decompose F(x)=((nx^n )/(x^(2n) +1)) inside C(x) and R(x) (n≥2) and determine ∫


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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
$another way for ∫_0 ^∞ F(x)dx we do the changement x^(2n) =t ⇒ x=t^(1/(2n)) ⇒ ∫_0 ^∞ n(x^n /(x^(2n) +1))dx =n∫_0 ^∞ (t^(1/2) /(t+1))×(1/(2n))t^((1/(2n))−1) dt =(1/2)∫_0 ^∞ (t^((1/(2n))+(1/2)−1) /(t+1))dt =(1/2) ×(π/(sin(π((1/(2n))+(1/2))))) =(π/(2sin((π/(2n))+(π/2)))) =(π/(2cos((π/(2n))))) { ∫_0 ^∞ (t^(a−1) /(1+t))dt =(π/(sin(πa)))=Γ(a)Γ(1−a)) with 0<a<1)}$
$a n o t h e r w a y f o r \int_{0}^{\infty} F (x) d x w e d o t h e c h a n g e m e n t x^{2 n} = t \Rightarrow$ $x = t^{\frac{1}{2 n}} \Rightarrow \int_{0}^{\infty} n \frac{x^{n}}{x^{2 n} + 1} d x = n \int_{0}^{\infty} \frac{t^{\frac{1}{2}}}{t + 1} \times \frac{1}{2 n} t^{\frac{1}{2 n} - 1} d t$ $= \frac{1}{2} \int_{0}^{\infty} \frac{t^{\frac{1}{2 n} + \frac{1}{2} - 1}}{t + 1} d t = \frac{1}{2} \times \frac{π}{s i n (π (\frac{1}{2 n} + \frac{1}{2}))} = \frac{π}{2 s i n (\frac{π}{2 n} + \frac{π}{2})}$ $= \frac{π}{2 c o s (\frac{π}{2 n})} {\int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{s i n (π a)} = Γ (a) Γ (1 - a)) w i t h 0 < a < 1)}$
	Answered by mind is power last updated on 22/Jan/20
	$over C(x) x^(2n) +1=0⇒x^(2n) =−1=e^(iπ+2kπ) x_k =e^(((i(1+2k)π)/(2n)) ) ,k∈[0,2n−1] x_(2n−k−1) =e^(i(((1+2(2n−k−1))/(2n)))π) =e^(−i(((1+2k)/(2n)))π) ⇒x_k =x_(2n−1−k) ^− ,E ((nx^n )/(x^(2n) +1))=Σ(a_k /(x−x_k )) a_k =((nx_k )/(2nx_k ^(2n−1) ))=(1/(2x_k ^(2n−2) ))=−(x_k ^2 /2) ((nx^n )/(x^(2n) +1))=−(1/2)Σ_(k=0) ^(2n−1) (x_k ^2 /((x−x_k ))) over R(X) −(1/2)Σ_(k=0) ^(2n−1) (x_k ^2 /((x−x_k )))=−(1/2).{Σ_(k=0) ^(n−1) (x_k ^2 /(x−x_k ))+Σ_(k=n) ^(2n−1) (x_k ^2 /(x−x_k ))} =−(1/2){Σ_(k=0) ^(n−1) [(x_k ^2 /(x−x_k ))+(x_(2n−1−k) ^2 /(x−x_(2n−1−k) ))]}ByE⇒ =−(1/2){Σ_(k=0) ^(n−1) [(x_k ^2 /(x−x_k ))+(x_k ^(−2) /(x−x_k ^− ))]} =−(1/2)Σ_(k=0) ^(n−1) [((x(x_k ^2 +x_k ^(−2) )−x_k ^− x_k ^2 −x_k x_k ^(−2) )/(x^2 −2Re(x_k )x+∣x_k ∣^2 )) =−(1/2)Σ_(k=0) ^(n−1) ((2x(Re(x_k ^2 ))−2Re(x_k ))/(x^2 −2Re(x_k )+1)) =−(1/2)Σ_(k=0) ^(n−1) ((2x(cos(((1+2k)/n)π))−2cos(((1+2k)/(2n))π))/(x^2 −2cos(((1+2k)/(2n))π)+1)) over IR(x) ∫_0 ^(+∞) n(x^n /(x^(2n) +1))dx x=tg^(1/n) (t)⇒dx=(1/n)(1+tg^2 (t))tg^((1/n)−1) (t)dt =∫_0 ^(π/2) tg(t)(((1+tg^2 (t))tg^((1/n)−1) )/(1+tg^2 (t)))dt =∫_0 ^(π/2) tg^(1/n) (t)dt=∫_0 ^(π/2) sin^(1/n) (t)cos^(−(1/n)) (t)dt =(1/2)β(((1+n)/(2n)),((n−1)/(2n)))=(1/2).((Γ(((1+n)/(2n)))Γ(((n−1)/(2n))))/(Γ(((n+1)/(2n))+((n−1)/(2n)))))=((Γ(((n+1)/(2n))).Γ(1−((n+1)/(2n))))/(2Γ(1))) Γ(x)Γ(1−x)=(π/(sin(πx)))⇒Γ(((n+1)/(2n)))Γ(1−((n+1)/(2n)))=(π/(sin(((n+1)/(2n))π))),Γ(1)=1 =(π/(2sin(((n+1)/(2n))π)))=(π/(2cos((π/(2n)))))$
