Question Number 139811 by mnjuly1970 last updated on 01/May/21
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{{x}}\:{ln}\left(\frac{\mathrm{1}}{\mathrm{1}β{x}}\right){dx} \\ $$$$\:\:\:\:\:{solution}: \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{{x}}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}}\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\::=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\::=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}β\frac{\mathrm{1}}{{n}+\frac{\mathrm{3}}{\mathrm{2}}}\right) \\ $$$$\:\:\:\:\:\:\::=\frac{\mathrm{2}}{\mathrm{3}}\left\{\gamma\:β\gamma+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}β\frac{\mathrm{1}}{{n}+\frac{\mathrm{3}}{\mathrm{2}}}\right)\:\right\} \\ $$$$\:\:\:\:\::=\:\frac{\mathrm{2}}{\mathrm{3}}\:\gamma\:+\frac{\mathrm{2}}{\mathrm{3}}\:\psi\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{2}}{\mathrm{3}}\:\gamma\:+\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\mathrm{3}}+\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\right) \\ $$$$\:\:\:\:\::=\frac{\mathrm{2}}{\mathrm{3}}\:\gamma\:+\frac{\mathrm{4}}{\mathrm{9}}\:+\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}+\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\:\:\:\:\::=\frac{\mathrm{2}}{\mathrm{3}}\:\gamma\:+\frac{\mathrm{16}}{\mathrm{9}}\:β\frac{\mathrm{2}}{\mathrm{3}}\:\gammaβ\frac{\mathrm{4}}{\mathrm{3}}\:{ln}\left(\mathrm{2}\right)\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\::=\frac{\mathrm{16}}{\mathrm{9}}\:β\frac{\mathrm{4}}{\mathrm{3}}\:{ln}\left(\mathrm{2}\right)\:…..\checkmark\:\: \\ $$
Answered by mathmax by abdo last updated on 01/May/21
$$\Phi=\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{\mathrm{x}}\mathrm{log}\left(\frac{\mathrm{1}}{\mathrm{1}β\mathrm{x}}\right)\mathrm{dx}\:\Rightarrow\Phi=β\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{\mathrm{x}}\mathrm{log}\left(\mathrm{1}β\mathrm{x}\right)\mathrm{dx} \\ $$$$=_{\sqrt{\mathrm{x}}=\mathrm{t}} \:\:\:β\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{t}\:\mathrm{log}\left(\mathrm{1}β\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{2t}\right)\mathrm{dt} \\ $$$$=β\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{t}^{\mathrm{2}} \:\mathrm{log}\left(\mathrm{1}β\mathrm{t}^{\mathrm{2}} \right)\mathrm{dt} \\ $$$$=β\mathrm{2}\left\{\:\:\left[\frac{\mathrm{t}^{\mathrm{3}} β\mathrm{1}}{\mathrm{3}}\mathrm{log}\left(\mathrm{1}β\mathrm{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} β\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{t}^{\mathrm{3}} β\mathrm{1}}{\mathrm{3}}Γ\frac{β\mathrm{2t}}{\mathrm{1}β\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\right\} \\ $$$$=β\frac{\mathrm{4}}{\mathrm{3}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{t}^{\mathrm{4}} β\mathrm{t}}{\mathrm{1}β\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\:=\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{t}\left(\mathrm{t}^{\mathrm{3}} β\mathrm{1}\right)}{\left(\mathrm{t}β\mathrm{1}\right)\left(\mathrm{t}+\mathrm{1}\right)}\mathrm{dt} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{t}\left(\mathrm{t}^{\mathrm{2}} +\mathrm{t}+\mathrm{1}\right)}{\mathrm{t}+\mathrm{1}}\mathrm{dt}\:=\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{t}^{\mathrm{3}} \:+\mathrm{t}^{\mathrm{2}} \:+\mathrm{t}}{\mathrm{t}+\mathrm{1}}\mathrm{dt} \\ $$$$=_{\mathrm{t}+\mathrm{1}=\mathrm{y}} \:\:\:\:\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\left(\mathrm{y}β\mathrm{1}\right)^{\mathrm{3}} \:+\left(\mathrm{y}β\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{y}β\mathrm{1}}{\mathrm{y}}\mathrm{dy} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{y}^{\mathrm{3}} β\mathrm{3y}^{\mathrm{2}} \:+\mathrm{3y}β\mathrm{1}+\mathrm{y}^{\mathrm{2}} β\mathrm{2y}+\mathrm{1}+\mathrm{y}β\mathrm{1}}{\mathrm{y}}\mathrm{dy} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{y}^{\mathrm{3}} β\mathrm{2y}^{\mathrm{2}} +\mathrm{2y}β\mathrm{1}}{\