Please-integrate-0-1-1-1-x-c-dx-where-c-is-a-constant-

Question Number 118674 by mathace last updated on 19/Oct/20

P l e a s e i n t e g r a t e

\int_{0}^{1} \frac{1}{1 + x^{c}} d x w h e r e c i s a c o n s t a n t .

Answered by Dwaipayan Shikari last updated on 19/Oct/20

\int_{0}^{1} \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . x^{n c} d x

\underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{1} x^{n c} d x [\frac{1}{1 + x^{c}} = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . x^{n c}]

\underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} {[\frac{x^{n c + 1}}{n c + 1}]}_{0}^{1}

\underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . \frac{1}{n c + 1}

Commented by Dwaipayan Shikari last updated on 19/Oct/20

W h e n

c = 2

\underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{1}{2 n + 1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots = \frac{π}{4}

\int_{0}^{1} \frac{1}{1 + x^{2}} d x = {[{t a n}^{- 1} x]}_{0}^{1} = \frac{π}{4}

Commented by mathace last updated on 19/Oct/20

C a n y o u e x p l a i n d e t a i l s h o w i t c o m e s

\frac{1}{1 + x^{c}} = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . x^{n c} s i r ?

Commented by Dwaipayan Shikari last updated on 19/Oct/20

1 + x + x^{2} + x^{3} + x^{4} + \dots . . = \frac{1}{1 - x} = \frac{T_{1}}{1 - R} (T_{1} = f i r s t t e r m R = c o m m o n R a t i o)

\underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} x^{n} = 1 - x + x^{2} - x^{3} + x^{4} - x^{5} + \dots = \frac{1}{1 + x} (H e r e c o m m o n R a t i o (- x))

1 - x^{c} + x^{2 c} - x^{3 c} + \dots = \frac{1}{1 + x^{c}} (C o m m o n R a t i o (- x^{c})

\underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} x^{n c} = \frac{1}{1 + x^{c}}

Commented by mathace last updated on 19/Oct/20

W o w! g r e a t . T h a n k y o u . V a l i d f o r ∣ x ∣< 1

Commented by Dwaipayan Shikari last updated on 19/Oct/20

y e s!

Commented by Ar Brandon last updated on 19/Oct/20

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