Advanced-Calculus-Q-Find-the-value-of-determinant-i-0-1-Ln-2-x-dx-ii-

Question Number 145602 by mnjuly1970 last updated on 06/Jul/21

.....Advanced .........Calculus..... Q:: Find the value of :: determinant ((( i :: 𝛗 := ∫_0 ^( 1) Ln ( Γ ( 2 + x ) )dx = ? )),(( ii :: Ω := Σ_(n=1) ^∞ (( 1)/( n ( 2n + 3 ))) = ?))) .....m.n.july.1970..... ■

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..\mathrm{Advanced}\:………\mathrm{Calculus}….. \\ $$$$\:\:\:\mathrm{Q}::\:\:\:\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\::: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{array}{|c|c|}{\:{i}\:::\:\:\:\boldsymbol{\phi}\::=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{Ln}\:\left(\:\Gamma\:\left(\:\mathrm{2}\:+\:{x}\:\right)\:\right){dx}\:=\:?\:\:\:\:}\\{\:{ii}\:::\:\:\:\Omega\::=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\mathrm{1}}{\:{n}\:\left(\:\mathrm{2}{n}\:+\:\mathrm{3}\:\right)}\:=\:?}\\\hline\end{array} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..{m}.{n}.{july}.\mathrm{1970}…..\:\:\:\:\blacksquare \\ $$$$ \\ $$

Answered by ajfour last updated on 06/Jul/21

(ii) S=Σ_(n=1) ^∞ (1/(n(2n+3))) ((3S)/2)=Σ_(n=1) ^∞ ((2n+3−2n)/(2n(2n+3))) =Σ_(n=1) ^∞ ((1/(2n))−(1/(2n+3))) ln (1−x)=−(x+(x^2 /2)+(x^3 /3)+..) Σ_(n=1) ^∞ (1/(2n))=−(1/2)lim_(x→1) ln (1−x) ln (1+x)−ln (1−x) =2(x+(x^3 /3)+(x^5 /5)+...) lim_(x→1) [(1/2){ln (1+x)−ln (1−x)} −x−(x^3 /3)] =Σ_(n=1) ^∞ (1/(2n+3)) ((3S)/2)=−(1/2)lim ln (1−x)− lim_(x→1) [(1/2){ln (1+x)−ln (1−x)} −x−(x^3 /3)] ((3S)/2)= (4/3)−((ln 2)/2) S=(8/9)−((ln 2)/3)

$$\left({ii}\right) \\ $$$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$$\:\:\frac{\mathrm{3}{S}}{\mathrm{2}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}+\mathrm{3}−\mathrm{2}{n}}{\mathrm{2}{n}\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$$\:\:\:\:\:\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}\right) \\ $$$$\mathrm{ln}\:\left(\mathrm{1}−{x}\right)=−\left({x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+..\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}}=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\mathrm{ln}\:\left(\mathrm{1}−{x}\right) \\ $$$$\mathrm{ln}\:\left(\mathrm{1}+{x}\right)−\mathrm{ln}\:\left(\mathrm{1}−{x}\right) \\ $$$$\:\:\:=\mathrm{2}\left({x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}}+…\right) \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)−\mathrm{ln}\:\left(\mathrm{1}−{x}\right)\right\}\right. \\ $$$$\left.\:\:−{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right]\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}} \\ $$$$\frac{\mathrm{3}{S}}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}\:\mathrm{ln}\:\left(\mathrm{1}−{x}\right)− \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)−\mathrm{ln}\:\left(\mathrm{1}−{x}\right)\right\}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right] \\ $$$$\:\:\frac{\mathrm{3}{S}}{\mathrm{2}}=\:\frac{\mathrm{4}}{\mathrm{3}}−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}} \\ $$$$\:\:{S}=\frac{\mathrm{8}}{\mathrm{9}}−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\: \\ $$

Commented by mnjuly1970 last updated on 06/Jul/21

$${bravo}\:{mr}\:{ajfor}\:{your}\:{work}\: \\ $$$$\:{is}\:{admirable}… \\ $$

Answered by Olaf_Thorendsen last updated on 06/Jul/21

Ω = Σ_(n=1) ^∞ (1/(n(2n+3)))= (1/2)Σ_(n=0) ^∞ (1/((n+1)(n+(5/2)))) Ω = (1/2).((ψ((5/2))−ψ(1))/((5/2)−1)) = (1/3)(ψ((5/2))+γ) ψ((5/2)) = ψ(1+(3/2)) = ψ((3/2))+(1/(3/2)) = ψ((3/2))+(2/3) ψ((3/2)) = ψ(1+(1/2)) = ψ((1/2))+(1/(1/2)) = −2ln(2)−γ+2 Ω = (1/3)(−2ln(2)−γ+2+(2/3)+γ) = (2/3)((4/3)−ln(2))

