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Question Number 114740 by mathdave last updated on 20/Sep/20
question proposed by m.n july 1970 show that ∫_0 ^∞ ((ln(1+x))/(x(1+x^2 )))dx=((5π^2 )/(48))
$${question}\:{proposed}\:{by}\:{m}.{n}\:{july}\:\mathrm{1970} \\ $$$${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$
Answered by mathdave last updated on 20/Sep/20
solution let I=∫_0 ^∞ ((ln(1+x))/(x(1+x^2 )))dx=∫_0 ^1 ((ln(1+x))/(x(1+x^2 )))dx+∫_1 ^∞ ((ln(1+x))/(x(1+x^2 )))dx put x=(1/x) for the second part I=∫_0 ^1 ((ln(1+x))/(x(1+x^2 )))dx+∫_1 ^0 ((xln(1+(1/x)))/((1+(1/x^2 ))))×−(1/x^2 )dx I=∫_0 ^1 ((ln(1+x))/(x(1+x^2 )))dx+∫_0 ^1 ((xln(1+x))/((1+x^2 )))dx−∫_0 ^1 ((xlnx)/((1+x^2 )))dx I=∫_0 ^1 (((1+x^2 )ln(1+x))/(x(1+x^2 )))dx−∫^1 _0 ((xlnx)/((1+x^2 )))dx I=∫_0 ^1 ((ln(1+x))/x)−∫_0 ^1 ((xlnx)/(1+x^2 ))dx=A−B let A=∫_0 ^1 ((ln(1+x))/x)dx put x=−x A=∫_0 ^(−1) ((ln(1−x))/x)dx=−Li_2 (−1)=−(−(π^2 /(12))) A=(π^2 /(12)).........(1) then B=∫_0 ^1 ((xlnx)/(1+x^2 ))dx=(−1)^n Σ_(n=0) ^∞ ∫_0 ^1 xlnx.x^(2n) dx B=(−1)^n Σ_(n=0) ^∞ ∫_0 ^1 x^(2n+1) lnxdx B=(∂/∂a)∣_(a=1) ((−1)^n Σ_(n=0) ^∞ ∫_0 ^1 x^(2n+1) .x^(a−1) dx) B=(∂/∂a)∣_(a=1) ((−1)^n Σ_(n=0) ^∞ ∫_0 ^1 x^(2n+a) dx)=(∂/∂a)∣_(a=1) ((−1)^n Σ_(n=0) ^∞ ((1/(2n+a+1)))) B=−Σ_(n=0) ^∞ (((−1)^n )/((2n+2)^2 ))=−(1/4)Σ_(n=0) ^∞ (((−1)^n )/((n+1)^2 )) B=−(1/4)Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 )=−(1/4)((π^2 /(12)))=−(π^2 /(48)) but I=A−B=(π^2 /(12))−(−(π^2 /(12)))=((5π^2 )/(48)) ∵∫_0 ^∞ ((ln(1+x))/(x(1+x^2 )))dx=((5π^2 )/(48)) by mathdave(20/09/2020)
$${solution} \\ $$$${let} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}+\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$${put}\:{x}=\frac{\mathrm{1}}{{x}}\:\:{for}\:{the}\:{second}\:{part} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}+\int_{\mathrm{1}} ^{\mathrm{0}} \frac{{x}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)}{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}×−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}\mathrm{ln}\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}\mathrm{ln}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}−\underset{\mathrm{0}} {\int}^{\mathrm{1}} \frac{{x}\mathrm{ln}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}={A}−{B} \\ $$$${let}\:{A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}\:\:\:\:\:\:{put}\:\:{x}=−{x} \\ $$$${A}=\int_{\mathrm{0}} ^{−\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx}=−{Li}_{\mathrm{2}} \left(−\mathrm{1}\right)=−\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right) \\ $$$${A}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}………\left(\mathrm{1}\right)\:\:\:{then} \\ $$$${B}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}\mathrm{ln}{x}.{x}^{\mathrm{2}{n}} {dx} \\ $$$${B}=\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} \mathrm{ln}{xdx} \\ $$$${B}=\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \left(\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} .{x}^{{a}−\mathrm{1}} {dx}\right) \\ $$$${B}=\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \left(\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}+{a}} {dx}\right)=\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \left(\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}+{a}+\mathrm{1}}\right)\right) \\ $$$${B}=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{2}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:\: \\ $$$${B}=−\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$${but}\:{I}={A}−{B}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right)=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$$\because\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$${by}\:{mathdave}\left(\mathrm{20}/\mathrm{09}/\mathrm{2020}\right) \\ $$
Commented by mnjuly1970 last updated on 21/Sep/20
thank you mr dave
$${thank}\:{you}\:{mr}\:{dave} \\ $$
Commented by Ar Brandon last updated on 21/Sep/20
welldone ��
Commented by Tawa11 last updated on 06/Sep/21
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$
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))
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}\:+\int_{\mathrm{1}} ^{+\infty} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}\left(\rightarrow\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}\:\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{t}\frac{\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}}\right)}{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)}\left(\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} }\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}×\left(\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)−\mathrm{lnx}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{2}} \mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{xln}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{xln}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \:\Rightarrow\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}} \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \:\mathrm{x}^{\mathrm{n}} }{\mathrm{n}}\Rightarrow\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\mathrm{x}^{\mathrm{n}−\mathrm{1}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}^{\mathrm{2}} }\:=−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} } \\ $$$$=−\left\{\mathrm{2}^{\mathrm{1}−\mathrm{2}} −\mathrm{1}\right\}\xi\left(\mathrm{2}\right)\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:\mathrm{also}\:\mathrm{we}\:\mathrm{have} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{xln}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{xlnx}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{2n}} \:\mathrm{dx} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{2n}+\mathrm{1}} \:\mathrm{ln}\left(\mathrm{x}\right)\mathrm{dx}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{u}_{\mathrm{n}} \\ $$$$\mathrm{by}\:\mathrm{parts}\:\mathrm{u}_{\mathrm{n}} =\left[\frac{\mathrm{x}^{\mathrm{2n}+\mathrm{2}} }{\mathrm{2n}+\mathrm{2}}\mathrm{ln}\left(\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{2n}+\mathrm{2}} }{\mathrm{2n}+\mathrm{2}}\frac{\mathrm{dx}}{\mathrm{x}}\:=−\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{2}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{xlnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{4}\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{4}}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right)\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{48}}\:\Rightarrow \\ $$$$\mathrm{I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\pi^{\mathrm{2}} }{\mathrm{48}}\:=\frac{\mathrm{4}\pi^{\mathrm{2}} \:+\pi^{\mathrm{2}} }{\mathrm{48}}\:=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 21/Sep/20
thank you so much
$${thank}\:{you}\:{so}\:{much} \\ $$
Commented by mathmax by abdo last updated on 21/Sep/20
you are welcome
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

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