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Question Number 62220 by maxmathsup by imad last updated on 17/Jun/19

l e t f (x) = \int_{0}^{\infty} \frac{t^{2}}{x^{6} + t^{6}} d t w i t h x > 0

1) c a l c u l a t e f (x)

2) c a l c u l a t e g (x) = \int_{0}^{\infty} \frac{t^{2}}{{(x^{6} + t^{6})}^{2}} d t

3) f i n d v a l u e s o f i n t e g r a l s \int_{0}^{\infty} \frac{t^{2}}{t^{6} + 8} d t a n d \int_{0}^{\infty} \frac{t^{2}}{{(t^{6} + 8)}^{2}} d t .

Commented by maxmathsup by imad last updated on 19/Jun/19

1) f (x) = \int_{0}^{\infty} \frac{t^{2}}{x^{6} + t^{6}} d t c h a n g e m e n t t = x u g i v e

f (x) = \int_{0}^{\infty} \frac{x^{2} u^{2}}{x^{6} + x^{6} u^{6}} x d u = \frac{1}{x^{3}} \int_{0}^{\infty} \frac{u^{2}}{1 + u^{6}} d u c h a n g e m e n t u^{6} = α g i v e u = α^{\frac{1}{6}}

x^{3} f (x) = \int_{0}^{\infty} \frac{{(α^{\frac{1}{6}})}^{2}}{1 + α} \frac{1}{6} α^{\frac{1}{6} - 1} d u = \int_{0}^{\infty} \frac{α^{\frac{1}{3} + \frac{1}{6} - 1}}{1 + α} d u = \frac{1}{6} \int_{0}^{\infty} \frac{α^{\frac{1}{2} - 1}}{1 + α} d α

= \frac{1}{6} \frac{π}{s i n (\frac{π}{2})} = \frac{π}{6} \Rightarrow f (x) = \frac{π}{6 x^{3}} (x > 0)

2) w e h a v e f^{^{'}} (x) = - \int_{0}^{\infty} \frac{6 x^{5} t^{2}}{{(x^{6} + t^{6})}^{2}} d t = - 6 x^{5} \int_{0}^{\infty} \frac{t^{2}}{{(x^{6} + t^{6})}^{2}} d t = - 6 x^{5} g (x) \Rightarrow

g (x) = - \frac{f^{^{'}} (x)}{6 x^{5}} b u t f (x) = \frac{π}{6} x^{- 3} \Rightarrow f^{^{'}} (x) = \frac{π}{6} (- 3) x^{- 4} = - \frac{π}{2 x^{4}} \Rightarrow

g (x) = - \frac{1}{6 x^{5}} (- \frac{π}{2 x^{4}}) = \frac{π}{12 x^{9}}

3) \int_{0}^{\infty} \frac{t^{2}}{t^{6} + 8} d t = \int_{0}^{\infty} \frac{t^{2}}{{(\sqrt{2})}^{6} + t^{6}} f (\sqrt{2}) = \frac{π}{6 {(\sqrt{2})}^{3}} = \frac{π}{12 \sqrt{2}} = \frac{π \sqrt{2}}{24}

\int_{0}^{\infty} \frac{t^{2}}{{(t^{6} + 8)}^{2}} d t = g (\sqrt{2}) = \frac{π}{12 {(\sqrt{2})}^{9}} = \frac{π}{12 . \sqrt{2} (2^{4})} = \frac{π}{12 \times 16 \times \sqrt{2}} = \dots

Commented by maxmathsup by imad last updated on 19/Jun/19

i h a v e u s e d t h e r e s u l t \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{s i n (π a)} i f 0 < a < 1