mathrm{y}}\mathrm{dy} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{y}^{\mathrm{2}} β\mathrm{2y}+\mathrm{2}β\frac{\mathrm{1}}{\mathrm{y}}\right)\mathrm{dy} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\left[\frac{\mathrm{y}^{\mathrm{3}} }{\mathrm{3}}β\mathrm{y}^{\mathrm{2}} \:+\mathrm{2y}β\mathrm{log}\mid\mathrm{y}\mid\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\left\{\frac{\mathrm{8}}{\mathrm{3}}β\mathrm{4}+\mathrm{4}β\mathrm{log2}β\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}β\mathrm{2}\right\} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\left\{\frac{\mathrm{7}}{\mathrm{3}}β\mathrm{1}β\mathrm{log2}\right\}\:=\frac{\mathrm{4}}{\mathrm{3}}\left(\frac{\mathrm{4}}{\mathrm{3}}β\mathrm{log2}\right)\:=\frac{\mathrm{16}}{\mathrm{9}}β\frac{\mathrm{4}}{\mathrm{3}}\mathrm{log2} \\ $$
Commented by mnjuly1970 last updated on 01/May/21
$${thx}\:{sir}\:{max}\:… \\ $$
Commented by mathmax by abdo last updated on 02/May/21
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir} \\ $$
Answered by Ar Brandon last updated on 16/May/21
$$\phi=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{x}}\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{1}β\mathrm{x}}\right)\mathrm{dx}=β\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{x}}\mathrm{ln}\left(\mathrm{1}β\mathrm{x}\right)\mathrm{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{m}β\mathrm{1}} \left(\mathrm{1}β\mathrm{x}\right)^{\mathrm{n}β\mathrm{1}} \mathrm{dx}=\beta\left(\mathrm{m},\mathrm{n}\right)=\frac{\Gamma\left(\mathrm{m}\right)\Gamma\left(\mathrm{n}\right)}{\Gamma\left(\mathrm{m}+\mathrm{n}\right)} \\ $$$$\mathrm{ln}\beta\left(\mathrm{m},\mathrm{n}\right)=\mathrm{ln}\Gamma\left(\mathrm{m}\right)+\mathrm{ln}\Gamma\left(\mathrm{n}\right)β\mathrm{ln}\Gamma\left(\mathrm{m}+\mathrm{n}\right) \\ $$$$\frac{\mathrm{1}}{\beta\left(\mathrm{m},\mathrm{n}\right)}\centerdot\frac{\partial\beta\left(\mathrm{m},\mathrm{n}\right)}{\partial\mathrm{n}}=\psi\left(\mathrm{n}\right)β\psi\left(\mathrm{m}+\mathrm{n}\right) \\ $$$$\Rightarrow\frac{\partial\beta\left(\frac{\mathrm{3}}{\mathrm{2}},\mathrm{1}\right)}{\partial\mathrm{n}}=\beta\left(\frac{\mathrm{3}}{\mathrm{2}},\mathrm{1}\right)\left\{\psi\left(\mathrm{1}\right)β\psi\left(\frac{\mathrm{5}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{5}}{\mathrm{2}}\right)}\left\{β\gammaβ\left(\frac{\mathrm{2}}{\mathrm{3}}+\mathrm{2}β\gammaβ\mathrm{2ln2}\right)\right\} \\ $$$$\phi=β\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2ln2}β\frac{\mathrm{8}}{\mathrm{3}}\right)=\frac{\mathrm{16}}{\mathrm{9}}β\frac{\mathrm{4ln2}}{\mathrm{3}} \\ $$
Commented by mnjuly1970 last updated on 01/May/21
$$\:\:{grateful}\:{mr}\:{brandon}… \\ $$
Commented by SLVR last updated on 01/May/21
$${could}\:{you}\:{explain}\:{what}\psi\left(\mathrm{1}\right)?\psi\left(\mathrm{5}/\mathrm{2}\right) \\ $$$${Thanks}\:{mr}.{Brandon}\:{sir} \\ $$
Commented by Ar Brandon last updated on 01/May/21
$$\mathrm{You}'\mathrm{re}\:\mathrm{welcome}\:\mathrm{Sir} \\ $$
Commented by Ar Brandon last updated on 01/May/21
$$\Gamma'\left(\mathrm{x}\right)=\Gamma\left(\mathrm{x}\right)\psi\left(\mathrm{x}\right)\Rightarrow\frac{\Gamma'\left(\mathrm{x}\right)}{\Gamma\left(\mathrm{x}\right)}=\psi\left(\mathrm{x}\right) \\ $$$$\psi\left(\mathrm{s}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{s}}+\psi\left(\mathrm{s}\right)\:,\:\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=β\gammaβ\mathrm{2ln2} \\ $$$$\psi\left(\mathrm{x}\right)=β\gamma+\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}β\frac{\mathrm{1}}{\mathrm{n}+\mathrm{x}}\right)\:\Rightarrow\psi\left(\mathrm{1}\right)=β\gamma \\ $$$$\psi\left(\mathrm{x}\right)\:\mathrm{represents}\:\mathrm{digamma}\:\mathrm{function}. \\ $$
Commented by SLVR last updated on 01/May/21
$${So}….{kind}\:{you}..{sir}\:\:{Great}…{explanation} \\ $$$${God}…{bless}\:{you}\:{for}\:{ever} \\ $$
Commented by Ar Brandon last updated on 01/May/21
$$\mathrm{Thanks}! \\ $$