - -

Φ \overset{I . B . P}{=} {[\frac{1}{2 x^{2}} {l n}^{3} (1 - x)]}_{0}^{1} + \frac{3}{2} \int_{0}^{1} \frac{{l n}^{2} (1 - x)}{x^{2} (1 - x)} d x

= \frac{1}{2} {l i m}_{ξ \to 1^{-}} \frac{{l n}^{3} (1 - ξ)}{ξ^{2}}

+ \frac{3}{2} [\int_{0}^{1} \frac{{l n}^{2} (1 - x)}{x} d x = 2 ζ (3)]

+ \frac{3}{2} [\int_{0}^{1} \frac{{l n}^{2} (1 - x)}{x^{2}} d x = \frac{π^{2}}{3} = 2 ζ (2)]

+ \frac{3}{2} \int_{0}^{1} \frac{{l n}^{2} (1 - x)}{1 - x} d x}

= \frac{1}{2} {l i m}_{ξ \to 1^{-}} {\frac{{l n}^{3} (1 - ξ)}{ξ^{2}} - {l n}^{3} (1 - ξ)}

+ \frac{3}{2} (2 ζ (3)) + \frac{3}{2} (2 ζ (2))

= 3 (ζ (3) + 3 ζ (2)) m . n

Commented by Gbenga last updated on 06/Dec/21

c a n y o u t e a c h m e t h i s s i r

Facebook Tweet Pin