Question Number 111719 by mnjuly1970 last updated on 04/Sep/20
$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:….{advanced}\:\:{mathematics}….\: \\ $$$$ \\ $$$${please}\:\:{demonstrate}\:{that}:: \\ $$$$\: \\ $$$$\Phi\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xlog}\left(\mathrm{1}−{x}\right).{log}\left(\mathrm{1}+{x}\right)=\:\frac{\mathrm{1}}{\mathrm{4}}\:−\:{log}\left(\mathrm{2}\right)\:\:… \\ $$$$ \\ $$$$\:\:\:\:\:\:{m}.{n}.{july}\:\mathrm{1970}\:# \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 05/Sep/20
![let take a try let A =∫_0 ^1 xln(1−x)ln(1+x)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)) +c(c=0) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^n ⇒ A =∫_0 ^1 xln(1−x)Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^n dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) ∫_0 ^1 x^(n+1) ln(1−x)dx let U_n =∫_0 ^1 x^(n+1) ln(1−x)dx by parts U_n =[((x^(n+2) −1)/(n+2)) ln(1−x)]_0 ^1 −∫_0 ^1 ((x^(n+2) −1)/(n+2))(((−1)/(1−x)))dx =0 +(1/(n+2)) ∫_0 ^1 ((x^(n+2) −1)/(1−x)) dx =−(1/(n+2)) ∫_0 ^1 ((x^(n+2) −1)/(x−1)) dx =−(1/(n+2)) ∫_0 ^1 (((x−1)(1+x+x^2 +...+x^(n+1) )/(x−1)) dx =−(1/(n+2)) ∫ (1+x+x^2 +...+x^(n+1) )dx =−(1/(n+2))[x+(x^2 /2) +(x^3 /3)+....+(x^(n+2) /(n+2))]_0 ^1 =−(1/(n+2))(1+(1/2)+(1/3)+....+(1/(n+2))) =−(H_(n+2) /(n+2)) (H_n =Σ_(k=1) ^n (1/k)) ⇒ A =−Σ_(n=1) ^∞ (((−1)^(n−1) )/n).(H_(n+2) /(n+2)) =Σ_(n=1) ^∞ (−1)^n ((1/n)−(1/(n+2)))H_(n+2) =Σ_(n=1) ^∞ (((−1)^(n ) H_(n+2) )/n) −Σ_(n=1) ^∞ (((−1)^n H_(n+2) )/(n+2)) Σ_(n=1) ^∞ (((−1)^n H_(n+2) )/n) =_(n+2=p) Σ_(p=3) ^∞ (((−1)^(p−2) H_p )/(p−2)) =Σ_(n=3) ^∞ (((−1)^n H_n )/(n−2)) Σ_(n=1) ^∞ (((−1)^n H_(n+2) )/(n+2)) =Σ_(n=3) ^∞ (((−1)^n H_n )/n) ⇒ A =Σ_(n=3) ^∞ (((−1)^n )/(n−2)) H_n −Σ_(n=3) ^∞ (((−1)^n )/n) H_n rest to find vslues of those series ....be continued...](https://www.tinkutara.com/question/Q111803.png)
$$\mathrm{let}\:\mathrm{take}\:\mathrm{a}\:\mathrm{try}\:\mathrm{let}\:\:\mathrm{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{xln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{dx}\:\mathrm{we}\:\mathrm{have} \\ $$$$\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty\:} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{n}} \:\Rightarrow\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\:+\mathrm{c}\left(\mathrm{c}=\mathrm{0}\right) \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\:\mathrm{x}^{\mathrm{n}} \:\Rightarrow\:\mathrm{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{xln}\left(\mathrm{1}−\mathrm{x}\right)\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\:\mathrm{x}^{\mathrm{n}} \:\mathrm{dx} \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{n}+\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx}\:\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{n}+\mathrm{1}} \:\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx} \\ $$$$\mathrm{by}\:\mathrm{parts}\:\mathrm{U}_{\mathrm{n}} =\left[\frac{\mathrm{x}^{\mathrm{n}+\mathrm{2}} −\mathrm{1}}{\mathrm{n}+\mathrm{2}}\:\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{n}+\mathrm{2}} −\mathrm{1}}{\mathrm{n}+\mathrm{2}}\left(\frac{−\mathrm{1}}{\mathrm{1}−\mathrm{x}}\right)\mathrm{dx} \\ $$$$=\mathrm{0}\:+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{n}+\mathrm{2}} −\mathrm{1}}{\mathrm{1}−\mathrm{x}}\:\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{n}+\mathrm{2}} −\mathrm{1}}{\mathrm{x}−\mathrm{1}}\:\mathrm{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} +…+\mathrm{x}^{\mathrm{n}+\mathrm{1}} \right.}{\mathrm{x}−\mathrm{1}}\:\mathrm{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\:\int\:\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \:+…+\mathrm{x}^{\mathrm{n}+\mathrm{1}} \right)\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\left[\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+….