calculate-0-x-2-lnx-1-x-6-dx-

Question Number 130136 by mathmax by abdo last updated on 22/Jan/21

calculate \int_{0}^{\infty} \frac{x^{2} lnx}{1 + x^{6}} dx

Answered by Dwaipayan Shikari last updated on 22/Jan/21

I^{'} (2) = \frac{- π^{2}}{6^{2}} . \frac{c o s (\frac{2 + 1}{6} π)}{{s i n}^{2} (\frac{2 + 1}{6} π)} = 0 (P r o v e d E a r l i e r)

Answered by Ar Brandon last updated on 22/Jan/21

I = \int_{0}^{1} \frac{x^{2} lnx}{1 + x^{6}} dx + \int_{1}^{\infty} \frac{x^{2} lnx}{1 + x^{6}} dx

= \int_{0}^{1} \frac{x^{2} lnx}{1 + x^{6}} dx - \int_{0}^{1} \frac{t^{2} lnt}{1 + t^{2}} dt = 0

Answered by mathmax by abdo last updated on 23/Jan/21

I = \int_{0}^{\infty} \frac{x^{2} lnx}{1 + x^{6}} dx we do the changement x^{6} = t \Rightarrow

I = \frac{1}{6} \int_{0}^{\infty} \frac{{(t^{\frac{1}{6}})}^{2} \ln (t^{\frac{1}{6}})}{1 + t} t^{\frac{1}{6} - 1} dt = \frac{1}{36} \int_{0}^{\infty} \frac{t^{\frac{1}{3} + \frac{1}{6} - 1} \ln (t)}{1 + t} dt

= \frac{1}{36} \int_{0}^{\infty} \frac{t^{- \frac{1}{2}} \ln (t)}{1 + t} dt but \int_{0}^{\infty} \frac{t^{a - 1} lnt}{1 + t} dt = - \frac{π^{2} \cos (π a)}{\sin^{2} (π a)} \Rightarrow

\int_{0}^{\infty} \frac{t^{- \frac{1}{2}} \ln (t)}{1 + t} dt = - \frac{π^{2} \cos (\frac{π}{2})}{\sin^{2} (\frac{π}{2})} = 0 \Rightarrow I = 0