Evaluate I=∫_0 ^( ∞) (1/(x^5 +x^4 +x^3 +x^2 +x+1))dx


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Question Number 193377 by Humble last updated on 12/Jun/23

$E v a l u a t e$ $I = \int_{0}^{\infty} \frac{1}{x^{5} + x^{4} + x^{3} + x^{2} + x + 1} d x$
Commented by Frix last updated on 12/Jun/23

$x^{5} + x^{4} + x^{3} + x^{2} + x + 1 =$ $= (x + 1) (x^{2} - x + 1) (x^{2} + x + 1)$ $Just decompose and solve it!$
	Answered by Mathspace last updated on 16/Jun/23
	$I=∫_0 ^∞ (dx/(1+x+x^2 +x^3 +x^4 +x^5 )) I=∫_0 ^∞ (dx/((1−x^6 )/(1−x)))=∫_0 ^∞ ((1−x)/(1−x^6 ))dx =∫_0 ^1 ((1−x)/(1−x^6 ))dx +∫_1 ^∞ ((1−x)/(1−x^6 ))dx(→x=(1/t)) =∫_0 ^1 ((1−x)/(1−x^6 ))dx−∫_0 ^1 ((1−(1/t))/(1−(1/t^6 )))×((−dt)/t^2 ) =∫_0 ^1 ((1−x)/(1−x^6 ))dx+∫_0 ^1 (1/t^3 ).((t−1)/(t^6 −1)).t^6 dt =∫_0 ^1 ((1−x)/(1−x^6 ))dx+∫_0 ^1 ((t^4 −t^3 )/(t^6 −1))dt =∫_0 ^1 ((1−x)/(1−x^6 ))dx+∫_0 ^1 ((x^3 −x^4 )/(1−x^6 ))dx =∫_0 ^1 ((1−x+x^3 −x^4 )/(1−x^6 ))dx =∫_0 ^1 (1−x+x^3 −x^4 )Σ_(n=0) ^∞ x^(6n) dx =Σ_(n=0) ^∞ ∫_0 ^1 (x^(6n) −x^(6n+1) +x^(6n+3) −x^(6n+4) )dx =Σ_(n=0) ^∞ ((1/(6n+1))−(1/(6n+2))+(1/(6n+4))−(1/(6n+5))) =Σ_(n=0) ^∞ (1/((6n+1)(6n+2))) +Σ_(n=0) ^∞ (1/((6n+4)(6n+5))) we have Σ_(n=0) ^∞ (1/((6n+1)(6n+2))) =(1/(36))Σ_(n=0) ^∞ (1/((n+(1/6))(n+(1/3)))) =(1/(36)).((Ψ((1/3))−Ψ((1/6)))/((1/3)−(1/6))) =(1/6){Ψ((1/3))−Ψ((1/6))} Σ_(n=0) ^∞ (1/((6n+4)(6n+5))) =(1/(36))Σ_(n=0) ^∞ (1/((n+(2/3))(n+(5/6)))) =(1/(36)).((Ψ((5/6))−Ψ((2/3)))/((5/6)−(2/3))) =(1/6)(Ψ((5/6))−Ψ((2/3))) ⇒I=(1/6){Ψ((1/3))−Ψ((1/6))+Ψ((5/6))−Ψ((2/3))} after use special values of Ψ....$
	$I = \int_{0}^{\infty} \frac{d x}{1 + x + x^{2} + x^{3} + x^{4} + x^{5}}$ $I = \int_{0}^{\infty} \frac{d x}{\frac{1 - x^{6}}{1 - x}} = \int_{0}^{\infty} \frac{1 - x}{1 - x^{6}} d x$ $= \int_{0}^{1} \frac{1 - x}{1 - x^{6}} d x + \int_{1}^{\infty} \frac{1 - x}{1 - x^{6}} d x (\to x = \frac{1}{t})$ $= \int_{0}^{1} \frac{1 - x}{1 - x^{6}} d x - \int_{0}^{1} \frac{1 - \frac{1}{t}}{1 - \frac{1}{t^{6}}} \times \frac{- d t}{t^{2}}$ $= \int_{0}^{1} \frac{1 - x}{1 - x^{6}} d x + \int_{0}^{1} \frac{1}{t^{3}} . \frac{t - 1}{t^{6} - 1} . t^{6} d t$ $= \int_{0}^{1} \frac{1 - x}{1 - x^{6}} d x + \int_{0}^{1} \frac{t^{4} - t^{3}}{t^{6} - 1} d t$ $= \int_{0}^{1} \frac{1 - x}{1 - x^{6}} d x + \int_{0}^{1} \frac{x^{3} - x^{4}}{1 - x^{6}} d x$ $= \int_{0}^{1} \frac{1 - x + x^{3} - x^{4}}{1 - x^{6}} d x$ $= \int_{0}^{1} (1 - x + x^{3} - x^{4}) \sum_{n = 0}^{\infty} x^{6 n} d x$ $= \sum_{n = 0}^{\infty} \int_{0}^{1} (x^{6 n} - x^{6 n + 1} + x^{6 n + 3} - x^{6 n + 4}) d x$ $= \sum_{n = 0}^{\infty} (\frac{1}{6 n + 1} - \frac{1}{6 n + 2} + \frac{1}{6 n + 4} - \frac{1}{6 n + 5})$ $= \sum_{n = 0}^{\infty} \frac{1}{(6 n + 1) (6 n + 2)}$ $+ \sum_{n = 0}^{\infty} \frac{1}{(6 n + 4) (6 n + 5)} w e h a v e$ $\sum_{n = 0}^{\infty} \frac{1}{(6 n + 1) (6 n + 2)}$ $= \frac{1}{36} \sum_{n = 0}^{\infty} \frac{1}{(n + \frac{1}{6}) (n + \frac{1}{3})}$ $= \frac{1}{36} . \frac{Ψ (\frac{1}{3}) - Ψ (\frac{1}{6})}{\frac{1}{3} - \frac{1}{6}}$ $= \frac{1}{6} {Ψ (\frac{1}{3}) - Ψ (\frac{1}{6})}$ $\sum_{n = 0}^{\infty} \frac{1}{(6 n + 4) (6 n + 5)}$ $= \frac{1}{36} \sum_{n = 0}^{\infty} \frac{1}{(n + \frac{2}{3}) (n + \frac{5}{6})}$ $= \frac{1}{36} . \frac{Ψ (\frac{5}{6}) - Ψ (\frac{2}{3})}{\frac{5}{6} - \frac{2}{3}}$ $= \frac{1}{6} (Ψ (\frac{5}{6}) - Ψ (\frac{2}{3}))$ $\Rightarrow I = \frac{1}{6} {Ψ (\frac{1}{3}) - Ψ (\frac{1}{6}) + Ψ (\frac{5}{6}) - Ψ (\frac{2}{3})}$ $a f t e r u s e s p e c i a l v a l u e s o f Ψ . . . .$
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