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Question-90158

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Question Number 90158 by I want to learn more last updated on 21/Apr/20
Commented by I want to learn more last updated on 21/Apr/20
The question is prove that
$$\mathrm{The}\:\mathrm{question}\:\mathrm{is}\:\mathrm{prove}\:\mathrm{that} \\ $$
Commented by abdomathmax last updated on 22/Apr/20
we have Γ(x).Γ(1−x)=(π/(sin(πx))) ⇒ ∫_0 ^1 lnΓ(x)dx +∫_0 ^1 lnΓ(1−x)dx =∫^1 _0 ln((π/(sin(πx))))dx but ∫_0 ^1 lnΓ(1−x)dx =_(1−x=t) dt∫_0 ^1 lnΓ(t)dt ⇒ 2∫_0 ^1 ln(Γ(x))dx =ln(π)−∫_0 ^1 ln(sin(πx))dx ∫_0 ^1 ln(sin(πx))dx =_(πx=t) (1/π)∫_0 ^π ln(sint) dt =(1/π) { ∫_0 ^(π/2) ln(sint)dt +∫_(π/2) ^π ln(sint)dt→(t=(π/2)+u)} =(1/π){ ∫_0 ^(π/2) ln(sint)dt +∫_0 ^(π/2) ln(cosu)du} =(1/π){−(π/2)ln(2)−(π/2)ln(2) =(1/π)(−πln(2)) −ln(2) ⇒2∫_0 ^1 ln(Γ(x)dx =ln(π)+ln(2)=ln(2π) ⇒ ∫_0 ^1 ln(Γ(x))dx =(1/2)ln(2π)=ln((√(2π)))
$${we}\:{have}\:\Gamma\left({x}\right).\Gamma\left(\mathrm{1}−{x}\right)=\frac{\pi}{{sin}\left(\pi{x}\right)}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\Gamma\left({x}\right){dx}\:+\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\Gamma\left(\mathrm{1}−{x}\right){dx}\:=\underset{\mathrm{0}} {\int}^{\mathrm{1}} {ln}\left(\frac{\pi}{{sin}\left(\pi{x}\right)}\right){dx} \\ $$$${but}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\Gamma\left(\mathrm{1}−{x}\right){dx}\:=_{\mathrm{1}−{x}={t}} \:\:{dt}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\Gamma\left({t}\right){dt}\:\Rightarrow \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx}\:={ln}\left(\pi\right)−\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({sin}\left(\pi{x}\right)\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({sin}\left(\pi{x}\right)\right){dx}\:=_{\pi{x}={t}} \:\:\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} {ln}\left({sint}\right)\:{dt} \\ $$$$=\frac{\mathrm{1}}{\pi}\:\left\{\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sint}\right){dt}\:+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:{ln}\left({sint}\right){dt}\rightarrow\left({t}=\frac{\pi}{\mathrm{2}}+{u}\right)\right\}\: \\ $$$$=\frac{\mathrm{1}}{\pi}\left\{\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sint}\right){dt}\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cosu}\right){du}\right\} \\ $$$$=\frac{\mathrm{1}}{\pi}\left\{−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:=\frac{\mathrm{1}}{\pi}\left(−\pi{ln}\left(\mathrm{2}\right)\right)\right. \\ $$$$−{ln}\left(\mathrm{2}\right)\:\Rightarrow\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left({x}\right){dx}\:={ln}\left(\pi\right)+{ln}\left(\mathrm{2}\right)={ln}\left(\mathrm{2}\pi\right)\:\Rightarrow\right. \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\pi\right)={ln}\left(\sqrt{\mathrm{2}\pi}\right) \\ $$$$ \\ $$
Answered by maths mind last updated on 21/Apr/20
∫_a ^b f(x)dx=∫_a ^b f(a+b−x)dx ∫_0 ^1 ln(Γ(x))dx=∫_0 ^1 ln(Γ(1−x))dx ⇒∫_0 ^1 ln(Γ(x))dx=(1/2)∫_0 ^1 ln(Γ(1−x)Γ(x))dx =(1/2)∫_0 ^1 ln((π/(sin(πx))))dx=ln((√π))−(1/2)∫_0 ^1 ln(sin(πx))dx..E ∫_0 ^1 ln(sin(πx))dx =(1/π)∫_0 ^π ln(sin(r))dr ∫_0 ^π ln(sin(r))dr=∫_0 ^(π/2) ln(((sin(2r))/2))dr =∫_0 ^π ln(sin(x))(dx/2)−(π/2)ln(2) ⇒∫_0 ^π ln(sin(x))dx=−πln(2) E=ln((√π))−(1/(2π)).−πln(2)=ln((√π))+ln((√2)) =ln((√(2π)))=∫_0 ^1 ln(Γ(x))dx
$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({a}+{b}−{x}\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left(\mathrm{1}−{x}\right)\right){dx} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left(\mathrm{1}−{x}\right)\Gamma\left({x}\right)\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\frac{\pi}{{sin}\left(\pi{x}\right)}\right){dx}={ln}\left(\sqrt{\pi}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({sin}\left(\pi{x}\right)\right){dx}..{E} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({sin}\left(\pi{x}\right)\right){dx} \\ $$$$=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} {ln}\left({sin}\left({r}\right)\right){dr} \\ $$$$\int_{\mathrm{0}} ^{\pi} {ln}\left({sin}\left({r}\right)\right){dr}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{{sin}\left(\mathrm{2}{r}\right)}{\mathrm{2}}\right){dr} \\ $$$$=\int_{\mathrm{0}} ^{\pi} {ln}\left({sin}\left({x}\right)\right)\frac{{dx}}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\pi} {ln}\left({sin}\left({x}\right)\right){dx}=−\pi{ln}\left(\mathrm{2}\right) \\ $$$${E}={ln}\left(\sqrt{\pi}\right)−\frac{\mathrm{1}}{\mathrm{2}\pi}.−\pi{ln}\left(\mathrm{2}\right)={ln}\left(\sqrt{\pi}\right)+{ln}\left(\sqrt{\mathrm{2}}\right) \\ $$$$={ln}\left(\sqrt{\mathrm{2}\pi}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by I want to learn more last updated on 21/Apr/20
Thanks sir
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

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