find-the-value-of-0-xlogx-1-x-3-2-dx-

Question Number 142426 by Mathspace last updated on 31/May/21

f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{x l o g x}{{(1 + x^{3})}^{2}} d x

Answered by mindispower last updated on 01/Jun/21

x \to \frac{1}{x}

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

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

- \frac{1}{9} \int_{0}^{\infty} \frac{t^{\frac{- 2}{3}}}{1 + t} d t - \frac{1}{27} \int_{0}^{\infty} \frac{t^{- \frac{2}{3}} l n (t)}{1 + t} d t

β (x, y) = \int_{0}^{\infty} \frac{t^{x - 1}}{{(1 + t)}^{x + y}}

- \frac{1}{9} β (\frac{1}{3}, \frac{2}{3}) - \frac{1}{27} β_{x} (\frac{1}{3}, \frac{2}{3})

= - \frac{2 π}{9 \sqrt{3}} - \frac{1}{27} β (\frac{1}{3}, \frac{2}{3}) (Ψ (1) - Ψ (\frac{1}{3}))

= - \frac{2 π}{9 \sqrt{3}} (1 + \frac{1}{3} (- γ + γ + \frac{π}{2 \sqrt{3}} + \frac{3}{2} l n (3))

= - \frac{2 π}{9 \sqrt{3}} - \frac{π^{2}}{81} - \frac{π l n (3)}{9 \sqrt{3}}