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Question Number 137093 by mnjuly1970 last updated on 29/Mar/21

.....advanced calculus.... please evaluate:: 1: Σ_(n=1 ) ^∞ (H_n /((2n+1)^2 )) =? 2: ∫_0 ^( 1) ((xln(x)ln(1−x))/(1−x))dx=? note:H_n =1+(1/2)+(1/3)+...+(1/n)=∫_0 ^( 1) ((1−x^n )/(1−x))dx

$$\:\:\:\:\:\:\:\:\:.....{advanced}\:\:\:\:{calculus}.... \\ $$$$\:\:\:\:\:{please}\:\:{evaluate}::\:\: \\ $$$$\:\:\:\:\:\mathrm{1}:\:\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=? \\ $$$$\:\:\:\:\mathrm{2}:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{xln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}}{dx}=? \\ $$$$\:\:{note}:{H}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+...+\frac{\mathrm{1}}{{n}}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}}{dx} \\ $$

Answered by Ar Brandon last updated on 31/Mar/21

Λ=∫_0 ^1 ((xln(x)ln(1−x))/(1−x))dx ={xln(x)∙((ln^2 (1−x))/(−2))+(1/2)∫(1+lnx)ln^2 (1−x)dx}_0 ^1 =(1/2)∫_0 ^1 (1+lnx)ln^2 (1−x)dx=(1/2)∫_0 ^1 (1+ln(1−x))ln^2 xdx =(1/2){∫_0 ^1 ln^2 xdx+∫_0 ^1 ln(1−x)ln^2 xdx} =(1/2){(∂^2 /∂α^2 )∣_(α=0) ∫_0 ^1 x^α dx−(∂^2 /∂ρ^2 )∣_(ρ=0) ∫_0 ^1 Σ_(n=1) ^∞ (x^(n+ρ) /n)} =(1/2){(∂^2 /∂α^2 )∣_(α=0) (1/(α+1))−(∂^2 /∂ρ^2 )∣_(ρ=0) Σ_(n=1) ^∞ (1/(n(n+ρ+1)))} =(1/2){2−Σ_(n=1) ^∞ (2/(n(n+1)^3 ))}=1−Σ_(n=1) ^∞ (1/(n(n+1)^3 )) =1−Σ_(n=1) ^∞ {(1/n)−(1/(n+1))−(1/((n+1)^2 ))−(1/((n+1)^3 ))} =1−1+Σ_(n=1) ^∞ (1/((n+1)^2 ))+Σ_(n=1) ^∞ (1/((n+1)^3 )) =Σ_(n=1) ^∞ (1/n^2 )−1+Σ_(n=1) ^∞ (1/n^3 )−1=ζ(2)+ζ(3)−2=(π^2 /6)+ζ(3)−2

$$\Lambda=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{xln}\left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{1}−\mathrm{x}}\mathrm{dx} \\ $$$$\:\:\:=\left\{\mathrm{xln}\left(\mathrm{x}\right)\centerdot\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)}{−\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+\mathrm{lnx}\right)\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx}\right\}_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\mathrm{lnx}\right)\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\right)\mathrm{ln}^{\mathrm{2}} \mathrm{xdx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}^{\mathrm{2}} \mathrm{xdx}+\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{ln}^{\mathrm{2}} \mathrm{xdx}\right\} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\partial^{\mathrm{2}} }{\partial\alpha^{\mathrm{2}} }\mid_{\alpha=\mathrm{0}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\alpha} \mathrm{dx}−\frac{\partial^{\mathrm{2}} }{\partial\rho^{\mathrm{2}} }\mid_{\rho=\mathrm{0}} \int_{\mathrm{0}} ^{\mathrm{1}} \underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{x}^{\mathrm{n}+\rho} }{\mathrm{n}}\right\} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\partial^{\mathrm{2}} }{\partial\alpha^{\mathrm{2}} }\mid_{\alpha=\mathrm{0}} \frac{\mathrm{1}}{\alpha+\mathrm{1}}−\frac{\partial^{\mathrm{2}} }{\partial\rho^{\mathrm{2}} }\mid_{\rho=\mathrm{0}} \underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\rho+\mathrm{1}\right)}\right\} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{2}−\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} }\right\}=\mathrm{1}−\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\:\:\:=\mathrm{1}−\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left\{\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} }\right\} \\ $$$$\:\:\:=\mathrm{1}−\mathrm{1}+\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }+\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\:\:\:=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }−\mathrm{1}+\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }−\mathrm{1}=\zeta\left(\mathrm{2}\right)+\zeta\left(\mathrm{3}\right)−\mathrm{2}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\zeta\left(\mathrm{3}\right)−\mathrm{2} \\ $$

