Question Number 167048 by mnjuly1970 last updated on 05/Mar/22
$$ \\ $$$$\:\:\:\:\:{prove}\:{that} \\ $$$$ \\ $$$$\Phi=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}.\psi\:\left(\mathrm{2}+{x}\:\right)=\:\mathrm{2}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{8}\pi\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:−−− \\ $$
Answered by shikaridwan last updated on 05/Mar/22
![ψ(x+2)=ψ(x+1)+(1/(x+1))=ψ(x)+(1/x)+(1/(x+1)) ∫_0 ^1 xψ(2+x)dx=∫_0 ^1 2−(1/(x+1))+xψ(x)dx =2−log (2)+[xlog(Γ(x))]_0 ^1 −∫_0 ^1 log(Γ(x))dx =2−log (2)−log((√(2π))) =2−(1/2)log (8π)](Q167053.png)
$$\psi\left({x}+\mathrm{2}\right)=\psi\left({x}+\mathrm{1}\right)+\frac{\mathrm{1}}{{x}+\mathrm{1}}=\psi\left({x}\right)+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}\psi\left(\mathrm{2}+{x}\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}−\frac{\mathrm{1}}{{x}+\mathrm{1}}+{x}\psi\left({x}\right){dx} \\ $$$$=\mathrm{2}−\mathrm{log}\:\left(\mathrm{2}\right)+\left[{xlog}\left(\Gamma\left({x}\right)\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left(\Gamma\left({x}\right)\right){dx} \\ $$$$=\mathrm{2}−\mathrm{log}\:\left(\mathrm{2}\right)−{log}\left(\sqrt{\mathrm{2}\pi}\right) \\ $$$$=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\left(\mathrm{8}\pi\right) \\ $$
Commented by shikaridwan last updated on 05/Mar/22
$${Hi}\:{sir}!\:{Can}\:{you}\:{recognize}\:{me}? \\ $$
Commented by mnjuly1970 last updated on 05/Mar/22
$${thsnk}\:{you}\:{so}\:{much} \\ $$$${Mr}.{Dwaypayan}\:...{thanks}\:{alot} \\ $$$$\:\:\:{and}\:\:{Welcome}\:{again}\:\:.... \\ $$$$\:\:\:\: \\ $$