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Question Number 62856 by mathmax by abdo last updated on 26/Jun/19

l e t f (λ) = \int_{0}^{+ \infty} \frac{x^{4}}{x^{6} + λ^{6}} d x w i t h λ > 0

1) c a l c u l a t e f (λ)

2) c a l c u l a t e a l s o g (λ) = \int_{0}^{\infty} \frac{x^{4}}{{(x^{6} + λ^{6})}^{2}} d x

3) f i n d t h e v a l u e s o f \int_{0}^{\infty} \frac{x^{4}}{x^{6} + 1} d x, \int_{0}^{\infty} \frac{x^{4}}{x^{6} + 8} d x a n d \int_{0}^{\infty} \frac{x^{4}}{{(x^{6} + 8)}^{2}} d x .

Commented by mathmax by abdo last updated on 26/Jun/19

1) w e u s e 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 (r e s u l t p r o v e d)

c h a n g e m e n t x = λ t g i v e f (λ) = \int_{0}^{\infty} \frac{{(λ t)}^{4}}{{(λ t)}^{6} + λ^{6}} λ d t = \frac{λ^{5}}{λ^{6}} \int_{0}^{\infty} \frac{t^{4}}{1 + t^{6}} d t

= \frac{1}{λ} \int_{0}^{\infty} \frac{t^{4} d t}{1 + t^{6}} =_{t = u^{\frac{1}{6}}} \frac{1}{λ} \int_{0}^{\infty} \frac{{(u^{\frac{1}{6}})}^{4}}{1 + u} \frac{1}{6} u^{\frac{1}{6} - 1} d u

= \frac{1}{6 λ} \int_{0}^{\infty} \frac{u^{\frac{2}{3} + \frac{1}{6} - 1}}{1 + u} d u = \frac{1}{6 λ} \int_{0}^{\infty} \frac{u^{\frac{5}{6} - 1}}{1 + u} d u = \frac{1}{6 λ} \frac{π}{s i n (\frac{5 π}{6})} = \frac{π}{6 λ s i n (π - \frac{π}{6})}

= \frac{π}{6 λ s i n (\frac{π}{6})} = \frac{π}{6 λ \frac{1}{2}} = \frac{π}{3 λ} \Rightarrow ★ f (λ) = \frac{π}{3 λ} ★

2) l e t d e r i v a t e f (λ) w e h a v e f^{^{'}} (λ) = - \int_{0}^{\infty} \frac{6 λ^{5} x^{4}}{{(x^{6} + λ^{6})}^{2}} d x

= - 6 λ^{5} \int_{0}^{\infty} \frac{x^{4}}{{(x^{6} + λ^{6})}^{2}} d x = - 6 λ^{5} g (λ) \Rightarrow g (λ) = - \frac{1}{6 λ^{5}} f^{^{'}} (λ)

w e h a v e f^{^{'}} (λ) = - \frac{π}{3} \frac{1}{λ^{2}} \Rightarrow g (λ) = - \frac{1}{6 λ^{5}} (- \frac{π}{3 λ^{2}}) = \frac{π}{18 λ^{7}} \Rightarrow

★ g (λ) = \frac{π}{18 λ^{7}} ★

Commented by mathmax by abdo last updated on 26/Jun/19

3) \int_{0}^{\infty} \frac{x^{4}}{1 + x^{6}} d x = f (1) = \frac{π}{3} l e t c a l c u l a t e

\int_{0}^{\infty} \frac{x^{4}}{x^{6} + 8} d x h e r e λ^{6} = 8 \Rightarrow λ = 8^{\frac{1}{6}} = {(2^{3})}^{\frac{1}{6}} = 2^{\frac{1}{2}} = 2 \Rightarrow

\int_{0}^{\infty} \frac{x^{4}}{x^{6} + 8} d x = f (\sqrt{2}) = \frac{π}{3 \sqrt{2}}

\int_{0}^{\infty} \frac{x^{4}}{{(x^{6} + 8)}^{2}} d x = g (\sqrt{2}) = \frac{π}{18 {(\sqrt{2})}^{7}} = \frac{π}{18 . 2^{\frac{7}{2}}} = \frac{π}{18.3 . \sqrt{2}} = \frac{π}{54 \sqrt{2}} .

Commented by mathmax by abdo last updated on 26/Jun/19

e r r o r o f t y p o 2^{\frac{1}{2}} = \sqrt{2}