decompose-F-x-nx-n-x-2n-1-inside-C-x-and-R-x-n-2-and-determine-0-F-x-dx-

Question Number 79108 by mathmax by abdo last updated on 22/Jan/20

d e c o m p o s e F (x) = \frac{{n x}^{n}}{x^{2 n} + 1} i n s i d e C (x) a n d R (x) (n ⩾ 2)

a n d d e t e r m i n e \int_{0}^{+ \infty} F (x) d x

Commented by mathmax by abdo last updated on 31/Jan/20

a n o t h e r w a y f o r \int_{0}^{\infty} F (x) d x w e d o t h e c h a n g e m e n t x^{2 n} = t \Rightarrow

x = t^{\frac{1}{2 n}} \Rightarrow \int_{0}^{\infty} n \frac{x^{n}}{x^{2 n} + 1} d x = n \int_{0}^{\infty} \frac{t^{\frac{1}{2}}}{t + 1} \times \frac{1}{2 n} t^{\frac{1}{2 n} - 1} d t

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

= \frac{π}{2 c o s (\frac{π}{2 n})} {\int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{s i n (π a)} = Γ (a) Γ (1 - a)) w i t h 0 < a < 1)}

Answered by mind is power last updated on 22/Jan/20

o v e r C (x)

x^{2 n} + 1 = 0 \Rightarrow x^{2 n} = - 1 = e^{i π + 2 k π}

x_{k} = e^{\frac{i (1 + 2 k) π}{2 n}}, k \in [0, 2 n - 1]

x_{2 n - k - 1} = e^{i (\frac{1 + 2 (2 n - k - 1)}{2 n}) π} = e^{- i (\frac{1 + 2 k}{2 n}) π}

\Rightarrow x_{k} = {\bar{x}}_{2 n - 1 - k}, E

\frac{{n x}^{n}}{x^{2 n} + 1} = Σ \frac{a_{k}}{x - x_{k}}

a_{k} = \frac{{n x}_{k}}{2 {n x}_{k}^{2 n - 1}} = \frac{1}{2 x_{k}^{2 n - 2}} = - \frac{x_{k}^{2}}{2}

\frac{{n x}^{n}}{x^{2 n} + 1} = - \frac{1}{2} \underset{k = 0}{\sum^{2 n - 1}} \frac{x_{k}^{2}}{(x - x_{k})}

o v e r R (X)

- \frac{1}{2} \underset{k = 0}{\sum^{2 n - 1}} \frac{x_{k}^{2}}{(x - x_{k})} = - \frac{1}{2} . {\underset{k = 0}{\sum^{n - 1}} \frac{x_{k}^{2}}{x - x_{k}} + \underset{k = n}{\sum^{2 n - 1}} \frac{x_{k}^{2}}{x - x_{k}}}

= - \frac{1}{2} {\underset{k = 0}{\sum^{n - 1}} [\frac{x_{k}^{2}}{x - x_{k}} + \frac{x_{2 n - 1 - k}^{2}}{x - x_{2 n - 1 - k}}]} B y E \Rightarrow

= - \frac{1}{2} {\underset{k = 0}{\sum^{n - 1}} [\frac{x_{k}^{2}}{x - x_{k}} + \frac{{\overset{- 2}{x}}_{k}}{x - {\bar{x}}_{k}}]}

= - \frac{1}{2} \underset{k = 0}{\sum^{n - 1}} [\frac{x (x_{k}^{2} + {\overset{- 2}{x}}_{k}) - {\bar{x}}_{k} x_{k}^{2} - x_{k} {\overset{- 2}{x}}_{k}}{x^{2} - 2 R e (x_{k}) x + ∣ x_{k} ∣^{2}}

= - \frac{1}{2} \underset{k = 0}{\sum^{n - 1}} \frac{2 x (R e (x_{k}^{2})) - 2 R e (x_{k})}{x^{2} - 2 R e (x_{k}) + 1}

= - \frac{1}{2} \underset{k = 0}{\sum^{n - 1}} \frac{2 x (c o s (\frac{1 + 2 k}{n} π)) - 2 c o s (\frac{1 + 2 k}{2 n} π)}{x^{2} - 2 c o s (\frac{1 + 2 k}{2 n} π) + 1} o v e r I R (x)

\int_{0}^{+ \infty} n \frac{x^{n}}{x^{2 n} + 1} d x

x = {t g}^{\frac{1}{n}} (t) \Rightarrow d x = \frac{1}{n} (1 + {t g}^{2} (t)) {t g}^{\frac{1}{n} - 1} (t) d t

= \int_{0}^{\frac{π}{2}} t g (t) \frac{(1 + {t g}^{2} (t)) {t g}^{\frac{1}{n} - 1}}{1 + {t g}^{2} (t)} d t

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

= \frac{1}{2} β (\frac{1 + n}{2 n}, \frac{n - 1}{2 n}) = \frac{1}{2} . \frac{Γ (\frac{1 + n}{2 n}) Γ (\frac{n - 1}{2 n})}{Γ (\frac{n + 1}{2 n} + \frac{n - 1}{2 n})} = \frac{Γ (\frac{n + 1}{2 n}) . Γ (1 - \frac{n + 1}{2 n})}{2 Γ (1)}

Γ (x) Γ (1 - x) = \frac{π}{s i n (π x)} \Rightarrow Γ (\frac{n + 1}{2 n}) Γ (1 - \frac{n + 1}{2 n}) = \frac{π}{s i n (\frac{n + 1}{2 n} π)}, Γ (1) = 1

= \frac{π}{2 s i n (\frac{n + 1}{2 n} π)} = \frac{π}{2 c o s (\frac{π}{2 n})}

Commented by mathmax by abdo last updated on 23/Jan/20

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

Commented by mind is power last updated on 23/Jan/20

y^{'} r e W e l c o m