question proposed by m.n july 1970 show that ∫_0 ^∞ ((ln(1+x))/(x(1+x^2 )))dx=((5π^2 )/(48))


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Question Number 114740 by mathdave last updated on 20/Sep/20

$q u e s t i o n p r o p o s e d b y m . n j u l y 1970$ $s h o w t h a t$ $\int_{0}^{\infty} \frac{\ln (1 + x)}{x (1 + x^{2})} d x = \frac{5 π^{2}}{48}$
	Answered by mathdave last updated on 20/Sep/20

	$s o l u t i o n$ $l e t$ $I = \int_{0}^{\infty} \frac{\ln (1 + x)}{x (1 + x^{2})} d x = \int_{0}^{1} \frac{\ln (1 + x)}{x (1 + x^{2})} d x + \int_{1}^{\infty} \frac{\ln (1 + x)}{x (1 + x^{2})} d x$ $p u t x = \frac{1}{x} f o r t h e s e c o n d p a r t$ $I = \int_{0}^{1} \frac{\ln (1 + x)}{x (1 + x^{2})} d x + \int_{1}^{0} \frac{x \ln (1 + \frac{1}{x})}{(1 + \frac{1}{x^{2}})} \times - \frac{1}{x^{2}} d x$ $I = \int_{0}^{1} \frac{\ln (1 + x)}{x (1 + x^{2})} d x + \int_{0}^{1} \frac{x \ln (1 + x)}{(1 + x^{2})} d x - \int_{0}^{1} \frac{x \ln x}{(1 + x^{2})} d x$ $I = \int_{0}^{1} \frac{(1 + x^{2}) \ln (1 + x)}{x (1 + x^{2})} d x - {\int_{0}}^{1} \frac{x \ln x}{(1 + x^{2})} d x$ $I = \int_{0}^{1} \frac{\ln (1 + x)}{x} - \int_{0}^{1} \frac{x \ln x}{1 + x^{2}} d x = A - B$ $l e t A = \int_{0}^{1} \frac{\ln (1 + x)}{x} d x p u t x = - x$ $A = \int_{0}^{- 1} \frac{\ln (1 - x)}{x} d x = - {L i}_{2} (- 1) = - (- \frac{π^{2}}{12})$ $A = \frac{π^{2}}{12} . . . . . . . . . (1) t h e n$ $B = \int_{0}^{1} \frac{x \ln x}{1 + x^{2}} d x = {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} x \ln x . x^{2 n} d x$ $B = {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} x^{2 n + 1} \ln x d x$ $B = \frac{\partial}{\partial a} ∣_{a = 1} ({(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} x^{2 n + 1} . x^{a - 1} d x)$ $B = \frac{\partial}{\partial a} ∣_{a = 1} ({(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} x^{2 n + a} d x) = \frac{\partial}{\partial a} ∣_{a = 1} ({(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} (\frac{1}{2 n + a + 1}))$ $B = - \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 2)}^{2}} = - \frac{1}{4} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}}$ $B = - \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n^{2}} = - \frac{1}{4} (\frac{π^{2}}{12}) = - \frac{π^{2}}{48}$ $b u t I = A - B = \frac{π^{2}}{12} - (- \frac{π^{2}}{12}) = \frac{5 π^{2}}{48}$ $∵ \int_{0}^{\infty} \frac{\ln (1 + x)}{x (1 + x^{2})} d x = \frac{5 π^{2}}{48}$ $b y m a t h d a v e (20 / 09 / 2020)$
	Commented by mnjuly1970 last updated on 21/Sep/20

	$t h a n k y o u m r d a v e$
	Commented by Ar Brandon last updated on 21/Sep/20
	welldone ��
	Commented by Tawa11 last updated on 06/Sep/21