+\frac{\mathrm{x}^{\mathrm{n}+\mathrm{2}} }{\mathrm{n}+\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+….+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\right)\:=−\frac{\mathrm{H}_{\mathrm{n}+\mathrm{2}} }{\mathrm{n}+\mathrm{2}}\:\:\left(\mathrm{H}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}}\right)\:\Rightarrow \\ $$$$\mathrm{A}\:=−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}.\frac{\mathrm{H}_{\mathrm{n}+\mathrm{2}} }{\mathrm{n}+\mathrm{2}}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \left(\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\right)\mathrm{H}_{\mathrm{n}+\mathrm{2}} \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}\:} \mathrm{H}_{\mathrm{n}+\mathrm{2}} }{\mathrm{n}}\:−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{H}_{\mathrm{n}+\mathrm{2}} }{\mathrm{n}+\mathrm{2}} \\ $$$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{H}_{\mathrm{n}+\mathrm{2}} }{\mathrm{n}}\:=_{\mathrm{n}+\mathrm{2}=\mathrm{p}} \:\:\sum_{\mathrm{p}=\mathrm{3}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{p}−\mathrm{2}} \:\mathrm{H}_{\mathrm{p}} }{\mathrm{p}−\mathrm{2}}\:=\sum_{\mathrm{n}=\mathrm{3}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{H}_{\mathrm{n}} }{\mathrm{n}−\mathrm{2}} \\ $$$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{H}_{\mathrm{n}+\mathrm{2}} }{\mathrm{n}+\mathrm{2}}\:=\sum_{\mathrm{n}=\mathrm{3}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{H}_{\mathrm{n}} }{\mathrm{n}}\:\Rightarrow \\ $$$$\mathrm{A}\:=\sum_{\mathrm{n}=\mathrm{3}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}−\mathrm{2}}\:\mathrm{H}_{\mathrm{n}} −\sum_{\mathrm{n}=\mathrm{3}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}}\:\mathrm{H}_{\mathrm{n}} \\ $$$$\mathrm{rest}\:\mathrm{to}\:\mathrm{find}\:\mathrm{vslues}\:\mathrm{of}\:\mathrm{those}\:\mathrm{series}\:….\mathrm{be}\:\mathrm{continued}… \\ $$
Answered by maths mind last updated on 07/Sep/20
![log(1−x)log(1+x)=(1/4)[(log(1−x)+log(1+x))^2 −(log(1−x)−log(1+x))^2 ] =((log^2 (1−x^2 ))/4)−((log^2 (((1−x)/(1+x))))/4) Φ=(1/4)∫_0 ^1 xlog^2 (1−x^2 )dx−(1/4)∫_0 ^1 xlog^2 (((1−x)/(1+x)))dx Φ=(1/4)(I−J) I=∫_0 ^1 xlog^2 (1−x^2 )dx let u=x^2 ⇒xdx=(du/2) I=(1/2)∫_0 ^1 log^2 (1−u)du β(x,y)=∫_0 ^1 t^(x−1) (1−t)^(y−1) ⇒∂^2 yβ(1,1)=2I J=∫_0 ^1 xlog^2 (((1−x)/(1+x)))dx u=((1−x)/(1+x))⇒x=((1−u)/(1+u)) dx=−(2/((1+u)^2 ))du =2∫_0 ^1 (((1−u))/((1+u)^3 )).log^2 (u)du =2∫_0 ^1 (−(1/((1+u)^2 ))+(2/((1+u)^3 )))log^2 (u)du =lim_(y→0) 2[_y ^1 ((1/(1+u))−(1/((1+u)^2 ))).]log^2 (u)_(=0) −4∫_0 ^1 [((log(u))/(u(1+u)))−((log(u))/((1+u)^2 u))]du =−4∫_0 ^1 ((log(u))/((1+u)^2 ))du=4lim_(x→0) ([((log(u))/(1+u))]_x ^1 −∫_x ^1 (du/(u(1+u)))) =4lim_(x→0) [−((log(x))/(1+x))+log(x)+log(2)−log(1+x)] =4log(2) ∂_y β(x,y)=β(x,y){Ψ(y)−Ψ(x+y)} ∂^2 yβ(x,y)=β(x,y)[{Ψ(y)−Ψ(x+y)}^2 +(Ψ′(y)−Ψ′(x+y))] ={Ψ(1)−Ψ(2)}^2 +(Ψ′(1)−Ψ′(2))=(−γ−(−γ−1))+(ζ(2)−(ζ(2)−1))^2 =2 I=(1/2)∂^2 yβ(1,1)=1 Φ=(1/4)(I−J)=(1/4)(1−4log(2))=(1/4)−log(2)](https://www.tinkutara.com/question/Q112270.png)
$${log}\left(\mathrm{1}−{x}\right){log}\left(\mathrm{1}+{x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left[\left({log}\left(\mathrm{1}−{x}\right)+\mathrm{log}\left(\mathrm{1}+\mathrm{x}\right)\right)^{\mathrm{2}} −\left(\mathrm{log}\left(\mathrm{1}−\mathrm{x}\right)−\mathrm{log}\left(\mathrm{1}+\mathrm{x}\right)\right)^{\mathrm{2}} \right] \\ $$$$=\frac{\mathrm{log}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{4}}−\frac{\mathrm{log}^{\mathrm{2}} \left(\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}\right)}{\mathrm{4}} \\ $$$$\Phi=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} {xlog}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx}−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} {xlog}^{\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right){dx} \\ $$$$\Phi=\frac{\mathrm{1}}{\mathrm{4}}\left({I}−{J}\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {xlog}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx}\:{let}\:{u}={x}^{\mathrm{2}} \Rightarrow{xdx}=\frac{{du}}{\mathrm{2}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left(\mathrm{1}−{u}\right){du} \\ $$$$\beta\left({x},{y}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{x}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{y}−\mathrm{1}} \Rightarrow\partial^{\mathrm{2}} {y}\beta\left(\mathrm{1},\mathrm{1}\right)=\mathrm{2}{I} \\ $$$${J}=\int_{\mathrm{0}} ^{\mathrm{1}} {xlog}^{\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right){dx} \\ $$$${u}=\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\Rightarrow{x}=\frac{\mathrm{1}−{u}}{\mathrm{1}+{u}} \\ $$$${dx}=−\frac{\mathrm{2}}{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }{du} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{u}\right)}{\left(\mathrm{1}+{u}\right)^{\mathrm{3}} }.{log}^{\mathrm{2}} \left({u}\right){du} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\frac{\mathrm{1}}{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }+\frac{\mathrm{2}}{\left(\mathrm{1}+{u}\right)^{\mathrm{3}} }\right){log}^{\mathrm{2}} \left({u}\right){du} \\ $$$$=\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}2}\left[_{{y}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{u}}−\frac{\mathrm{1}}{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }\right).\right]{log}^{\mathrm{2}} \left({u}\underset{=\mathrm{0}} {\right)} \\ $$$$−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \left[\frac{{log}\left({u}\right)}{{u}\left(\mathrm{1}+{u}\right)}−\frac{{log}\left({u}\right)}{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} {u}}\right]{du} \\ $$$$=−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left({u}\right)}{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }{du}=\mathrm{4}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\left[\frac{{log}\left({u}\right)}{\mathrm{1}+{u}}\right]_{{x}} ^{\mathrm{1}} −\int_{{x}} ^{\mathrm{1}} \frac{{du}}{{u}\left(\mathrm{1}+{u}\right)}\right) \\ $$$$=\mathrm{4}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[−\frac{{log}\left({x}\right)}{\mathrm{1}+{x}}+{log}\left({x}\right)+{log}\left(\mathrm{2}\right)−{log}\left(\mathrm{1}+{x}\right)\right] \\ $$$$=\mathrm{4}{log}\left(\mathrm{2}\right) \\ $$$$\partial_{{y}} \beta\left({x},{y}\right)=\beta\left({x},{y}\right)\left\{\Psi\left({y}\right)−\Psi\left({x}+{y}\right)\right\} \\ $$$$\partial^{\mathrm{2}} {y}\beta\left({x},{y}\right)=\beta\left({x},{y}\right)\left[\left\{\Psi\left({y}\right)−\Psi\left({x}+{y}\right)\right\}^{\mathrm{2}} +\left(\Psi'\left({y}\right)−\Psi'\left({x}+{y}\right)\right)\right] \\ $$$$=\left\{\Psi\left(\mathrm{1}\right)−\Psi\left(\mathrm{2}\right)\right\}^{\mathrm{2}} +\left(\Psi'\left(\mathrm{1}\right)−\Psi'\left(\mathrm{2}\right)\right)=\left(−\gamma−\left(−\gamma−\mathrm{1}\right)\right)+\left(\zeta\left(\mathrm{2}\right)−\left(\zeta\left(\mathrm{2}\right)−\mathrm{1}\right)\right)^{\mathrm{2}} =\mathrm{2} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\partial^{\mathrm{2}} {y}\beta\left(\mathrm{1},\mathrm{1}\right)=\mathrm{1} \\ $$$$\Phi=\frac{\mathrm{1}}{\mathrm{4}}\left({I}−{J}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\mathrm{4}{log}\left(\mathrm{2}\right)\right)=\frac{\mathrm{1}}{\mathrm{4}}−{log}\left(\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$