	$o v e r C (x)$ $x^{2 n} + 1 = 0 \Rightarrow x^{2 n} = - 1 = e^{i π + 2 k π}$ $x_{k} = e^{\frac{i (1 + 2 k) π}{2 n}}, k \in [0, 2 n - 1]$ $x_{2 n - k - 1} = e^{i (\frac{1 + 2 (2 n - k - 1)}{2 n}) π} = e^{- i (\frac{1 + 2 k}{2 n}) π}$ $\Rightarrow x_{k} = {\bar{x}}_{2 n - 1 - k}, E$ $\frac{{n x}^{n}}{x^{2 n} + 1} = Σ \frac{a_{k}}{x - x_{k}}$ $a_{k} = \frac{{n x}_{k}}{2 {n x}_{k}^{2 n - 1}} = \frac{1}{2 x_{k}^{2 n - 2}} = - \frac{x_{k}^{2}}{2}$ $\frac{{n x}^{n}}{x^{2 n} + 1} = - \frac{1}{2} \underset{k = 0}{\sum^{2 n - 1}} \frac{x_{k}^{2}}{(x - x_{k})}$ $o v e r R (X)$ $- \frac{1}{2} \underset{k = 0}{\sum^{2 n - 1}} \frac{x_{k}^{2}}{(x - x_{k})} = - \frac{1}{2} . {\underset{k = 0}{\sum^{n - 1}} \frac{x_{k}^{2}}{x - x_{k}} + \underset{k = n}{\sum^{2 n - 1}} \frac{x_{k}^{2}}{x - x_{k}}}$ $= - \frac{1}{2} {\underset{k = 0}{\sum^{n - 1}} [\frac{x_{k}^{2}}{x - x_{k}} + \frac{x_{2 n - 1 - k}^{2}}{x - x_{2 n - 1 - k}}]} B y E \Rightarrow$ $= - \frac{1}{2} {\underset{k = 0}{\sum^{n - 1}} [\frac{x_{k}^{2}}{x - x_{k}} + \frac{{\overset{- 2}{x}}_{k}}{x - {\bar{x}}_{k}}]}$ $= - \frac{1}{2} \underset{k = 0}{\sum^{n - 1}} [\frac{x (x_{k}^{2} + {\overset{- 2}{x}}_{k}) - {\bar{x}}_{k} x_{k}^{2} - x_{k} {\overset{- 2}{x}}_{k}}{x^{2} - 2 R e (x_{k}) x + ∣ x_{k} ∣^{2}}$ $= - \frac{1}{2} \underset{k = 0}{\sum^{n - 1}} \frac{2 x (R e (x_{k}^{2})) - 2 R e (x_{k})}{x^{2} - 2 R e (x_{k}) + 1}$ $= - \frac{1}{2} \underset{k = 0}{\sum^{n - 1}} \frac{2 x (c o s (\frac{1 + 2 k}{n} π)) - 2 c o s (\frac{1 + 2 k}{2 n} π)}{x^{2} - 2 c o s (\frac{1 + 2 k}{2 n} π) + 1} o v e r I R (x)$ $\int_{0}^{+ \infty} n \frac{x^{n}}{x^{2 n} + 1} d x$ $x = {t g}^{\frac{1}{n}} (t) \Rightarrow d x = \frac{1}{n} (1 + {t g}^{2} (t)) {t g}^{\frac{1}{n} - 1} (t) d t$ $= \int_{0}^{\frac{π}{2}} t g (t) \frac{(1 + {t g}^{2} (t)) {t g}^{\frac{1}{n} - 1}}{1 + {t g}^{2} (t)} d t$ $= \int_{0}^{\frac{π}{2}} {t g}^{\frac{1}{n}} (t) d t = \int_{0}^{\frac{π}{2}} {s i n}^{\frac{1}{n}} (t) {c o s}^{- \frac{1}{n}} (t) d t$ $= \frac{1}{2} β (\frac{1 + n}{2 n}, \frac{n - 1}{2 n}) = \frac{1}{2} . \frac{Γ (\frac{1 + n}{2 n}) Γ (\frac{n - 1}{2 n})}{Γ (\frac{n + 1}{2 n} + \frac{n - 1}{2 n})} = \frac{Γ (\frac{n + 1}{2 n}) . Γ (1 - \frac{n + 1}{2 n})}{2 Γ (1)}$ $Γ (x) Γ (1 - x) = \frac{π}{s i n (π x)} \Rightarrow Γ (\frac{n + 1}{2 n}) Γ (1 - \frac{n + 1}{2 n}) = \frac{π}{s i n (\frac{n + 1}{2 n} π)}, Γ (1) = 1$ $= \frac{π}{2 s i n (\frac{n + 1}{2 n} π)} = \frac{π}{2 c o s (\frac{π}{2 n})}$
	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

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