Answered by Dwaipayan Shikari last updated on 29/Mar/21

ð(a,b)=∫_0 ^1 x^(α−1) (1−x)^(β−1) dx (∂/(∂a∂b))ð(a,b)=∫_0 ^1 log(x)log(1−x)x^(α−1) (1−x)^(β−1) dx ((∂ð(a,b))/(∂a∂b))=((∂(((Γ(a)Γ(b))/(Γ(a+b)))))/(∂a∂b))=(∂/∂a)(((Γ(a)Γ(a+b)Γ′(b)−Γ(b)Γ(a)Γ′(a+b))/(Γ^2 (a+b)))) (∂/∂a)(((Γ(a)Γ′(b))/(Γ(a+b)))−((Γ(b)Γ(a)ψ(a+b))/(Γ(a+b)))) =((Γ(a+b)Γ′(a)Γ′(b)−Γ′(a+b)Γ(a)Γ′(b))/(Γ^2 (a+b)))−((Γ′(b)ψ(a+b)Γ(a)+ψ′(a+b)Γ(a)Γ(b)−ψ^2 (a+b)Γ(a)Γ(b)Γ(b+a))/(Γ^2 (a+b))) a=1 b=0

$$\eth\left({a},{b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\alpha−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{\beta−\mathrm{1}} {dx} \\ $$$$\frac{\partial}{\partial{a}\partial{b}}\eth\left({a},{b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left({x}\right){log}\left(\mathrm{1}−{x}\right){x}^{\alpha−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{\beta−\mathrm{1}} {dx} \\ $$$$\frac{\partial\eth\left({a},{b}\right)}{\partial{a}\partial{b}}=\frac{\partial\left(\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)}\right)}{\partial{a}\partial{b}}=\frac{\partial}{\partial{a}}\left(\frac{\Gamma\left({a}\right)\Gamma\left({a}+{b}\right)\Gamma'\left({b}\right)−\Gamma\left({b}\right)\Gamma\left({a}\right)\Gamma'\left({a}+{b}\right)}{\Gamma^{\mathrm{2}} \left({a}+{b}\right)}\right) \\ $$$$\frac{\partial}{\partial{a}}\left(\frac{\Gamma\left({a}\right)\Gamma'\left({b}\right)}{\Gamma\left({a}+{b}\right)}−\frac{\Gamma\left({b}\right)\Gamma\left({a}\right)\psi\left({a}+{b}\right)}{\Gamma\left({a}+{b}\right)}\right) \\ $$$$=\frac{\Gamma\left({a}+{b}\right)\Gamma'\left({a}\right)\Gamma'\left({b}\right)−\Gamma'\left({a}+{b}\right)\Gamma\left({a}\right)\Gamma'\left({b}\right)}{\Gamma^{\mathrm{2}} \left({a}+{b}\right)}−\frac{\Gamma'\left({b}\right)\psi\left({a}+{b}\right)\Gamma\left({a}\right)+\psi'\left({a}+{b}\right)\Gamma\left({a}\right)\Gamma\left({b}\right)−\psi^{\mathrm{2}} \left({a}+{b}\right)\Gamma\left({a}\right)\Gamma\left({b}\right)\Gamma\left({b}+{a}\right)}{\Gamma^{\mathrm{2}} \left({a}+{b}\right)} \\ $$$${a}=\mathrm{1}\:\:{b}=\mathrm{0} \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 29/Mar/21

