0-1-x-10-1-dx-2pi-5-5-1-pi-5-

Question Number 96034 by M±th+et+s last updated on 29/May/20

\int_{0}^{\infty} \frac{1}{x^{10} + 1} d x = \frac{2 π}{5 (\sqrt{5} - 1)} = \frac{π ϕ}{5}

Commented by M±th+et+s last updated on 30/May/20

t h a n k s f o r s o l u t i o n s

Answered by abdomathmax last updated on 29/May/20

we do the changement x^{10} = t \Rightarrow x = t^{\frac{1}{10}}

\Rightarrow \int_{0}^{\infty} \frac{dx}{1 + x^{10}} = \frac{1}{10} \int_{0}^{\infty} \frac{t^{\frac{1}{10} - 1}}{1 + t} dt

= \frac{1}{10} \times \frac{π}{\sin (\frac{π}{10})}

\sin^{2} (\frac{π}{10}) = \frac{1 - \cos (\frac{π}{5})}{2} = \frac{1 - \frac{1 + \sqrt{5}}{4}}{2} = \frac{3 - \sqrt{5}}{8}

\Rightarrow \sin (\frac{π}{10}) = \frac{\sqrt{3 - \sqrt{5}}}{2 \sqrt{2}} \Rightarrow

\int_{0}^{\infty} \frac{dx}{1 + x^{10}} = \frac{π}{10} \times \frac{\sqrt{3 - \sqrt{5}}}{2 \sqrt{2}} = \frac{π}{20 \sqrt{2}} \times \sqrt{3 - \sqrt{5}}

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Answered by Sourav mridha last updated on 29/May/20

\int_{0}^{\infty} \frac{1}{1 + {(x^{5})}^{2}} dx substitute (x^{5}) by \tan Φ

and after little manipulation you get

= \frac{1}{5} \int_{0}^{\frac{π}{2}} {(s i n Φ)}^{- \frac{4}{5}} . {(\cos Φ)}^{\frac{4}{5}} d Φ

= \frac{1}{10} \frac{Γ (\frac{1}{10}) . Γ (\frac{9}{10})}{Γ (1)} = \frac{1}{10} Γ (\frac{1}{10}) Γ (1 - \frac{1}{10})

= \frac{1}{10} \frac{π}{\sin (\frac{π}{10})}

now putting the value of (\sin (18^{°}))

= \frac{\sqrt{5} - 1}{4} . . we get

\int_{0}^{\infty} \frac{1}{x^{10} + 1} d x = \frac{2 π}{5 (\sqrt{5} - 1)} = \frac{π \emptyset}{5}