show-that-0-x-1-x-6-dx-pi-3-3-

Question Number 54585 by peter frank last updated on 07/Feb/19

s h o w t h a t \int_{0}^{\infty} \frac{x}{1 + x^{6}} d x = \frac{π}{3 \sqrt{3}}

Commented by maxmathsup by imad last updated on 07/Feb/19

l e t I = \int_{0}^{\infty} \frac{x}{1 + x^{6}} d x c h a n g e m e n t x^{6} = t g i v e x = t^{\frac{1}{6}} \Rightarrow

I = \int_{0}^{\infty} \frac{t^{\frac{1}{6}}}{1 + t} \frac{1}{6} t^{\frac{1}{6} - 1} d t = \frac{1}{6} \int_{0}^{\infty} \frac{t^{\frac{1}{3} - 1}}{1 + t} d t b u t w e h a v e p r o v e d t h a t

\int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{s i n (π a)} w i t h 0 < a < 1 \Rightarrow I = \frac{1}{6} \frac{π}{s i n (\frac{π}{3})} = \frac{π}{6 \frac{\sqrt{3}}{2}} = \frac{π}{3 \sqrt{3}} .

Answered by arvinddayama00@gmail.com last updated on 08/Feb/19

x^{2} = t x = 0, t = 0

2 x d x = d t x = \infty, t = \infty

\frac{1}{2} \int_{0}^{\infty} \frac{d t}{1 + t^{3}}

∵ \int_{0}^{\infty} \frac{d x}{1 + x^{α}} = \frac{π}{α s i n (\frac{π}{α})} ★ ★

s o \frac{1}{2} \int_{0}^{\infty} \frac{d t}{1 + t^{3}} = \frac{1}{2} . \frac{π}{3 . \frac{\sqrt{3}}{2}} = \frac{π}{3 \sqrt{3}}

p l s . . c h e a k \dots . .

Commented by peter frank last updated on 07/Feb/19

r i g h t s i r

Commented by rahul 19 last updated on 08/Feb/19

S i r, h o w \int_{0}^{\infty} \frac{d x}{1 + x^{α}} = \frac{π}{α \sin (\frac{π}{α})} ?

Commented by tanmay.chaudhury50@gmail.com last updated on 08/Feb/19

x^{α} = {t a n}^{2} θ x = {(t a n θ)}^{\frac{2}{α}}

d x = \frac{2}{α} {(t a n θ)}^{\frac{2}{α} - 1} ({s e c}^{2} θ) d θ

\int_{0}^{\frac{π}{2}} \frac{2}{α} {(t a n θ)}^{\frac{2}{α} - 1} d θ

\frac{2}{α} \int_{0}^{\frac{π}{2}} {(s i n θ)}^{\frac{2}{α} - 1} {(c o s θ)}^{1 - \frac{2}{α}} d θ

n o w g a m m a b e t a f u n c t i o n f o r m u l a

2 \int_{0}^{\frac{π}{2}} {(s i n θ)}^{2 p - 1} {(c o s θ)}^{2 q - 1} d θ = \frac{⌈ (p) ⌈ q)}{⌈ (p + q)}

n o w h e r e 2 p - 1 = \frac{2}{α} - 1 s o p = \frac{1}{α}

2 q - 1 = 1 - \frac{2}{α} s o q = (1 - \frac{1}{α})

n o w a n s w e r i s = \frac{1}{α} \times \frac{⌈ (\frac{1}{α}) ⌈ (1 - \frac{1}{α})}{⌈ (1)} = \frac{1}{α} \times \frac{π}{s i n (\frac{π}{α})}

[a g a i n f o r m u l a ⌈ (a) ⌈ (1 - a) = \frac{π}{s i n (a π)}]