calculate-0-log-3-x-1-x-3-dx-

Question Number 142425 by Mathspace last updated on 31/May/21

c a l c u l a t e \int_{0}^{\infty} \frac{{l o g}^{3} x}{1 + x^{3}} d x

Answered by mathmax by abdo last updated on 02/Jun/21

Φ = \int_{0}^{\infty} \frac{\log^{3} x}{1 + x^{3}} dx changement x^{3} = t give

Φ = \int_{0}^{\infty} \frac{{(\log (t^{\frac{1}{3}}))}^{3}}{1 + t} \frac{1}{3} t^{\frac{1}{3} - 1} dt = \frac{1}{3^{4}} \int_{0}^{\infty} t^{\frac{1}{3} - 1} \frac{\log^{3} t}{1 + t} dt

let f (a) = \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} dt with 0 < a < 1 we have f (a) = \frac{π}{\sin (π a)}

and f^{^{'}} (a) = \int_{0}^{\infty} t^{a - 1} \frac{logt}{1 + t} dt \Rightarrow f^{,,} (a) = \int_{0}^{\infty} \frac{t^{a - 1} \log^{2} t}{1 + t} dt \Rightarrow

f^{(3)} (a) = \int_{0}^{\infty} \frac{t^{a - 1} \log^{3} t}{1 + t} dt \Rightarrow f^{(3)} (\frac{1}{3}) = \int_{0}^{\infty} \frac{t^{\frac{1}{3} - 1} \log^{3} t}{1 + t} dt

= 3^{4} Φ but f (a) = \frac{π}{\sin (π a)} \Rightarrow f^{^{'}} (a) = - π^{2} \frac{\cos (π a)}{\sin^{2} (π a)} \Rightarrow

f^{(2)} (a) = - π^{2} . \frac{- π \sin (π a) - 2 π \sin (π a) \cos (π a)}{\sin^{4} (π a)}

= π^{3} . \frac{1 + 2 \cos (π a)}{\sin^{3} (π a)} \Rightarrow

f^{(3)} (a) = π^{3} . \frac{- 2 π \sin (π a) . \sin^{3} (π a) - 3 π \sin^{2} (π a) \cos (π a)}{\sin^{6} (π a)}

= π^{4} . \frac{- {2 \sin}^{2} (π a) - 3 \cos (π a)}{\sin^{4} (π a)} \Rightarrow

f^{(3)} (\frac{1}{3}) = - π^{4} \times \frac{2 {(\frac{\sqrt{3}}{2})}^{2} + 3 . \frac{1}{2}}{{(\frac{\sqrt{3}}{2})}^{4}} = - π^{4} . \frac{\frac{3}{2} + \frac{3}{2}}{\frac{9}{16}}

= (- π^{4}) . (\frac{16}{9}) (3) = - \frac{16 π^{4}}{3} \Rightarrow Φ = - \frac{16 π^{4}}{3^{5}}