Question Number 136733 by Ñï= last updated on 25/Mar/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}\frac{\mathrm{1}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\pi^{\mathrm{3}} }{\mathrm{48}}+\frac{\pi}{\mathrm{4}}{ln}^{\mathrm{2}} \mathrm{2} \\ $$
Answered by mindispower last updated on 25/Mar/21
![start by (1/( (√(1−4x))))=Σ_(n≥0) (((2n)),(n) )x^n ⇒(1/( (√(1−4x^2 ))))Σ (((2n)),(n) )x^(2n) ⇒(1/2)arcsin(2x)=Σ (((2n)),(n) ).(x^(2n+1) /(2n+1)) ⇒(1/2)∫_0 ^t ((arcsin(2x))/x)dx=Σ (((2n)),(n) ).(t^(2n+1) /((2n+1)^2 )) we want ∫_0 ^(1/2) (1/x)∫_0 ^x ((arcsin(2t))/t)dtdx=Σ (((2n)),(n) ).(1/(4^n (2n+1)^3 )) by part[ ln(x)∫_0 ^x ((arcsin(2t))/t)dt]_0 ^(1/2) −∫_0 ^(1/2) ((ln(x)arcsin(2x))/x)dx =ln((1/2))∫_0 ^(1/2) ((arcsin(2x))/x)dx−[(1/2)ln^2 (x)arcsin(2x)]_0 ^(1/2) +∫_0 ^(1/2) ((ln^2 (x))/( (√(1−4x^2 ))))dx ∫_0 ^(1/2) ((arcsin(2x))/x)dx=∫_0 ^1 ((arcsin(t))/t)dt=−∫_0 ^1 ((ln(t))/( (√(1−t^2 ))))dt =−∫_0 ^(π/2) ln(sin(t))dt=((πln(2))/2) ∫_0 ^(1/2) ((ln^2 (x))/( (√(1−4x^2 ))))dx=(1/2)∫_0 ^1 ((ln^2 (t)−2ln(2)ln(t)+ln^2 (2))/( (√(1−t^2 )))) =(1/2){∫_0 ^1 ((ln^2 (t))/( (√(1−t^2 ))))−ln(2).((πln(2))/2)+ln^2 (2)(π/2)} =(1/2)A ∫_0 ^1 ((ln^2 (t))/( (√(1−t^2 ))))dt=∫_0 ^(π/2) ln^2 (sin(t))dt=A β(a,b)=2∫_0 ^(π/2) sin^(2a−1) (t)cos^(2b−1) (t)dt A=(1/8)(∂^2 /∂a^2 )β((3/2),(1/2)) ∂_a β=β(a,b)(Ψ(a)−Ψ(a+b)} ∂_a ^2 β=β(a,b){(Ψ(a)−Ψ(a+b))^2 +Ψ′(a)−Ψ′(a+b)} A=(1/8)β((3/2),(1/2)){(Ψ((1/2))−Ψ(2))^2 +Ψ′((1/2))−Ψ′(2)} A=(1/8)(Γ((3/2))Γ((1/2))){(π^2 /2)−(π^2 /6)+1+(−2ln(2)−γ−1+γ)^2 } A=(1/8).(1/2).π{(π^2 /3)+1+(−2ln(2)−1)^2 } S=Σ_(n≥0) (((2n)),(n) ).(1/(4^n (2n+1)^2 ))=−((πln^2 (2))/2)−((πln^2 (2))/4) +(π/(32))(4ln^2 (2)+2+4ln(2)+(π^2 /3))...continued later](https://www.tinkutara.com/question/Q136736.png)
$${start}\:{by}\: \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}}}=\underset{{n}\geqslant\mathrm{0}} {\sum}\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}{x}^{{n}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}\Sigma\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}{x}^{\mathrm{2}{n}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}{arcsin}\left(\mathrm{2}{x}\right)=\Sigma\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}.\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{t}} \frac{{arcsin}\left(\mathrm{2}{x}\right)}{{x}}{dx}=\Sigma\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}.\frac{{t}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${we}\:{want}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{{x}} \frac{{arcsin}\left(\mathrm{2}{t}\right)}{{t}}{dtdx}=\Sigma\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}.\frac{\mathrm{1}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${by}\:{part}\left[\:{ln}\left({x}\right)\int_{\mathrm{0}} ^{{x}} \frac{{arcsin}\left(\mathrm{2}{t}\right)}{{t}}{dt}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{ln}\left({x}\right){arcsin}\left(\mathrm{2}{x}\right)}{{x}}{dx} \\ $$$$={ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{arcsin}\left(\mathrm{2}{x}\right)}{{x}}{dx}−\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left({x}\right){arcsin}\left(\mathrm{2}{x}\right)\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$+\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}{dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{arcsin}\left(\mathrm{2}{x}\right)}{{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arcsin}\left({t}\right)}{{t}}{dt}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$$$=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({t}\right)\right){dt}=\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({t}\right)−\mathrm{2}{ln}\left(\mathrm{2}\right){ln}\left({t}\right)+{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({t}\right)}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}−{ln}\left(\mathrm{2}\right).\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\frac{\pi}{\mathrm{2}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{A} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({t}\right)}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}^{\mathrm{2}} \left({sin}\left({t}\right)\right){dt}={A} \\ $$$$\beta\left({a},{b}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{a}−\mathrm{1}} \left({t}\right){cos}^{\mathrm{2}{b}−\mathrm{1}} \left({t}\right){dt} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{8}}\frac{\partial^{\mathrm{2}} }{\partial{a}^{\mathrm{2}} }\beta\left(\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\partial_{{a}} \beta=\beta\left({a},{b}\right)\left(\Psi\left({a}\right)−\Psi\left({a}+{b}\right)\right\} \\ $$$$\partial_{{a}} ^{\mathrm{2}} \beta=\beta\left({a},{b}\right)\left\{\left(\Psi\left({a}\right)−\Psi\left({a}+{b}\right)\right)^{\mathrm{2}} +\Psi'\left({a}\right)−\Psi'\left({a}+{b}\right)\right\} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{8}}\beta\left(\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)\left\{\left(\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\Psi\left(\mathrm{2}\right)\right)^{\mathrm{2}} +\Psi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\Psi'\left(\mathrm{2}\right)\right\} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{8}}\left(\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\left\{\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\mathrm{1}+\left(−\mathrm{2}{ln}\left(\mathrm{2}\right)−\gamma−\mathrm{1}+\gamma\right)^{\mathrm{2}} \right\} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{8}}.\frac{\mathrm{1}}{\mathrm{2}}.\pi\left\{\frac{\pi^{\mathrm{2}} }{\mathrm{3}}+\mathrm{1}+\left(−\mathrm{2}{ln}\left(\mathrm{2}\right)−\mathrm{1}\right)^{\mathrm{2}} \right\} \\ $$$${S}=\underset{{n}\geqslant\mathrm{0}} {\sum}\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}.\frac{\mathrm{1}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\pi{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\pi{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{4}} \\ $$$$+\frac{\pi}{\mathrm{32}}\left(\mathrm{4}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{2}+\mathrm{4}{ln}\left(\mathrm{2}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\right)…{continued}\:{later}\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Ñï= last updated on 25/Mar/21
$${thank}\:{you}\:{sir}! \\ $$
Commented by mindispower last updated on 26/Mar/21
$${pleasur} \\ $$