Question Number 112114 by mnjuly1970 last updated on 06/Sep/20
![solution of Φ=∫_0 ^1 xH_x dx =^(γ+ψ(x+1)) ∫_0 ^( 1) x(γ+ψ(x+1))dx =^(ψ(x+1)=(1/x)+ψ(x)) ∫_0 ^( 1) x(γ+(1/x)+ψ(x))dx =(γ/2)+1+ ∫_0 ^( 1) x.(d/dx)(ln(Γ(x))) =(γ/2)+1+[xln(Γ(x))]_0 ^1 −∫_0 ^( 1) ln(Γ(x))dx we know (why?) ∫_0 ^( 1) ln(Γ(x))dx=ln((√(2π)) ) and lim_(x→0^+ ) (xln(Γ(x)))=0 (why??) finally Φ =∫_0 ^( 1) xH_x dx=(γ/2)+1−ln((√(2π))) ✓ m.n.july 1970....](Q112114.png)
$$\:\:{solution}\:{of} \\ $$$$\Phi=\int_{\mathrm{0}} ^{\mathrm{1}} {xH}_{{x}} {dx}\:\overset{\gamma+\psi\left({x}+\mathrm{1}\right)} {=}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}\left(\gamma+\psi\left({x}+\mathrm{1}\right)\right){dx}\: \\ $$$$\:\overset{\psi\left({x}+\mathrm{1}\right)=\frac{\mathrm{1}}{{x}}+\psi\left({x}\right)} {=}\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}\left(\gamma+\frac{\mathrm{1}}{{x}}+\psi\left({x}\right)\right){dx} \\ $$$$=\frac{\gamma}{\mathrm{2}}+\mathrm{1}+\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}.\frac{{d}}{{dx}}\left({ln}\left(\Gamma\left({x}\right)\right)\right) \\ $$$$=\frac{\gamma}{\mathrm{2}}+\mathrm{1}+\left[{xln}\left(\Gamma\left({x}\right)\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx} \\ $$$$\:\:{we}\:\:{know}\:\left({why}?\right)\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx}={ln}\left(\sqrt{\mathrm{2}\pi}\:\right) \\ $$$$\:{and}\:\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \left({xln}\left(\Gamma\left({x}\right)\right)\right)=\mathrm{0}\:\left({why}??\right) \\ $$$${finally}\:\Phi\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xH}_{{x}} {dx}=\frac{\gamma}{\mathrm{2}}+\mathrm{1}−{ln}\left(\sqrt{\mathrm{2}\pi}\right)\:\checkmark \\ $$$$\:\:{m}.{n}.{july}\:\mathrm{1970}.... \\ $$$$\: \\ $$