$$\Omega\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left(\mathrm{2}{n}+\mathrm{3}\right)}=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\frac{\mathrm{5}}{\mathrm{2}}\right)} \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\frac{\psi\left(\frac{\mathrm{5}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}\right)}{\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\psi\left(\frac{\mathrm{5}}{\mathrm{2}}\right)+\gamma\right) \\ $$$$\psi\left(\frac{\mathrm{5}}{\mathrm{2}}\right)\:=\:\psi\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\:\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{3}/\mathrm{2}}\:=\:\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)+\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\:\psi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{1}/\mathrm{2}}\:=\:−\mathrm{2ln}\left(\mathrm{2}\right)−\gamma+\mathrm{2} \\ $$$$\: \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(−\mathrm{2ln}\left(\mathrm{2}\right)−\gamma+\mathrm{2}+\frac{\mathrm{2}}{\mathrm{3}}+\gamma\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{ln}\left(\mathrm{2}\right)\right) \\ $$

Commented by mnjuly1970 last updated on 06/Jul/21

$${grateful}\:{mr}\:{olaf}\: \\ $$$${very}\:{nice}\:{as}\:{always}..{mercey} \\ $$

Answered by qaz last updated on 06/Jul/21

$$\Gamma\left(\mathrm{x}\right)\Gamma\left(\mathrm{1}−\mathrm{x}\right)=\frac{\pi}{\mathrm{sin}\:\pi\mathrm{x}} \\ $$

Answered by Kamel last updated on 06/Jul/21

i) use t=1−x and: Γ(t)Γ(1−t)=(π/(sin(πt))) ii.. Ω=Σ_(n=1) ^(+∞) (1/(n(2n+3)))=(2/3)Σ_(n=1) ^(+∞) ((1/(2n))−(1/(2n+3)))=(2/3)(Σ_(n=1) ^(+∞) ((1/(2n))−(1/(2n+1)))+Σ_(n=1) ^(+∞) ((1/(2n+1))−(1/(2(n+1)+1)))) =(2/3)(Σ_(n=1) ^(+∞) (((−1)^n )/n)+1+(1/3))=(2/3)((4/3)−Ln(2))=(8/9)−((2Ln(2))/3)

$$\left.{i}\right)\:{use}\:{t}=\mathrm{1}−{x}\:{and}:\:\Gamma\left({t}\right)\Gamma\left(\mathrm{1}−{t}\right)=\frac{\pi}{{sin}\left(\pi{t}\right)} \\ $$$${ii}..\:\Omega=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}\left(\mathrm{2}{n}+\mathrm{3}\right)}=\frac{\mathrm{2}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}\right)=\frac{\mathrm{2}}{\mathrm{3}}\left(\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)+\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)+\mathrm{1}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\left(\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{4}}{\mathrm{3}}−{Ln}\left(\mathrm{2}\right)\right)=\frac{\mathrm{8}}{\mathrm{9}}−\frac{\mathrm{2}{Ln}\left(\mathrm{2}\right)}{\mathrm{3}} \\ $$

Commented by mnjuly1970 last updated on 06/Jul/21

$${tashakor}\:{mr}\:{kamel} \\ $$

Answered by ArielVyny last updated on 06/Jul/21

∅=∫_0 ^1 ln(Γ(2+x))dx =∫_0 ^1 ln((x+1)!)dx x+1=t→dt=dx ∅=∫_1 ^2 ln(t!)dt=∫_1 ^2 ln[((t/e))^t (√(2πt))] ∫_1 ^2 ln((t/e))^t +∫_1 ^2 ln((√(2πt)))dt ∫_1 ^2 ln(e^(tln((t/e))) )+∫_1 ^2 ln((√(2π)))dt+(1/2)∫_1 ^2 ln(t)dt ∫_1 ^2 tln((t/e))+ln((√(2π)))+(1/2)[xln(x)−x]_1 ^2 ∫_1 ^2 tln(t)−∫_1 ^2 tdt+ln((√(2π)))+(1/2)(2ln(2)+1) ∫_1 ^2 tln(t)dt−(3/2)+(1/2)ln(2π)+((2ln2)/2)+(1/2) ∫_1 ^2 tln(t)dt du=ln(t) v=t u=tln(t)−t dv=1 ∫_1 ^2 tln(t)=[t^2 ln(t)−t]−∫tlntdt+∫tdt 2∫_1 ^2 tln(t)dt=[t^2 ln(t)−t+(1/2)t^2 ]_1 ^2 =4ln(2)−2+(1/2)4+1−(1/2) ∫_1 ^2 tln(t)dt=2ln(2)−1+1+(1/2)−(1/4) =2ln(2)+(1/4) ∅=2ln(2)+(1/4)−(3/2)+((ln(2π))/2)+((2ln(2))/2)+(1/2) ∅=2ln(2)+(1/4)−1+((ln(8π))/2) ∅=−(3/4)+((2ln(4))/2)+((ln(8π))/2) ∅=−(3/4)+(1/2)ln(8^3 π)

