show-that-0-x-pi-5-1-1-x-2pi-dx-

Question Number 80924 by M±th+et£s last updated on 08/Feb/20

s h o w t h a t

\int_{0}^{\infty} \frac{x^{\frac{π}{5} - 1}}{1 + x^{2 π}} d x = ϕ

Commented by abdomathmax last updated on 09/Feb/20

c h a n g e m e n t x^{2 π} = t g i v e x = t^{\frac{1}{2 π}} \Rightarrow

\int_{0}^{\infty} \frac{x^{\frac{π}{5} - 1}}{1 + x^{2 π}} d x = \int_{0}^{\infty} \frac{{(t^{\frac{1}{2 π}})}^{\frac{π}{5} - 1}}{1 + t} \times \frac{1}{2 π} t^{\frac{1}{2 π} - 1} d t

= \frac{1}{2 π} \int_{0}^{\infty} \frac{t^{\frac{1}{10} - \frac{1}{2 π} + \frac{1}{2 π} - 1}}{1 + t} d t = \frac{1}{2 π} \int_{0}^{\infty} \frac{t^{\frac{1}{10} - 1}}{1 + t} d t

= \frac{1}{2 π} \times \frac{π}{s i n (\frac{π}{10})} = \frac{1}{2 s i n (\frac{π}{10})}

w e k n o w c o s (\frac{π}{5}) = \frac{1 + \sqrt{5}}{4}

{s i n}^{2} (\frac{π}{10}) = \frac{1 - c o s (\frac{π}{5})}{2} = \frac{1 - \frac{1 + \sqrt{5}}{4}}{2}

= \frac{3 - \sqrt{5}}{8} \Rightarrow s i n (\frac{π}{10}) = \sqrt{\frac{3 - \sqrt{5}}{8}}

{(\frac{\sqrt{5} - 1}{4})}^{2} = \frac{6 - 2 \sqrt{5}}{16} = \frac{3 - \sqrt{5}}{8} \Rightarrow s i n (\frac{π}{10}) = \frac{\sqrt{5} - 1}{4} \Rightarrow

I = \frac{1}{\frac{\sqrt{5} - 1}{2}} = φ

Commented by M±th+et£s last updated on 09/Feb/20

g o d b l e s s y o u s i r

Commented by mathmax by abdo last updated on 10/Feb/20

y o u a r e w e l v o m e s i r .

Answered by mind is power last updated on 09/Feb/20

A = \int_{0}^{+ \infty} \frac{x^{\frac{π}{5} - 1}}{1 + x^{2 π}} d x

x = {t g}^{\frac{1}{π}} (θ) \Rightarrow

A = \int_{0}^{\frac{π}{2}} \frac{{t g}^{\frac{1}{5} - \frac{1}{π}}}{1 + {t g}^{2} (θ)} . \frac{(1 + {t g}^{2} (θ))}{π} {t g}^{\frac{1}{π} - 1} (θ) d θ

A = \frac{1}{π} \int_{0}^{\frac{π}{2}} {t g}^{- \frac{4}{5}} (θ) d θ

= \frac{1}{π} \int_{0}^{\frac{π}{2}} {s i n}^{- \frac{4}{5}} (θ) {c o s}^{\frac{4}{5}} (θ) d θ

= \frac{1}{2 π} β (\frac{1}{10}, \frac{9}{10}) = \frac{Γ (\frac{1}{10}) Γ (\frac{9}{10})}{2 π Γ (1)} = \frac{Γ (\frac{1}{10}) Γ (1 - \frac{1}{10})}{2 π} = \frac{π}{2 π s i n (\frac{π}{10})}

A = \frac{1}{2 s i n (\frac{π}{10})}

s i n (\frac{π}{10}) = \frac{\sqrt{5} - 1}{4}

A = \frac{2}{\sqrt{5} - 1} = \frac{\sqrt{5} + 1}{2} = \emptyset = \int_{0}^{\infty} \frac{x^{\frac{π}{5} - 1}}{1 + x^{2 π}} d x

Commented by M±th+et£s last updated on 09/Feb/20

g r e a t s i r t h a n k y o u

Commented by mind is power last updated on 09/Feb/20

W i t h e p l e a s u r h a v e y o u s o m m Q u a t i o n

a b o u t h y p e r g e o m e t r i c F u n c t i o n ?

Commented by M±th+et£s last updated on 09/Feb/20

y e s i h a v e t h i s i s a n e x a m p l e

\int \sqrt{x^{3} + 1} d x = - x_{2} F_{1} (\frac{- 1}{2}, \frac{1}{3}; \frac{4}{3}, - x^{3}) + c

Commented by M±th+et£s last updated on 09/Feb/20

i n e e d s o m e t i m e t o p o s t a l l o f m y e q u a t i o n s