0-1-ln-x-dx-

Question Number 91931 by M±th+et+s last updated on 03/May/20

$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right)\:{dx} \\ $$

Commented by mathmax by abdo last updated on 04/May/20

we have Γ(x).Γ(1−x) =(π/(sin(πx))) ⇒ ∫_0 ^1 ln(Γ(x))dx +∫_0 ^1 ln(Γ(1−x))dx =∫_0 ^1 ln((π/(sin(πx))))dx ∫_0 ^1 ln(Γ(1−x))dx =_(1−x =t) ∫_0 ^1 ln(Γ(t))dt ⇒ 2∫_0 ^1 ln(Γ(x))dx =ln(π)−∫_0 ^1 ln(sin(πx)dx ∫_0 ^1 ln(sin(πx))dx =_(πx =u) (1/π)∫_0 ^π ln(sinu)du ∫_0 ^π ln(sinu)du =∫_0 ^(π/2) ln(sinu)du + ∫_0 ^(π/2) ln(sinu)du but ∫_0 ^(π/2) ln(sinu)du =_(u =(π/2)−α) ∫_0 ^(π/2) ln(cosα)dα ⇒ ∫_0 ^π ln(sinu)du =−(π/2)ln(2)−(π/2)ln(2) =−πln(2) ⇒ ∫_0 ^1 ln(Γ(x))dx =((ln(π))/2) +(π/2)ln(2)

$${we}\:{have}\:\Gamma\left({x}\right).\Gamma\left(\mathrm{1}−{x}\right)\:=\frac{\pi}{{sin}\left(\pi{x}\right)}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx}\:+\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left(\mathrm{1}−{x}\right)\right){dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\frac{\pi}{{sin}\left(\pi{x}\right)}\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left(\mathrm{1}−{x}\right)\right){dx}\:=_{\mathrm{1}−{x}\:={t}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left({t}\right)\right){dt}\:\Rightarrow \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx}\:={ln}\left(\pi\right)−\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({sin}\left(\pi{x}\right){dx}\right. \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({sin}\left(\pi{x}\right)\right){dx}\:=_{\pi{x}\:={u}} \:\:\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} \:{ln}\left({sinu}\right){du} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:{ln}\left({sinu}\right){du}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinu}\right){du}\:+\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinu}\right){du}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinu}\right){du}\:=_{{u}\:=\frac{\pi}{\mathrm{2}}−\alpha} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({cos}\alpha\right){d}\alpha\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\pi} \:{ln}\left({sinu}\right){du}\:=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:=−\pi{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\Gamma\left({x}\right)\right){dx}\:=\frac{{ln}\left(\pi\right)}{\mathrm{2}}\:+\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\ $$

Commented by M±th+et+s last updated on 04/May/20

$${thank}\:{you}\:{sir} \\ $$

Commented by abdomathmax last updated on 04/May/20

$${you}\:{are}\:{welcome}. \\ $$

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