Σ_(n=1) ^∞ (H_n /((2n+1)^2 ))=∫_0 ^1 Σ_(n=1) ^∞ (H_n /(2n+1))x^(2n+1) Σ_(n=1) ^∞ H_n x^(2n+α) =−x^α ((log(1−x^2 ))/(1−x^2 )) Σ_(n=1) ^∞ (H_n /(2n+α+1))=−∫_0 ^1 ((x^α log(1−x^2 ))/(1−x^2 ))dx Σ_(n=1) ^∞ (H_n /((2n+α+1)^2 ))=(∂/∂α)∫_0 ^1 ((x^α log(1−x^2 ))/(1−x^2 ))dx =(∂/∂α)∫_0 ^1 ((u^((α−1)/2) log(1−u))/(1−u))du=(∂/∂α).(∂/∂ϑ)∣_(ϑ=0) (((Γ(((α+1)/2))Γ(ϑ))/(Γ((α/2)+(1/2)+ϑ))))∣

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{{n}} {x}^{\mathrm{2}{n}+\alpha} =−{x}^{\alpha} \frac{{log}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{\mathrm{2}{n}+\alpha+\mathrm{1}}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\alpha} {log}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{\left(\mathrm{2}{n}+\alpha+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\partial}{\partial\alpha}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\alpha} {log}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$$=\frac{\partial}{\partial\alpha}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{\frac{\alpha−\mathrm{1}}{\mathrm{2}}} {log}\left(\mathrm{1}−{u}\right)}{\mathrm{1}−{u}}{du}=\frac{\partial}{\partial\alpha}.\frac{\partial}{\partial\vartheta}\mid_{\vartheta=\mathrm{0}} \left(\frac{\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\vartheta\right)}{\Gamma\left(\frac{\alpha}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}+\vartheta\right)}\right)\mid \\ $$

Commented by mnjuly1970 last updated on 29/Mar/21

thanks alot...

$${thanks}\:{alot}... \\ $$

Answered by Ñï= last updated on 29/Mar/21

∫_0 ^1 ((xln(x)ln(1−x))/(1−x))dx=∫_0 ^1 (((1−x)ln(x)ln(1−x))/x)dx =∫_0 ^1 ((ln(x)ln(1−x))/x)dx−∫_0 ^1 ln(x)ln(1−x)dx =∫_0 ^1 ((Li_2 (x))/x)dx+∫_0 ^1 x(((ln(1−x))/x)−((lnx)/(1−x)))dx =Li_3 (1)+∫_0 ^1 ln(1−x)dx−∫_0 ^1 (1−x)((ln(1−x))/x)dx =Li_3 (1)+∫_0 ^1 ln(1−x)dx−∫_0 ^1 ((ln(1−x))/x)−ln(1−x)dx =Li_3 (1)+2∫_0 ^1 lnxdx+Li_2 (1) =Li_3 (1)+Li_2 (1)−2

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{x}\right){ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left({x}\right)}{{x}}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left(\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}−\frac{{lnx}}{\mathrm{1}−{x}}\right){dx} \\ $$$$={Li}_{\mathrm{3}} \left(\mathrm{1}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right){dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}\right)\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx} \\ $$$$={Li}_{\mathrm{3}} \left(\mathrm{1}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right){dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}−{ln}\left(\mathrm{1}−{x}\right){dx} \\ $$$$={Li}_{\mathrm{3}} \left(\mathrm{1}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {lnxdx}+{Li}_{\mathrm{2}} \left(\mathrm{1}\right) \\ $$$$={Li}_{\mathrm{3}} \left(\mathrm{1}\right)+{Li}_{\mathrm{2}} \left(\mathrm{1}\right)−\mathrm{2} \\ $$

Commented by mnjuly1970 last updated on 29/Mar/21

thanks alot ...

$${thanks}\:{alot}\:... \\ $$

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