	$great sir$
	Answered by mathmax by abdo last updated on 20/Sep/20
	$I =∫_0 ^∞ ((ln(1+x))/(x(1+x^2 )))dx ⇒I =∫_0 ^1 ((ln(1+x))/(x(1+x^2 )))dx +∫_1 ^(+∞) ((ln(1+x))/(x(1+x^2 )))dx(→x=(1/t)) =∫_0 ^1 ((ln(1+x))/(x(1+x^2 )))dx +∫_0 ^1 t((ln(1+(1/t)))/((1+(1/t^2 ))))((dt/t^2 )) =∫_0 ^1 ((ln(1+x))/(x(1+x^2 )))dx +∫_0 ^1 ((x×(ln(1+x)−lnx))/(1+x^2 )) dt =∫_0 ^1 ((ln(1+x))/(x(1+x^2 )))dx +∫_0 ^1 ((x^2 ln(1+x))/(x(1+x^2 )))dx−∫_0 ^1 ((xln(x))/(1+x^2 ))dx =∫_0 ^1 ((ln(1+x))/x)dx−∫_0 ^1 ((xln(x))/(1+x^2 ))dx we have (d/dx)ln(1+x) =(1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n ⇒ln(1+x)=Σ_(n=0) ^∞ (((−1)^n x^(n+1) )/(n+1)) =Σ_(n=1) ^∞ (((−1)^(n−1) x^n )/n)⇒((ln(1+x))/x) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^(n−1) ⇒ ∫_0 ^1 ((ln(1+x))/x)dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =−Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =−{2^(1−2) −1}ξ(2) =(π^2 /(12)) also we have ∫_0 ^1 ((xln(x))/(1+x^2 ))dx =∫_0 ^1 xlnxΣ_(n=0) ^∞ (−1)^n x^(2n) dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(2n+1) ln(x)dx =Σ_(n=0) ^∞ (−1)^n u_n by parts u_n =[(x^(2n+2) /(2n+2))ln(x)]_0 ^1 −∫_0 ^1 (x^(2n+2) /(2n+2))(dx/x) =−(1/((2n+2)^2 )) ⇒ ∫_0 ^1 ((xlnx)/(1+x^2 ))dx =−Σ_(n=0) ^∞ (((−1)^n )/(4(n+1)^2 )) =−(1/4)Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =(1/4)Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(1/4)(−(π^2 /(12))) =−(π^2 /(48)) ⇒ I =(π^2 /(12))+(π^2 /(48)) =((4π^2 +π^2 )/(48)) =((5π^2 )/(48))$
	$I = \int_{0}^{\infty} \frac{\ln (1 + x)}{x (1 + x^{2})} dx \Rightarrow I = \int_{0}^{1} \frac{\ln (1 + x)}{x (1 + x^{2})} dx + \int_{1}^{+ \infty} \frac{\ln (1 + x)}{x (1 + x^{2})} dx (\to x = \frac{1}{t})$ $= \int_{0}^{1} \frac{\ln (1 + x)}{x (1 + x^{2})} dx + \int_{0}^{1} t \frac{\ln (1 + \frac{1}{t})}{(1 + \frac{1}{t^{2}})} (\frac{dt}{t^{2}})$ $= \int_{0}^{1} \frac{\ln (1 + x)}{x (1 + x^{2})} dx + \int_{0}^{1} \frac{x \times (\ln (1 + x) - lnx)}{1 + x^{2}} dt$ $= \int_{0}^{1} \frac{\ln (1 + x)}{x (1 + x^{2})} dx + \int_{0}^{1} \frac{x^{2} \ln (1 + x)}{x (1 + x^{2})} dx - \int_{0}^{1} \frac{xln (x)}{1 + x^{2}} dx$ $= \int_{0}^{1} \frac{\ln (1 + x)}{x} dx - \int_{0}^{1} \frac{xln (x)}{1 + x^{2}} dx$ $we have \frac{d}{dx} \ln (1 + x) = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} \Rightarrow \ln (1 + x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{n + 1}}{n + 1}$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} x^{n}}{n} \Rightarrow \frac{\ln (1 + x)}{x} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n - 1} \Rightarrow$ $\int_{0}^{1} \frac{\ln (1 + x)}{x} dx = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} = - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}}$ $= - {2^{1 - 2} - 1} ξ (2) = \frac{π^{2}}{12} also we have$ $\int_{0}^{1} \frac{xln (x)}{1 + x^{2}} dx = \int_{0}^{1} xlnx \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n} dx$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{2 n + 1} \ln (x) dx = \sum_{n = 0}^{\infty} {(- 1)}^{n} u_{n}$ $by parts u_{n} = {[\frac{x^{2 n + 2}}{2 n + 2} \ln (x)]}_{0}^{1} - \int_{0}^{1} \frac{x^{2 n + 2}}{2 n + 2} \frac{dx}{x} = - \frac{1}{{(2 n + 2)}^{2}} \Rightarrow$ $\int_{0}^{1} \frac{xlnx}{1 + x^{2}} dx = - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 {(n + 1)}^{2}} = - \frac{1}{4} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}}$ $= \frac{1}{4} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = \frac{1}{4} (- \frac{π^{2}}{12}) = - \frac{π^{2}}{48} \Rightarrow$ $I = \frac{π^{2}}{12} + \frac{π^{2}}{48} = \frac{4 π^{2} + π^{2}}{48} = \frac{5 π^{2}}{48}$
	Commented by mnjuly1970 last updated on 21/Sep/20

	$t h a n k y o u s o m u c h$
	Commented by mathmax by abdo last updated on 21/Sep/20

	$you are welcome$
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