Question Number 137439 by mnjuly1970 last updated on 02/Apr/21
$$\:\:\:\:\:\:\:\:\:\:\:\:……{nice}\:\:{calculus}….. \\ $$$$\:\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\:\boldsymbol{\chi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}+\sqrt{\mathrm{1}β{x}^{\mathrm{2}} \:}\:\right)}{{x}}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\:…. \\ $$
Answered by mindispower last updated on 03/Apr/21
$${f}\left({t}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({tx}+\sqrt{\mathrm{1}β{x}^{\mathrm{2}} }\right)}{{x}}{dx},{We}\:{want}\:{f}\left(\mathrm{1}\right) \\ $$$${f}'\left({t}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{{tx}+\sqrt{\mathrm{1}β{x}^{\mathrm{2}} }} \\ $$$${x}={sin}\left({r}\right) \\ $$$${f}'\left({t}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({r}\right){dr}}{{tsin}\left({r}\right)+{cos}\left({r}\right)} \\ $$$${cos}\left({r}\right)={a}\left({tsin}\left({r}\right)+{cos}\left({r}\right)\right)+{b}\left({tcos}\left({r}\right)β{sin}\left({r}\right)\right) \\ $$$${at}β{b}=\mathrm{0} \\ $$$${a}+{bt}=\mathrm{1} \\ $$$${b}=\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${f}'\left({t}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dr}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{tcos}\left({r}\right)β{sin}\left({r}\right)}{{tsin}\left({r}\right)+{cos}\left({r}\right)}{dr} \\ $$$$=\frac{\pi}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{ln}\left({t}\right) \\ $$$${f}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}β{t}^{\mathrm{2}} \right)}{{t}}{dt}=\frac{\mathrm{1}}{\mathrm{2}}.β\underset{{n}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$=β\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=β\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}'\left({t}\right)={f}\left(\mathrm{1}\right)β{f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}}{arctan}\left(\mathrm{1}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}β\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(β\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{2}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}β\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right)=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$${f}\left(\mathrm{1}\right)=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}}+{f}\left(\mathrm{0}\right)=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}}β\frac{\pi^{\mathrm{2}} }{\mathrm{24}}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 03/Apr/21
$${thanks}\:{alot}\:{mr}\:{power}.. \\ $$
Commented by mnjuly1970 last updated on 03/Apr/21
Commented by mindispower last updated on 03/Apr/21
$${pleasur}\:{sir}\: \\ $$
Answered by mathmax by abdo last updated on 03/Apr/21
![Ο=β«_0 ^1 ((log(x+(β(1βx^2 ))))/x)dx let f(a)=β«_0 ^1 ((log(ax+(β(1βx^2 ))))/x)dx witha>0 f^β² (a)=β«_0 ^1 (x/(x(ax+(β(1βx^2 )))))dx =β«_0 ^1 (dx/(ax +(β(1βx^2 )))) =_(x=sint) β«_0 ^(Ο/2) ((cost )/(asint +cost))dt =β«_0 ^(Ο/2) (dt/(atant +1)) =_(tant=y) β«_0 ^β (1/(ay +1)) Γ(dy/((1+y^2 ))) let decompose F(y) =(1/((ay+1)(y^2 +1))) βF(y)=(Ξ±/(ay+1)) +((Ξ²y +Ξ΄)/(y^2 +1)) Ξ± =(1/((1/a^2 )+1)) =(a^2 /(a^2 +1)) lim_(yβ+β) yF(y)=0 =(Ξ±/a)+Ξ² βΞ²=β(a/(a^2 +1)) F(0)=1 =Ξ±+Ξ΄ βΞ΄=1β(a^2 /(a^2 +1)) =(1/(a^2 +1)) β F(y)=(a^2 /((a^2 +1)(ay+1)))+ ((β(a/(1+a^2 ))y+(1/(1+a^2 )))/(y^2 +1)) β β«_0 ^β F(y)dy =(1/(a^2 +1))β«_0 ^β ( ((a^2 dy)/(ay +1))β((ayβ1)/(y^2 +1)))dy =(1/(a^2 +1))β«_0 ^β ((a^2 /(ay+1))β(a/2)((2y)/(y^2 +1)) +(1/(y^2 +1)))dy =(1/(a^2 +1))[aln(ay+1)βaln(β(y^2 +1))]_0 ^β +(Ο/(2(1+a^2 ))) =(a/(a^2 +1))[ln(((ay+1)/( (β(y^2 +1)))))]_0 ^β +(Ο/(2(1+a^2 ))) =((aloga)/(a^2 +1))+(Ο/(2(a^2 +1))) =f^β² (a) β f(a)=β« ((aloga)/(a^2 +1))da +(Ο/2)arctan(a)+C f(1)βf(0) =β«_0 ^1 f^β² (a)da =β«_0 ^1 ((aloga)/(1+a^2 ))da +(Ο/2)β«_0 ^1 (da/(1+a^2 )) β«_0 ^1 (da/(1+a^2 )) =[arctana]_0 ^1 =(Ο/4) β«_0 ^1 ((aloga)/(1+a^2 ))da =[(1/2)log(1+a^2 )loga]_0 ^(1 ) β(1/2)β«_0 ^1 ((log(1+a^2 ))/a)da =β(1/2)β«_0 ^1 ((log(1+a^2 ))/a)da (log(1+u)^β² =(1/(1+u)) =Ξ£_(n=0) ^β (β1)^n u^n βlog(1+u)=Ξ£_(n=0) ^β (β1)^n (u^(n+1) /(n+1)) =Ξ£_(n=1) ^β (β1)^(nβ1) (u^n /n) βlog(1+x^2 ) =Ξ£_(n=1) ^β (β1)^(nβ1) (x^(2n) /n) β β«_0 ^1 ((log(1+x^2 ))/x)dx =Ξ£_(n=1) ^β (((β1)^(nβ1) )/(n(2n+1))) =u_n (u_n /2)=Ξ£_(n=1) ^β (((β1)^(nβ1) )/(2n(2n+1))) =Ξ£_(n=1) ^β (β1)^(nβ1) ((1/(2n))β(1/(2n+1))) =(1/2)Ξ£_(n=1) ^β (((β1)^(nβ1) )/n)βΞ£_(n=1) ^β (((β1)^n )/(2n+1)) =((log2)/2)β((Ο/4)β1) βf(1)βf(0)=β((log2)/4)+(1/2)((Ο/4)β1) =β((log2)/4)+(Ο/8)β(1/2) βf(1)=Ο =β((log2)/4)+(Ο/8)β(1/2) +f(0) f(0)=β«_0 ^1 ((log((β(1βx^2 ))))/x)dx rest to find f(0)....be continued...](https://www.tinkutara.com/question/Q137449.png)
$$\chi=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{x}+\sqrt{\mathrm{1}β\mathrm{x}^{\mathrm{2}} }\right)}{\mathrm{x}}\mathrm{dx}\:\:\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{ax}+\sqrt{\mathrm{1}β\mathrm{x}^{\mathrm{2}} }\right)}{\mathrm{x}}\mathrm{dx}\:\mathrm{witha}>\mathrm{0} \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{x}}{\mathrm{x}\left(\mathrm{ax}+\sqrt{\mathrm{1}β\mathrm{x}^{\mathrm{2}} }\right)}\mathrm{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\mathrm{ax}\:+\sqrt{\mathrm{1}β\mathrm{x}^{\mathrm{2}} }} \\ $$$$=_{\mathrm{x}=\mathrm{sint}} \:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{cost}\:}{\mathrm{asint}\:+\mathrm{cost}}\mathrm{dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{dt}}{\mathrm{atant}\:+\mathrm{1}} \\ $$$$=_{\mathrm{tant}=\mathrm{y}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{ay}\:+\mathrm{1}}\:Γ\frac{\mathrm{dy}}{\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)}\:\:\mathrm{let}\:\mathrm{decompose} \\ $$$$\mathrm{F}\left(\mathrm{y}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{ay}+\mathrm{1}\right)\left(\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow\mathrm{F}\left(\mathrm{y}\right)=\frac{\alpha}{\mathrm{ay}+\mathrm{1}}\:+\frac{\beta\mathrm{y}\:+\delta}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\alpha\:=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\mathrm{1}}\:=\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\mathrm{lim}_{\mathrm{y}\rightarrow+\infty} \:\mathrm{yF}\left(\mathrm{y}\right)=\mathrm{0}\:=\frac{\alpha}{\mathrm{a}}+\beta\:\Rightarrow\beta=β\frac{\mathrm{a}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\mathrm{F}\left(\mathrm{0}\right)=\mathrm{1}\:=\alpha+\delta\:\Rightarrow\delta=\mathrm{1}β\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{y}\right)=\frac{\mathrm{a}^{\mathrm{2}} }{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{ay}+\mathrm{1}\right)}+\:\frac{β\frac{\mathrm{a}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} }\mathrm{y}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} }}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{F}\left(\mathrm{y}\right)\mathrm{dy}\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \left(\:\frac{\mathrm{a}^{\mathrm{2}} \:\mathrm{dy}}{\mathrm{ay}\:+\mathrm{1}}β\frac{\mathrm{ay}β\mathrm{1}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\right)\mathrm{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{ay}+\mathrm{1}}β\frac{\mathrm{a}}{\mathrm{2}}\frac{\mathrm{2y}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\right)\mathrm{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}\left[\mathrm{aln}\left(\mathrm{ay}+\mathrm{1}\right)β\mathrm{aln}\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\right]_{\mathrm{0}} ^{\infty} \:+\frac{\pi}{\mathrm{2}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{a}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\left[\mathrm{ln}\left(\frac{\mathrm{ay}+\mathrm{1}}{\:\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}}\right)\right]_{\mathrm{0}} ^{\infty} +\frac{\pi}{\mathrm{2}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