.... nice calculus .... evaluate ::: Ω=^(???) ∫_(−∞) ^( ∞) (x^2 /((1+e^x )(1+e^(−x) )))dx


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Question Number 123060 by mnjuly1970 last updated on 22/Nov/20

$. . . . n i c e c a l c u l u s . . . .$ $e v a l u a t e :::$ $Ω \overset{? ? ?}{=} \int_{- \infty}^{\infty} \frac{x^{2}}{(1 + e^{x}) (1 + e^{- x})} d x$
	Answered by mindispower last updated on 22/Nov/20

	$= 2 \int_{0}^{\infty} \frac{e^{x} x^{2}}{{(1 + e^{x})}^{2}}$ $= 2 {[- \frac{x^{2}}{1 + e^{x}}]}_{0}^{\infty} + 4 \int_{0}^{\infty} \frac{x}{1 + e^{x}} d x$ $= 4 \int_{0}^{\infty} x Σ {(- 1 e^{- x})}^{k} e^{- x} d x$ $= 4 \sum_{k ⩾ 0} \int_{0}^{\infty} {(- 1)}^{k} e^{- (1 + k) x} x d x$ $= 4 \sum_{k ⩾ 0} \int_{0}^{\infty} \frac{{(- 1)}^{k}}{{(1 + k)}^{2}} {x e}^{- x} d x$ $= 4 Σ \frac{{(- 1)}^{k}}{{(1 + k)}^{2}} = 4 (\frac{ζ (2)}{2}) = \frac{π^{2}}{3}$
	Commented by mnjuly1970 last updated on 22/Nov/20

	$g r a t e f u l$ $m r m i n d s p o w e r . . .$
	Answered by Dwaipayan Shikari last updated on 22/Nov/20

	$2 \int_{0}^{\infty} \frac{x^{2} e^{x}}{{(1 + e^{x})}^{2}} d x$ $= 4 x \int_{0}^{\infty} \frac{e^{x}}{{(1 + e^{x})}^{2}} d x - \int_{0}^{\infty} 4 x \int \frac{e^{x}}{{(1 + e^{x})}^{2}}$ $= - {[\frac{2 x}{(1 + e^{x})}]}_{0}^{\infty} + 4 \int_{0}^{\infty} \frac{x}{(1 + e^{x})} d x$ $= 4 \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \int_{0}^{\infty} {x e}^{- n x} d x$ $= 4 \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \frac{1}{n^{2}} \int_{0}^{\infty} {u e}^{- u} d u$ $= 4 \sum^{\infty} \frac{{(- 1)}^{n + 1}}{n^{2}} Γ (2) = \frac{π^{2}}{12} . 4 = \frac{π^{2}}{3}$
	Commented by mnjuly1970 last updated on 22/Nov/20

