calculate-0-dx-x-n-8-3-withn-gt-1-

Question Number 66470 by mathmax by abdo last updated on 15/Aug/19

c a l c u l a t e \int_{0}^{\infty} \frac{d x}{{(x^{n} + 8)}^{3}} w i t h n > 1

Commented by ~ À ® @ 237 ~ last updated on 16/Aug/19

l e t s t a t e \forall a > 0 f (a) = \int_{0}^{\infty} \frac{d x}{{(x^{n} + a)}^{2}}

f (a) = \int_{0}^{\infty} \frac{d x}{a^{2} {[{(\frac{x}{a^{\frac{1}{n}}})}^{n} + 1]}^{2}}

w h e n c h a n g i n g x = a^{\frac{1}{n}} u w e g e t

f (a) = a^{\frac{1}{n} - 2} \int_{0}^{\infty} \frac{d u}{{(u^{n} + 1)}^{2}}

n o w c h a n g e v = \frac{1}{u^{n} + 1} \Rightarrow u = {(\frac{1}{v} - 1)}^{\frac{1}{n}} \Rightarrow d u = \frac{1}{n} . (\frac{- 1}{v^{2}}) {(\frac{1}{v} - 1)}^{\frac{1}{n} - 1} d v

t h e n

f (a) = \frac{1}{n} . a^{\frac{1}{n} - 2} \int_{0}^{1} v^{2} . (\frac{1}{v^{2}}) {(\frac{1}{v} - 1)}^{\frac{1}{n} - 1} d v

= \frac{a^{\frac{1}{n} - 2}}{n} \int_{0}^{1} v^{1 - \frac{1}{n}} {(1 - v)}^{\frac{1}{n} - 1} d v

= \frac{a^{\frac{1}{n} - 2}}{n} B (2 - \frac{1}{n}, \frac{1}{n})

= \frac{a^{\frac{1}{n} - 2}}{n} . \frac{Γ (2 - \frac{1}{n}) Γ (\frac{1}{n})}{Γ (2)}

a s Γ (x + 1) = x Γ (x) w e h a v e Γ (2 - \frac{1}{n}) = (1 - \frac{1}{n}) Γ (1 - \frac{1}{n}) a n d Γ (2) = Γ (1) = 1

S o f (a) = \frac{a^{\frac{1}{n} - 2}}{n} . (1 - \frac{1}{n}) Γ (1 - \frac{1}{n}) Γ (\frac{1}{n})

a s Γ (1 - z) Γ (z) = \frac{π}{s i n (π z)} w e f i n a l l y g e t

f (a) = \frac{1}{n} (1 - \frac{1}{n}) a^{\frac{1}{n} - 2} . \frac{π}{s i n (\frac{π}{n})}

N o w w e c a n d e d u c e f (8), f (3), f (n)

Commented by ~ À ® @ 237 ~ last updated on 16/Aug/19

i f \forall p > 2 g_{p} (a) = \int_{0}^{\infty} \frac{d x}{{(x^{n} + a)}^{p}}

s t a r t a s t h e p r e v i o u s a n d r e a c h t o

g_{p} (a) = \frac{a^{\frac{1}{n} - p}}{n} \int_{0}^{1} v^{p} . (\frac{1}{v^{2}}) {(\frac{1}{v} - 1)}^{\frac{1}{n} - 1} d v

= \frac{a^{\frac{1}{n} - p}}{n} \int_{0}^{1} v^{p - 2 + 1 - \frac{1}{n}} {(1 - v)}^{\frac{1}{n} - 1} d v

= \frac{a^{\frac{1}{n} - p}}{n} B (p - \frac{1}{n}, \frac{1}{n})

= \frac{a^{\frac{1}{n} - p}}{n} . \frac{Γ (p - \frac{1}{n}) Γ (\frac{1}{n})}{Γ (p)}

a s Γ (p + z) = z (z + 1) \dots \dots . (z + p - 1) Γ (z) w e h a v e Γ (p - \frac{1}{n}) = Γ (p - 1 + (1 - \frac{1}{n})) = \underset{k = 0}{\prod^{p - 2}} [(1 - \frac{1}{n}) + k] Γ (1 - \frac{1}{n})

N o w

g_{p} (a) = \frac{a^{\frac{1}{n} - p}}{n . (p - 1)!} . (\underset{k = 0}{\prod^{p - 2}} [(1 - \frac{1}{n}) + k]) Γ (1 - \frac{1}{n}) Γ (\frac{1}{n})

= \frac{a^{\frac{1}{n} - p}}{n . (p - 1)!} . \frac{π}{s i n (\frac{π}{n})} . \underset{k = 0}{\prod^{p - 2}} [(1 - \frac{1}{n}) + k]

n o w w e c a n d e d u c e g_{3} (8), \cdot \dots \dots .

Commented by mathmax by abdo last updated on 16/Aug/19

t h a n k y o u s i r .

Commented by mathmax by abdo last updated on 16/Aug/19

l e t g (a) = \int_{0}^{\infty} \frac{d x}{{(a + x^{n})}^{2}} w i t h a > 0 w e h a v e p r o v e d t h a t

g (a) = \frac{π (1 - \frac{1}{n}) a^{\frac{1}{n} - 2}}{n s i n (\frac{π}{n})} a l s o w e h a v e

g^{^{'}} (a) = - \int_{0}^{\infty} \frac{2 (a + x^{n})}{{(a + x^{n})}^{4}} d x = - 2 \int_{0}^{\infty} \frac{d x}{{(a + x^{n})}^{3}} \Rightarrow

\int_{0}^{\infty} \frac{d x}{{(a + x^{n})}^{3}} = - \frac{1}{2} g^{^{'}} (a) w e h a v e

g^{^{'}} (a) = \frac{π}{n s i n (\frac{π}{n})} (1 - \frac{1}{n}) (\frac{1}{n} - 2) a^{\frac{1}{n} - 3} \Rightarrow

\int_{0}^{\infty} \frac{d x}{{(a + x^{n})}^{3}} = \frac{2 π}{n s i n (\frac{π}{n})} (1 - \frac{1}{n}) (2 - \frac{1}{n}) a^{\frac{1}{n} - 3}

a = 8 \Rightarrow \int_{0}^{\infty} \frac{d x}{{(8 + x^{n})}^{3}} = \frac{2 π}{n s i n (\frac{π}{n})} (1 - \frac{1}{n}) (2 - \frac{1}{n}) 8^{\frac{1}{n} - 3}

Commented by mathmax by abdo last updated on 16/Aug/19

e r r o r a t f i n a l l i n e \int_{0}^{\infty} \frac{d x}{{(a + x^{n})}^{3}} = \frac{π}{2 n s i n (\frac{π}{n})} (1 - \frac{1}{n}) (2 - \frac{1}{n}) a^{\frac{1}{n} - 3}

a = 8 \Rightarrow \int_{0}^{\infty} \frac{d x}{{(8 + x^{n})}^{3}} = \frac{π}{2 n s i n (\frac{π}{n})} (1 - \frac{1}{n}) (2 - \frac{1}{n}) 8^{\frac{1}{n} - 3}