{aloga}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}+\frac{\pi}{\mathrm{2}\left(\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\mathrm{f}^{'} \left(\mathrm{a}\right)\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\int\:\frac{\mathrm{aloga}}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}\mathrm{da}\:+\frac{\pi}{\mathrm{2}}\mathrm{arctan}\left(\mathrm{a}\right)+\mathrm{C} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)β\mathrm{f}\left(\mathrm{0}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{f}^{'} \left(\mathrm{a}\right)\mathrm{da}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{aloga}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} }\mathrm{da}\:+\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{da}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{da}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} }\:=\left[\mathrm{arctana}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\pi}{\mathrm{4}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{aloga}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} }\mathrm{da}\:=\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)\mathrm{loga}\right]_{\mathrm{0}} ^{\mathrm{1}\:} β\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)}{\mathrm{a}}\mathrm{da} \\ $$$$=β\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)}{\mathrm{a}}\mathrm{da} \\ $$$$\left(\mathrm{log}\left(\mathrm{1}+\mathrm{u}\right)^{'} \:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{u}}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(β\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{u}^{\mathrm{n}} \:\Rightarrow\mathrm{log}\left(\mathrm{1}+\mathrm{u}\right)=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(β\mathrm{1}\right)^{\mathrm{n}} \:\frac{\mathrm{u}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\right. \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\left(β\mathrm{1}\right)^{\mathrm{n}β\mathrm{1}} \:\frac{\mathrm{u}^{\mathrm{n}} }{\mathrm{n}}\:\Rightarrow\mathrm{log}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(β\mathrm{1}\right)^{\mathrm{n}β\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{2n}} }{\mathrm{n}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}\mathrm{dx}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(β\mathrm{1}\right)^{\mathrm{n}β\mathrm{1}} }{\mathrm{n}\left(\mathrm{2n}+\mathrm{1}\right)}\:=\mathrm{u}_{\mathrm{n}} \\ $$$$\frac{\mathrm{u}_{\mathrm{n}} }{\mathrm{2}}=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(β\mathrm{1}\right)^{\mathrm{n}β\mathrm{1}} }{\mathrm{2n}\left(\mathrm{2n}+\mathrm{1}\right)}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\left(β\mathrm{1}\right)^{\mathrm{n}β\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2n}}β\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(β\mathrm{1}\right)^{\mathrm{n}β\mathrm{1}} }{\mathrm{n}}β\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(β\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{log2}}{\mathrm{2}}β\left(\frac{\pi}{\mathrm{4}}β\mathrm{1}\right)\:\Rightarrow\mathrm{f}\left(\mathrm{1}\right)β\mathrm{f}\left(\mathrm{0}\right)=β\frac{\mathrm{log2}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{4}}β\mathrm{1}\right) \\ $$$$=β\frac{\mathrm{log2}}{\mathrm{4}}+\frac{\pi}{\mathrm{8}}β\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{f}\left(\mathrm{1}\right)=\chi\:=β\frac{\mathrm{log2}}{\mathrm{4}}+\frac{\pi}{\mathrm{8}}β\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{f}\left(\mathrm{0}\right) \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\sqrt{\mathrm{1}β\mathrm{x}^{\mathrm{2}} }\right)}{\mathrm{x}}\mathrm{dx}\:\mathrm{rest}\:\mathrm{to}\:\mathrm{find}\:\mathrm{f}\left(\mathrm{0}\right)….\mathrm{be}\:\mathrm{continued}… \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 03/Apr/21
$${thanks}\:{alot}\:{mr}\:{max}…. \\ $$