Question Number 137592 by mnjuly1970 last updated on 04/Apr/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……{advanced}…..{calculus}…. \\ $$$$\:\:\:\:\boldsymbol{\Omega}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\psi''\left({n}\right)}{{n}}=??? \\ $$$$\:{I}\:{havefound}\:::\:\:\Omega=ā\frac{\pi^{\mathrm{4}} }{\mathrm{36}}\:\:…\:! \\ $$
Answered by Dwaipayan Shikari last updated on 04/Apr/21
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\psi''\left({n}\right)}{{n}} \\ $$$$=ā\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}.\frac{{log}^{\mathrm{2}} \left({x}\right){x}^{{n}ā\mathrm{1}} }{\mathrm{1}ā{x}} \\ $$$$=ā\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({x}\right)\frac{{x}^{{n}ā\mathrm{1}} }{{n}}}{\mathrm{1}ā{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({x}\right){log}\left(\mathrm{1}ā{x}\right)}{{x}\left(\mathrm{1}ā{x}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({x}\right){log}\left(\mathrm{1}ā{x}\right)}{{x}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({x}\right){log}\left(\mathrm{1}ā{x}\right)}{\mathrm{1}ā{x}}{dx} \\ $$$$=\left[{log}^{\mathrm{2}} \left({x}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{ā{x}^{{n}} }{{n}^{\mathrm{2}} }\right]+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left({x}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}ā\mathrm{1}} }{{n}^{\mathrm{2}} }{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}ā{x}\right){log}\left({x}\right)}{{x}}{dx} \\ $$$$=\mathrm{2}{log}\left({x}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{\mathrm{3}} }ā\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}ā\mathrm{1}} }{{n}^{\mathrm{3}} }{dx}+\aleph \\ $$$$=ā\frac{\pi^{\mathrm{4}} }{\mathrm{45}}+\aleph=ā\frac{\pi^{\mathrm{4}} }{\mathrm{36}} \\ $$$$\aleph=ā\frac{\pi^{\mathrm{4}} }{\mathrm{180}}\:\:\left({May}\:{be}\:{done}\:{previously}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 04/Apr/21
$${thank}\:{you}\:{so}\:{much}\:{mr}\:{payan}… \\ $$$$\:\:\:{yes}\:{you}\:{are}\:{right} \\ $$$$\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}\right).{ln}^{\mathrm{2}} \left(\mathrm{1}ā{x}\right)}{{x}}{dx}=\frac{ā\pi^{\mathrm{4}} }{\mathrm{180}} \\ $$$$\:\:\:\:\Downarrow\:\Downarrow\:\Downarrow \\ $$
Answered by mnjuly1970 last updated on 04/Apr/21
Answered by mnjuly1970 last updated on 04/Apr/21