Question Number 159447 by mnjuly1970 last updated on 17/Nov/21
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:#{calculate}# \\ $$$$\:\:\:\:\:\Omega\::=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\:{x}^{\:\frac{{t}}{\mathrm{2}}} −{x}^{\:{t}} }{\mathrm{1}\:−\:{x}}\:{dx}\:{dt}\:=\:? \\ $$$$\:\:\:\:\:\:\:\:\:−−−{m}.{n}−−− \\ $$
Commented by Ar Brandon last updated on 17/Nov/21
Answered by Ar Brandon last updated on 17/Nov/21
![Ω=∫_0 ^1 ∫_0 ^1 ((x^(t/2) −x^t )/(1−x))dxdt =∫_0 ^1 (ψ(t+1)−ψ((t/2)+1))dt =[ln(Γ(t+1))−2ln(Γ((t/2)+1))]_0 ^1 =ln(((Γ(2))/(Γ^2 ((3/2)))))−ln(((Γ(1))/(Γ^2 (1))))=ln((4/π))](Q159451.png)
$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\frac{{t}}{\mathrm{2}}} −{x}^{{t}} }{\mathrm{1}−{x}}{dxdt} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\psi\left({t}+\mathrm{1}\right)−\psi\left(\frac{{t}}{\mathrm{2}}+\mathrm{1}\right)\right){dt} \\ $$$$\:\:\:\:=\left[\mathrm{ln}\left(\Gamma\left({t}+\mathrm{1}\right)\right)−\mathrm{2ln}\left(\Gamma\left(\frac{{t}}{\mathrm{2}}+\mathrm{1}\right)\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:=\mathrm{ln}\left(\frac{\Gamma\left(\mathrm{2}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)}\right)−\mathrm{ln}\left(\frac{\Gamma\left(\mathrm{1}\right)}{\Gamma^{\mathrm{2}} \left(\mathrm{1}\right)}\right)=\mathrm{ln}\left(\frac{\mathrm{4}}{\pi}\right) \\ $$
Commented by mnjuly1970 last updated on 17/Nov/21
$${peace}\:{be}\:{upon}\:{you}\:{sir}\:{brandon}.. \\ $$
Commented by Ar Brandon last updated on 17/Nov/21
With you too, Sir.