Question Number 137501 by mnjuly1970 last updated on 03/Apr/21
$$\:\:\:\:\:\:\:….{advanced}\:….\:{calculus}…. \\ $$$$\:\:{prove}\:{that}:: \\ $$$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}\right).{ln}\left(\mathrm{1}β{x}\right)}{\mathrm{1}+{x}}{dx}=\frac{\mathrm{13}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)β\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\left(\mathrm{2}\right)…. \\ $$
Answered by ΓΓ―= last updated on 04/Apr/21
$$\phi=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}\:{ln}\left(\mathrm{1}β{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnxln}\left(\mathrm{1}β{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}}β\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnxln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnxln}\left(\mathrm{1}β{x}^{\mathrm{2}} \right)}{\mathrm{1}β{x}^{\mathrm{2}} }{dx}β\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xln}\:{xln}\:\left(\mathrm{1}β{x}^{\mathrm{2}} \right)}{\mathrm{1}β{x}^{\mathrm{2}} }{dx}β\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\:{xln}\:\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$=\frac{\partial}{\partial{a}\partial{b}}\mid_{{a}=\mathrm{0},{b}=β\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}} \left(\mathrm{1}β{x}^{\mathrm{2}} \right)^{{b}} {dx}β\frac{\partial}{\partial{a}\partial{b}}\mid_{{a}=\mathrm{1},{b}=β\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}} \left(\mathrm{1}β{x}^{\mathrm{2}} \right)^{{b}} {dx}β\left\{\int\frac{{ln}\left(\mathrm{1}+{x}\right)\left\{{ln}\left(\mathrm{1}+{x}\right)+{ln}\frac{{x}}{\mathrm{1}+{x}}\right\}}{\mathrm{1}+{x}}\right\}_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\partial}{\partial{a}\partial{b}}\mid_{{a}=\mathrm{0},{b}=β\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\left({a}β\mathrm{1}\right)/\mathrm{2}} \left(\mathrm{1}β{u}\right)^{{b}} {du}β\frac{\partial}{\partial{a}\partial{b}}\mid_{{a}=\mathrm{1},{b}=β\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\left({a}β\mathrm{1}\right)/\mathrm{2}} \left(\mathrm{1}β{u}\right)^{{b}} {du}β\left\{\int\frac{{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx}+\int\frac{{ln}\left(\mathrm{1}β\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)}{\mathrm{1}+{x}}{ln}\:\left(\mathrm{1}+{x}\right){dx}\right\}_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\partial}{\partial{a}\partial{b}}\mid_{{a}=\mathrm{0},{b}=β\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{{a}+\mathrm{1}}{\mathrm{2}},{b}+\mathrm{1}\right)β\frac{\partial}{\partial{a}\partial{b}}\mid_{{a}=\mathrm{1},{b}=β\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{{a}+\mathrm{1}}{\mathrm{2}},{b}+\mathrm{1}\right)β\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}+{x}\right)^{β{n}} }{{n}}\centerdot\frac{{ln}\:\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\zeta\left(\mathrm{3}\right)β\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\mathrm{2}β\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\left\{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}\right)^{β{n}β\mathrm{1}} {ln}\left(\mathrm{1}+{x}\right){dx}\right\} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\zeta\left(\mathrm{3}\right)β\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\mathrm{2}β\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\left\{\frac{\left(\mathrm{1}+{x}\right)^{β{n}} }{β{n}}{ln}\left(\mathrm{1}+{x}\right)β\int\frac{\left(\mathrm{1}+{x}\right)^{β{n}β\mathrm{1}} }{β{n}}{dx}\right\}_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\zeta\left(\mathrm{3}\right)β\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\mathrm{2}β\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}β\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{1}+{x}\right)^{β{n}} {ln}\left(\mathrm{1}+{x}\right)}{{n}^{\mathrm{2}} }\mid_{\mathrm{0}} ^{\mathrm{1}} +\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{1}+{x}\right)^{β{n}} }{{n}^{\mathrm{3}} }\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\zeta\left(\mathrm{3}\right)β\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\mathrm{2}β\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}β{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right){ln}\left(\mathrm{1}+{x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} β{Li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\zeta\left(\mathrm{3}\right)β\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\mathrm{2}β\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}β{ln}\mathrm{2}{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)β{Li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+{Li}_{\mathrm{3}} \left(\mathrm{1}\right) \\ $$$$={ans}… \\ $$
Commented by mnjuly1970 last updated on 04/Apr/21
$${great}\:…{thanks}\:{alot}… \\ $$