	$t h a n k y o u m r$ $p a y a n . .$
	Answered by mathmax by abdo last updated on 22/Nov/20
	$Ω =∫_(−∞) ^(+∞) (x^2 /((1+e^x )(1+e^(−x) )))dx changement e^(x ) =t give Ω =∫_0 ^∞ ((ln^2 t)/((1+t)(1+t^(−1) )))(dt/t) =∫_0 ^∞ ((ln^2 t)/((1+t)^2 ))dt =∫_0 ^1 ((ln^2 t)/((1+t)^2 ))dt +∫_1 ^∞ ((ln^2 t)/((1+t)^2 ))dt(→t=(1/x)) =2 ∫_0 ^1 ((ln^2 x)/((1+x)^2 ))dx we hsve for ∣x∣<1 (1/(1+x))=Σ_(n=0) ^∞ (−1)^n x^n ⇒ −(1/((1+x)^2 ))=Σ_(n=1) ^∞ n(−1)^n x^(n−1) =Σ_(n=0) ^∞ (n+1)(−1)^(n+1) x^n ⇒ (1/((1+x)^2 ))=Σ_(n=0) ^∞ (n+1)(−1)^n x^n ⇒ Ω =2 ∫_0 ^1 ln^2 x(Σ_(n=0) ^∞ (n+1)(−1)^n x^n )dx =2Σ_(n=0) ^∞ (n+1)(−1)^n ∫_0 ^1 x^n ln^2 x dx we have by parts ∫_0 ^1 x^n ln^2 xdx =[(x^(n+1) /(n+1))ln^2 x]_0 ^1 −∫_0 ^(1 ) (x^(n+1) /(n+1))((2lnx)/x)dx =−(2/(n+1))∫_0 ^1 x^(n ) lnx dx =−(2/(n+1)){[(x^(n+1) /(n+1))lnx]_0 ^1 −∫_0 ^1 (x^(n+1) /(n+1))(dx/x)} =(2/((n+1)^3 )) ⇒ Ω =2 Σ_(n=0) ^∞ (n+1)(−1)^n (2/((n+1)^3 )) =4 Σ_(n=0) ^∞ (((−1)^n )/((n+1)^2 )) =4 Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =−4 Σ_(n=1) ^∞ (((−1)^n )/n^2 ) let δ(x)=Σ_(n=1) ^∞ (((−1)^n )/n^x ) we know δ(x) =(2^(1−x) −1)ξ(x) ⇒ δ(2)=Σ_(n=1) ^∞ (((−1)^n )/n^2 )=(2^(1−2) −1)ξ(2) =−(1/2).(π^2 /6)=−(π^2 /(12)) ⇒ Ω =−4×(−(π^2 /(12))) ⇒Ω =(π^2 /3)$
	$Ω = \int_{- \infty}^{+ \infty} \frac{x^{2}}{(1 + e^{x}) (1 + e^{- x})} dx changement e^{x} = t give$ $Ω = \int_{0}^{\infty} \frac{\ln^{2} t}{(1 + t) (1 + t^{- 1})} \frac{dt}{t} = \int_{0}^{\infty} \frac{\ln^{2} t}{{(1 + t)}^{2}} dt = \int_{0}^{1} \frac{\ln^{2} t}{{(1 + t)}^{2}} dt + \int_{1}^{\infty} \frac{\ln^{2} t}{{(1 + t)}^{2}} dt (\to t = \frac{1}{x})$ $= 2 \int_{0}^{1} \frac{\ln^{2} x}{{(1 + x)}^{2}} dx we hsve for ∣ x ∣< 1 \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} \Rightarrow$ $- \frac{1}{{(1 + x)}^{2}} = \sum_{n = 1}^{\infty} n {(- 1)}^{n} x^{n - 1} = \sum_{n = 0}^{\infty} (n + 1) {(- 1)}^{n + 1} x^{n} \Rightarrow$ $\frac{1}{{(1 + x)}^{2}} = \sum_{n = 0}^{\infty} (n + 1) {(- 1)}^{n} x^{n} \Rightarrow$ $Ω = 2 \int_{0}^{1} \ln^{2} x (\sum_{n = 0}^{\infty} (n + 1) {(- 1)}^{n} x^{n}) dx$ $= 2 \sum_{n = 0}^{\infty} (n + 1) {(- 1)}^{n} \int_{0}^{1} x^{n} \ln^{2} x dx we have by parts$ $\int_{0}^{1} x^{n} \ln^{2} xdx = {[\frac{x^{n + 1}}{n + 1} \ln^{2} x]}_{0}^{1} - \int_{0}^{1} \frac{x^{n + 1}}{n + 1} \frac{2 lnx}{x} dx$ $= - \frac{2}{n + 1} \int_{0}^{1} x^{n} lnx dx = - \frac{2}{n + 1} {{[\frac{x^{n + 1}}{n + 1} lnx]}_{0}^{1} - \int_{0}^{1} \frac{x^{n + 1}}{n + 1} \frac{dx}{x}}$ $= \frac{2}{{(n + 1)}^{3}} \Rightarrow Ω = 2 \sum_{n = 0}^{\infty} (n + 1) {(- 1)}^{n} \frac{2}{{(n + 1)}^{3}}$ $= 4 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}} = 4 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} = - 4 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}}$ $let δ (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{x}} we know δ (x) = (2^{1 - x} - 1) ξ (x) \Rightarrow$ $δ (2) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = (2^{1 - 2} - 1) ξ (2) = - \frac{1}{2} . \frac{π^{2}}{6} = - \frac{π^{2}}{12} \Rightarrow$ $Ω = - 4 \times (- \frac{π^{2}}{12}) \Rightarrow Ω = \frac{π^{2}}{3}$
	Commented by mnjuly1970 last updated on 22/Nov/20

	$t h a n k y o u s o m u c h s i r m a x .$
	Commented by mathmax by abdo last updated on 23/Nov/20

	$you are welcome sir$
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