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Question Number 137439 by mnjuly1970 last updated on 02/Apr/21
            ......nice  calculus.....      prove that::        π›˜=∫_0 ^( 1) ((ln(x+(√(1βˆ’x^2  )) ))/x)dx=(Ο€^2 /(16)) ....
$$\:\:\:\:\:\:\:\:\:\:\:\:……{nice}\:\:{calculus}….. \\ $$$$\:\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\:\boldsymbol{\chi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}+\sqrt{\mathrm{1}βˆ’{x}^{\mathrm{2}} \:}\:\right)}{{x}}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\:…. \\ $$
Answered by mindispower last updated on 03/Apr/21
f(t)=∫_0 ^1 ((ln(tx+(√(1βˆ’x^2 ))))/x)dx,We want f(1)  fβ€²(t)=∫_0 ^1 (dx/(tx+(√(1βˆ’x^2 ))))  x=sin(r)  fβ€²(t)=∫_0 ^(Ο€/2) ((cos(r)dr)/(tsin(r)+cos(r)))  cos(r)=a(tsin(r)+cos(r))+b(tcos(r)βˆ’sin(r))  atβˆ’b=0  a+bt=1  b=(t/(t^2 +1))  a=(1/(1+t^2 ))  fβ€²(t)=∫_0 ^(Ο€/2) (dr/(1+t^2 ))+(t/(1+t^2 ))∫_0 ^(Ο€/2) ((tcos(r)βˆ’sin(r))/(tsin(r)+cos(r)))dr  =(Ο€/2).(1/(1+t^2 ))+(t/(1+t^2 ))ln(t)  f(0)=(1/2)∫_0 ^1 ((ln(1βˆ’t^2 ))/t)dt=(1/2).βˆ’Ξ£_(nβ‰₯0) ∫_0 ^1 (t^(2n+1) /(n+1))  =βˆ’(1/4)Ξ£_(nβ‰₯0) (1/((n+1)^2 ))=βˆ’(Ο€^2 /(24))  ∫_0 ^1 fβ€²(t)=f(1)βˆ’f(0)=(Ο€/2)arctan(1)+∫_0 ^1 ((tln(t))/(1+t^2 ))dt  =(Ο€^2 /8)βˆ’Ξ£_(nβ‰₯0) (((βˆ’1)^n )/((2n+2)^2 ))=(Ο€^2 /8)βˆ’(1/4)((Ο€^2 /(12)))=((5Ο€^2 )/(48))  f(1)=((5Ο€^2 )/(48))+f(0)=((5Ο€^2 )/(48))βˆ’(Ο€^2 /(24))=(Ο€^2 /(16))
$${f}\left({t}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({tx}+\sqrt{\mathrm{1}βˆ’{x}^{\mathrm{2}} }\right)}{{x}}{dx},{We}\:{want}\:{f}\left(\mathrm{1}\right) \\ $$$${f}'\left({t}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{{tx}+\sqrt{\mathrm{1}βˆ’{x}^{\mathrm{2}} }} \\ $$$${x}={sin}\left({r}\right) \\ $$$${f}'\left({t}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({r}\right){dr}}{{tsin}\left({r}\right)+{cos}\left({r}\right)} \\ $$$${cos}\left({r}\right)={a}\left({tsin}\left({r}\right)+{cos}\left({r}\right)\right)+{b}\left({tcos}\left({r}\right)βˆ’{sin}\left({r}\right)\right) \\ $$$${at}βˆ’{b}=\mathrm{0} \\ $$$${a}+{bt}=\mathrm{1} \\ $$$${b}=\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${f}'\left({t}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dr}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{tcos}\left({r}\right)βˆ’{sin}\left({r}\right)}{{tsin}\left({r}\right)+{cos}\left({r}\right)}{dr} \\ $$$$=\frac{\pi}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{ln}\left({t}\right) \\ $$$${f}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}βˆ’{t}^{\mathrm{2}} \right)}{{t}}{dt}=\frac{\mathrm{1}}{\mathrm{2}}.βˆ’\underset{{n}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$=βˆ’\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}'\left({t}\right)={f}\left(\mathrm{1}\right)βˆ’{f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}}{arctan}\left(\mathrm{1}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}βˆ’\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(βˆ’\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{2}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}βˆ’\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right)=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$${f}\left(\mathrm{1}\right)=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}}+{f}\left(\mathrm{0}\right)=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}}βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{24}}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 03/Apr/21
thanks alot mr power..
