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Question Number 207582 by sniper237 last updated on 19/May/24
∫_0 ^π ln(sinx)dx=−πln2 ∫_0 ^1 lnΓ(x)dx = ln(2π)
$$\int_{\mathrm{0}} ^{\pi} \:{ln}\left({sinx}\right){dx}=−\pi{ln}\mathrm{2} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\Gamma\left({x}\right){dx}\:=\:{ln}\left(\mathrm{2}\pi\right) \\ $$
Answered by mathzup last updated on 21/May/24
I=∫_0 ^π ln(sinx)dx=∫_0 ^(π/2) ln(sinx)dx +∫_(π/2) ^π ln(sinx)dx(→x=(π/2)+t) =−(π/2)ln(2)+∫_0 ^(π/2) ln(cost)dt =−(π/2)ln2−(π/2)ln2=−πln2
$${I}=\int_{\mathrm{0}} ^{\pi} {ln}\left({sinx}\right){dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinx}\right){dx} \\ $$$$+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} {ln}\left({sinx}\right){dx}\left(\rightarrow{x}=\frac{\pi}{\mathrm{2}}+{t}\right) \\ $$$$=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cost}\right){dt} \\ $$$$=−\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}−\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}=−\pi{ln}\mathrm{2} \\ $$
Answered by mathzup last updated on 21/May/24
Γ(x).Γ(1−x)=(π/(sin(πx))) ⇒ ln(Γ(x))+ln(Γ(1−x))=lnπ−ln(sin(πx)) ⇒∫_0 ^1 ln(Γ(x))dx+∫_0 ^1 ln(Γ(1−x))dx =ln(π)−∫_0 ^1 ln(sin(πx))dx(→πx=t) =ln(π)−∫_0 ^π ln(sint)(dt/π) =ln(π)−(1/π)(−πln2) =ln(π)+ln(2)=ln(2π) but ∫_0 ^1 ln(Γ(1−x))dx=_(1−x=t) ∫_1 ^0 ln(Γ(t))(−dt) =∫_0 ^1 ln(Γ(x)dx ⇒ ∫_0 ^1 ln(Γ(x))dx=(1/2)ln(2π) il ya erreur dans l enonce !...
$$\Gamma\left({x}\right).\Gamma\left(\mathrm{1}−{x}\right)=\frac{\pi}{{sin}\left(\pi{x}\right)}\:\Rightarrow \\ $$$${ln}\left(\Gamma\left({x}\right)\right)+{ln}\left(\Gamma\left(\mathrm{1}−{x}\right)\right)={ln}\pi−{ln}\left({sin}\left(\pi{x}\right)\right) \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left(\mathrm{1}−{x}\right)\right){dx} \\ $$$$={ln}\left(\pi\right)−\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({sin}\left(\pi{x}\right)\right){dx}\left(\rightarrow\pi{x}={t}\right) \\ $$$$={ln}\left(\pi\right)−\int_{\mathrm{0}} ^{\pi} {ln}\left({sint}\right)\frac{{dt}}{\pi} \\ $$$$={ln}\left(\pi\right)−\frac{\mathrm{1}}{\pi}\left(−\pi{ln}\mathrm{2}\right) \\ $$$$={ln}\left(\pi\right)+{ln}\left(\mathrm{2}\right)={ln}\left(\mathrm{2}\pi\right) \\ $$$${but}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left(\mathrm{1}−{x}\right)\right){dx}=_{\mathrm{1}−{x}={t}} \int_{\mathrm{1}} ^{\mathrm{0}} {ln}\left(\Gamma\left({t}\right)\right)\left(−{dt}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left({x}\right){dx}\:\Rightarrow\right. \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\pi\right) \\ $$$${il}\:{ya}\:{erreur}\:{dans}\:{l}\:{enonce}\:!… \\ $$

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