$$\emptyset=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left(\mathrm{2}+{x}\right)\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\left({x}+\mathrm{1}\right)!\right){dx} \\ $$$${x}+\mathrm{1}={t}\rightarrow{dt}={dx} \\ $$$$\emptyset=\int_{\mathrm{1}} ^{\mathrm{2}} {ln}\left({t}!\right){dt}=\int_{\mathrm{1}} ^{\mathrm{2}} {ln}\left[\left(\frac{{t}}{{e}}\right)^{{t}} \sqrt{\mathrm{2}\pi{t}}\right] \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} {ln}\left(\frac{{t}}{{e}}\right)^{{t}} +\int_{\mathrm{1}} ^{\mathrm{2}} {ln}\left(\sqrt{\mathrm{2}\pi{t}}\right){dt} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} {ln}\left({e}^{{tln}\left(\frac{{t}}{{e}}\right)} \right)+\int_{\mathrm{1}} ^{\mathrm{2}} {ln}\left(\sqrt{\mathrm{2}\pi}\right){dt}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{2}} {ln}\left({t}\right){dt} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} {tln}\left(\frac{{t}}{{e}}\right)+{ln}\left(\sqrt{\mathrm{2}\pi}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left[{xln}\left({x}\right)−{x}\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} {tln}\left({t}\right)−\int_{\mathrm{1}} ^{\mathrm{2}} {tdt}+{ln}\left(\sqrt{\mathrm{2}\pi}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{ln}\left(\mathrm{2}\right)+\mathrm{1}\right) \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} {tln}\left({t}\right){dt}−\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\pi\right)+\frac{\mathrm{2}{ln}\mathrm{2}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} {tln}\left({t}\right){dt} \\ $$$${du}={ln}\left({t}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:{v}={t} \\ $$$${u}={tln}\left({t}\right)−{t}\:\:\:\:\:\:\:\:\:{dv}=\mathrm{1} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} {tln}\left({t}\right)=\left[{t}^{\mathrm{2}} {ln}\left({t}\right)−{t}\right]−\int{tlntdt}+\int{tdt} \\ $$$$\mathrm{2}\int_{\mathrm{1}} ^{\mathrm{2}} {tln}\left({t}\right){dt}=\left[{t}^{\mathrm{2}} {ln}\left({t}\right)−{t}+\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} \right]_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{4}{ln}\left(\mathrm{2}\right)−\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{4}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} {tln}\left({t}\right){dt}=\mathrm{2}{ln}\left(\mathrm{2}\right)−\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\emptyset=\mathrm{2}{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{2}}+\frac{{ln}\left(\mathrm{2}\pi\right)}{\mathrm{2}}+\frac{\mathrm{2}{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\emptyset=\mathrm{2}{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}+\frac{{ln}\left(\mathrm{8}\pi\right)}{\mathrm{2}} \\ $$$$\emptyset=−\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{2}{ln}\left(\mathrm{4}\right)}{\mathrm{2}}+\frac{{ln}\left(\mathrm{8}\pi\right)}{\mathrm{2}} \\ $$$$\emptyset=−\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{8}^{\mathrm{3}} \pi\right) \\ $$

Commented by mathmax by abdo last updated on 06/Jul/21

t!∼((t/e))^t (√(2πt))is valable pour t assez grand(t→∞)

$$\mathrm{t}!\sim\left(\frac{\mathrm{t}}{\mathrm{e}}\right)^{\mathrm{t}} \sqrt{\mathrm{2}\pi\mathrm{t}}\mathrm{is}\:\mathrm{valable}\:\mathrm{pour}\:\mathrm{t}\:\mathrm{assez}\:\mathrm{grand}\left(\mathrm{t}\rightarrow\infty\right) \\ $$

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