$${thanks}\:{alot}\:{mr}\:{power}.. \\ $$
Commented by mnjuly1970 last updated on 03/Apr/21
Commented by mindispower last updated on 03/Apr/21
pleasur sir
$${pleasur}\:{sir}\: \\ $$
Answered by mathmax by abdo last updated on 03/Apr/21
Ο‡=∫_0 ^1  ((log(x+(√(1βˆ’x^2 ))))/x)dx  let f(a)=∫_0 ^1  ((log(ax+(√(1βˆ’x^2 ))))/x)dx witha>0  f^β€² (a)=∫_0 ^1   (x/(x(ax+(√(1βˆ’x^2 )))))dx =∫_0 ^1  (dx/(ax +(√(1βˆ’x^2 ))))  =_(x=sint)    ∫_0 ^(Ο€/2)  ((cost )/(asint +cost))dt =∫_0 ^(Ο€/2)  (dt/(atant +1))  =_(tant=y)    ∫_0 ^∞   (1/(ay +1)) Γ—(dy/((1+y^2 )))  let decompose  F(y) =(1/((ay+1)(y^2  +1))) β‡’F(y)=(Ξ±/(ay+1)) +((Ξ²y +Ξ΄)/(y^2  +1))  Ξ± =(1/((1/a^2 )+1)) =(a^2 /(a^2  +1))  lim_(yβ†’+∞)  yF(y)=0 =(Ξ±/a)+Ξ² β‡’Ξ²=βˆ’(a/(a^2  +1))  F(0)=1 =Ξ±+Ξ΄ β‡’Ξ΄=1βˆ’(a^2 /(a^2  +1)) =(1/(a^2  +1)) β‡’  F(y)=(a^2 /((a^2 +1)(ay+1)))+ ((βˆ’(a/(1+a^2 ))y+(1/(1+a^2 )))/(y^2  +1)) β‡’  ∫_0 ^∞  F(y)dy =(1/(a^2  +1))∫_0 ^∞ ( ((a^2  dy)/(ay +1))βˆ’((ayβˆ’1)/(y^2  +1)))dy  =(1/(a^2  +1))∫_0 ^∞ ((a^2 /(ay+1))βˆ’(a/2)((2y)/(y^2  +1)) +(1/(y^2  +1)))dy  =(1/(a^2 +1))[aln(ay+1)βˆ’aln(√(y^2 +1))]_0 ^∞  +(Ο€/(2(1+a^2 )))  =(a/(a^2  +1))[ln(((ay+1)/( (√(y^2 +1)))))]_0 ^∞ +(Ο€/(2(1+a^2 )))  =((aloga)/(a^2  +1))+(Ο€/(2(a^2  +1))) =f^β€² (a) β‡’  f(a)=∫ ((aloga)/(a^2 +1))da +(Ο€/2)arctan(a)+C  f(1)βˆ’f(0) =∫_0 ^1  f^β€² (a)da =∫_0 ^1  ((aloga)/(1+a^2 ))da +(Ο€/2)∫_0 ^1  (da/(1+a^2 ))  ∫_0 ^1  (da/(1+a^2 )) =[arctana]_0 ^1  =(Ο€/4)  ∫_0 ^1  ((aloga)/(1+a^2 ))da =[(1/2)log(1+a^2 )loga]_0 ^(1 ) βˆ’(1/2)∫_0 ^1  ((log(1+a^2 ))/a)da  =βˆ’(1/2)∫_0 ^1  ((log(1+a^2 ))/a)da  (log(1+u)^β€²  =(1/(1+u)) =Ξ£_(n=0) ^∞  (βˆ’1)^n  u^n  β‡’log(1+u)=Ξ£_(n=0) ^∞ (βˆ’1)^n  (u^(n+1) /(n+1))  =Ξ£_(n=1) ^∞  (βˆ’1)^(nβˆ’1)  (u^n /n) β‡’log(1+x^2 ) =Ξ£_(n=1) ^∞ (βˆ’1)^(nβˆ’1)  (x^(2n) /n) β‡’  ∫_0 ^1  ((log(1+x^2 ))/x)dx =Ξ£_(n=1) ^∞  (((βˆ’1)^(nβˆ’1) )/(n(2n+1))) =u_n   (u_n /2)=Ξ£_(n=1) ^∞  (((βˆ’1)^(nβˆ’1) )/(2n(2n+1))) =Ξ£_(n=1) ^∞  (βˆ’1)^(nβˆ’1) ((1/(2n))βˆ’(1/(2n+1)))  =(1/2)Ξ£_(n=1) ^∞  (((βˆ’1)^(nβˆ’1) )/n)βˆ’Ξ£_(n=1) ^∞  (((βˆ’1)^n )/(2n+1))  =((log2)/2)βˆ’((Ο€/4)βˆ’1) β‡’f(1)βˆ’f(0)=βˆ’((log2)/4)+(1/2)((Ο€/4)βˆ’1)  =βˆ’((log2)/4)+(Ο€/8)βˆ’(1/2) β‡’f(1)=Ο‡ =βˆ’((log2)/4)+(Ο€/8)βˆ’(1/2) +f(0)  f(0)=∫_0 ^1  ((log((√(1βˆ’x^2 ))))/x)dx rest to find f(0)....be continued...
$$\chi=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{x}+\sqrt{\mathrm{1}βˆ’\mathrm{x}^{\mathrm{2}} }\right)}{\mathrm{x}}\mathrm{dx}\:\:\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{ax}+\sqrt{\mathrm{1}βˆ’\mathrm{x}^{\mathrm{2}} }\right)}{\mathrm{x}}\mathrm{dx}\:\mathrm{witha}>\mathrm{0} \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{x}}{\mathrm{x}\left(\mathrm{ax}+\sqrt{\mathrm{1}βˆ’\mathrm{x}^{\mathrm{2}} }\right)}\mathrm{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\mathrm{ax}\:+\sqrt{\mathrm{1}βˆ’\mathrm{x}^{\mathrm{2}} }} \\ $$$$=_{\mathrm{x}=\mathrm{sint}} \:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{cost}\:}{\mathrm{asint}\:+\mathrm{cost}}\mathrm{dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{dt}}{\mathrm{atant}\:+\mathrm{1}} \\ $$$$=_{\mathrm{tant}=\mathrm{y}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{ay}\:+\mathrm{1}}\:Γ—\frac{\mathrm{dy}}{\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)}\:\:\mathrm{let}\:\mathrm{decompose} \\ $$$$\mathrm{F}\left(\mathrm{y}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{ay}+\mathrm{1}\right)\left(\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow\mathrm{F}\left(\mathrm{y}\right)=\frac{\alpha}{\mathrm{ay}+\mathrm{1}}\:+\frac{\beta\mathrm{y}\:+\delta}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\alpha\:=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\mathrm{1}}\:=\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\mathrm{lim}_{\mathrm{y}\rightarrow+\infty} \:\mathrm{yF}\left(\mathrm{y}\right)=\mathrm{0}\:=\frac{\alpha}{\mathrm{a}}+\beta\:\Rightarrow\beta=βˆ’\frac{\mathrm{a}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\mathrm{F}\left(\mathrm{0}\right)=\mathrm{1}\:=\alpha+\delta\:\Rightarrow\delta=\mathrm{1}βˆ’\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{y}\right)=\frac{\mathrm{a}^{\mathrm{2}} }{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{ay}+\mathrm{1}\right)}+\:\frac{βˆ’\frac{\mathrm{a}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} }\mathrm{y}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} }}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{F}\left(\mathrm{y}\right)\mathrm{dy}\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \left(\:\frac{\mathrm{a}^{\mathrm{2}} \:\mathrm{dy}}{\mathrm{ay}\:+\mathrm{1}}βˆ’\frac{\mathrm{ay}βˆ’\mathrm{1}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\right)\mathrm{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{ay}+\mathrm{1}}βˆ’\frac{\mathrm{a}}{\mathrm{2}}\frac{\mathrm{2y}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\right)\mathrm{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}\left[\mathrm{aln}\left(\mathrm{ay}+\mathrm{1}\right)βˆ’\mathrm{aln}\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\right]_{\mathrm{0}} ^{\infty} \:+\frac{\pi}{\mathrm{2}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{a}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\left[\mathrm{ln}\left(\frac{\mathrm{ay}+\mathrm{1}}{\:\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}}\right)\right]_{\mathrm{0}} ^{\infty} +\frac{\pi}{\mathrm{2}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{aloga}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}+\frac{\pi}{\mathrm{2}\left(\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\mathrm{f}^{'} \left(\mathrm{a}\right)\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\int\:\frac{\mathrm{aloga}}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}\mathrm{da}\:+\frac{\pi}{\mathrm{2}}\mathrm{arctan}\left(\mathrm{a}\right)+\mathrm{C} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)βˆ’\mathrm{f}\left(\mathrm{0}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{f}^{'} \left(\mathrm{a}\right)\mathrm{da}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{aloga}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} }\mathrm{da}\:+\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{da}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{da}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} }\:=\left[\mathrm{arctana}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\pi}{\mathrm{4}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{aloga}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} }\mathrm{da}\:=\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)\mathrm{loga}\right]_{\mathrm{0}} ^{\mathrm{1}\:} βˆ’\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)}{\mathrm{a}}\mathrm{da} \\ $$$$=βˆ’\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)}{\mathrm{a}}\mathrm{da} \\ $$$$\left(\mathrm{log}\left(\mathrm{1}+\mathrm{u}\right)^{'} \:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{u}}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(βˆ’\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{u}^{\mathrm{n}} \:\Rightarrow\mathrm{log}\left(\mathrm{1}+\mathrm{u}\right)=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(βˆ’\mathrm{1}\right)^{\mathrm{n}} \:\frac{\mathrm{u}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\right. \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\left(βˆ’\mathrm{1}\right)^{\mathrm{n}βˆ’\mathrm{1}} \:\frac{\mathrm{u}^{\mathrm{n}} }{\mathrm{n}}\:\Rightarrow\mathrm{log}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(βˆ’\mathrm{1}\right)^{\mathrm{n}βˆ’\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{2n}} }{\mathrm{n}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}\mathrm{dx}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(βˆ’\mathrm{1}\right)^{\mathrm{n}βˆ’\mathrm{1}} }{\mathrm{n}\left(\mathrm{2n}+\mathrm{1}\right)}\:=\mathrm{u}_{\mathrm{n}} \\ $$$$\frac{\mathrm{u}_{\mathrm{n}} }{\mathrm{2}}=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(βˆ’\mathrm{1}\right)^{\mathrm{n}βˆ’\mathrm{1}} }{\mathrm{2n}\left(\mathrm{2n}+\mathrm{1}\right)}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\left(βˆ’\mathrm{1}\right)^{\mathrm{n}βˆ’\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2n}}βˆ’\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(βˆ’\mathrm{1}\right)^{\mathrm{n}βˆ’\mathrm{1}} }{\mathrm{n}}βˆ’\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(βˆ’\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{log2}}{\mathrm{2}}βˆ’\left(\frac{\pi}{\mathrm{4}}βˆ’\mathrm{1}\right)\:\Rightarrow\mathrm{f}\left(\mathrm{1}\right)βˆ’\mathrm{f}\left(\mathrm{0}\right)=βˆ’\frac{\mathrm{log2}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{4}}βˆ’\mathrm{1}\right) \\ $$$$=βˆ’\frac{\mathrm{log2}}{\mathrm{4}}+\frac{\pi}{\mathrm{8}}βˆ’\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{f}\left(\mathrm{1}\right)=\chi\:=βˆ’\frac{\mathrm{log2}}{\mathrm{4}}+\frac{\pi}{\mathrm{8}}βˆ’\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{f}\left(\mathrm{0}\right) \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\sqrt{\mathrm{1}βˆ’\mathrm{x}^{\mathrm{2}} }\right)}{\mathrm{x}}\mathrm{dx}\:\mathrm{rest}\:\mathrm{to}\:\mathrm{find}\:\mathrm{f}\left(\mathrm{0}\right)….\mathrm{be}\:\mathrm{continued}… \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 03/Apr/21
thanks alot mr max....
$${thanks}\:{alot}\:{mr}\:{max}…. \\ $$

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