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Question Number 108697 by 150505R last updated on 18/Aug/20

	Answered by mathmax by abdo last updated on 18/Aug/20
	$I =∫_0 ^∞ (((lnx)^2 )/(1+x^2 )) dx ⇒ I =∫_0 ^1 (((lnx)^2 )/(1+x^2 ))dx +∫_1 ^(+∞) (((lnx)^2 )/(1+x^2 ))dx(→x=(1/t)) =∫_0 ^1 (((lnx)^2 )/(1+x^2 ))dx −∫_0 ^1 (((lnt)^2 )/(1+(1/t^2 )))(−(dt/t^2 )) =2∫_0 ^1 (((lnx)^2 )/(1+x^2 ))dx =2 ∫_0 ^1 (lnx)^2 Σ_(n=0) ^∞ (−1)^n x^(2n) dx =2Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(2n) (lnx)^2 dx =2Σ_(n=0) ^∞ (−1)^n A_n A_n =∫_0 ^1 x^(2n) (lnx)^2 dx =_(by psrts) [(x^(2n+1) /(2n+1))(lnx)^2 ]_0 ^1 −∫_0 ^1 (x^(2n+1) /(2n+1))((2ln(x))/x)dx =−(2/(2n+1)) ∫_0 ^1 x^(2n) lnxdx =−(2/(2n+1)){ [ (x^(2n+1) /(2n+1))lnx]_0 ^1 −∫_0 ^1 (x^(2n) /(2n+1))dx} =−(2/(2n+1)){−(1/((2n+1)^2 ))} =(2/((2n+1)^3 )) ⇒ I =4 Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^3 )) rest to find the value of this serie by fourier ...be continued....$
	$I = \int_{0}^{\infty} \frac{{(lnx)}^{2}}{1 + x^{2}} dx \Rightarrow I = \int_{0}^{1} \frac{{(lnx)}^{2}}{1 + x^{2}} dx + \int_{1}^{+ \infty} \frac{{(lnx)}^{2}}{1 + x^{2}} dx (\to x = \frac{1}{t})$ $= \int_{0}^{1} \frac{{(lnx)}^{2}}{1 + x^{2}} dx - \int_{0}^{1} \frac{{(lnt)}^{2}}{1 + \frac{1}{t^{2}}} (- \frac{dt}{t^{2}}) = 2 \int_{0}^{1} \frac{{(lnx)}^{2}}{1 + x^{2}} dx$ $= 2 \int_{0}^{1} {(lnx)}^{2} \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n} dx = 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{2 n} {(lnx)}^{2} dx$ $= 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} A_{n}$ $A_{n} = \int_{0}^{1} x^{2 n} {(lnx)}^{2} dx =_{by psrts} {[\frac{x^{2 n + 1}}{2 n + 1} {(lnx)}^{2}]}_{0}^{1} - \int_{0}^{1} \frac{x^{2 n + 1}}{2 n + 1} \frac{2 \ln (x)}{x} dx$ $= - \frac{2}{2 n + 1} \int_{0}^{1} x^{2 n} lnxdx = - \frac{2}{2 n + 1} {{[\frac{x^{2 n + 1}}{2 n + 1} lnx]}_{0}^{1} - \int_{0}^{1} \frac{x^{2 n}}{2 n + 1} dx}$ $= - \frac{2}{2 n + 1} {- \frac{1}{{(2 n + 1)}^{2}}} = \frac{2}{{(2 n + 1)}^{3}} \Rightarrow$ $I = 4 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} rest to find the value of this serie by fourier$ $. . . be continued . . . .$
	Commented by 150505R last updated on 18/Aug/20

	$t h a n k s a l o t s i r . . .$
	Commented by mathdave last updated on 18/Aug/20

	$u v e t r i e d l e t m e c o m p l e t d r e s t$
	Commented by abdomsup last updated on 18/Aug/20

	$y o u a r e w e l c o m e s i r$
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