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Question Number 137592 by mnjuly1970 last updated on 04/Apr/21
               ......advanced.....calculus....      š›€=Ī£_(n=1) ^āˆž ((Ļˆā€²ā€²(n))/n)=???   I havefound ::  Ī©=āˆ’(Ļ€^4 /(36))  ... !
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……{advanced}…..{calculus}…. \\ $$$$\:\:\:\:\boldsymbol{\Omega}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\psi''\left({n}\right)}{{n}}=??? \\ $$$$\:{I}\:{havefound}\:::\:\:\Omega=āˆ’\frac{\pi^{\mathrm{4}} }{\mathrm{36}}\:\:…\:! \\ $$
Answered by Dwaipayan Shikari last updated on 04/Apr/21
Ī£_(n=1) ^āˆž ((Ļˆā€²ā€²(n))/n)  =āˆ’āˆ«_0 ^1 Ī£_(n=1) ^āˆž (1/n).((log^2 (x)x^(nāˆ’1) )/(1āˆ’x))  =āˆ’Ī£_(n=1) ^āˆž āˆ«_0 ^1 ((log^2 (x)(x^(nāˆ’1) /n))/(1āˆ’x))dx  =āˆ«_0 ^1 ((log^2 (x)log(1āˆ’x))/(x(1āˆ’x)))dx  =āˆ«_0 ^1 ((log^2 (x)log(1āˆ’x))/x)+āˆ«_0 ^1 ((log^2 (x)log(1āˆ’x))/(1āˆ’x))dx  =[log^2 (x)Ī£_(n=1) ^āˆž ((āˆ’x^n )/n^2 )]+2āˆ«_0 ^1 log(x)Ī£_(n=1) ^āˆž (x^(nāˆ’1) /n^2 )dx+āˆ«_0 ^1 ((log^2 (1āˆ’x)log(x))/x)dx  =2log(x)Ī£_(n=1) ^āˆž (x^n /n^3 )āˆ’2Ī£_(n=1) ^āˆž āˆ«_0 ^1 (x^(nāˆ’1) /n^3 )dx+ā„µ  =āˆ’(Ļ€^4 /(45))+ā„µ=āˆ’(Ļ€^4 /(36))  ā„µ=āˆ’(Ļ€^4 /(180))  (May be done previously)
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\psi''\left({n}\right)}{{n}} \\ $$$$=āˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}.\frac{{log}^{\mathrm{2}} \left({x}\right){x}^{{n}āˆ’\mathrm{1}} }{\mathrm{1}āˆ’{x}} \\ $$$$=āˆ’\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({x}\right)\frac{{x}^{{n}āˆ’\mathrm{1}} }{{n}}}{\mathrm{1}āˆ’{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({x}\right){log}\left(\mathrm{1}āˆ’{x}\right)}{{x}\left(\mathrm{1}āˆ’{x}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({x}\right){log}\left(\mathrm{1}āˆ’{x}\right)}{{x}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({x}\right){log}\left(\mathrm{1}āˆ’{x}\right)}{\mathrm{1}āˆ’{x}}{dx} \\ $$$$=\left[{log}^{\mathrm{2}} \left({x}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{āˆ’{x}^{{n}} }{{n}^{\mathrm{2}} }\right]+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left({x}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}āˆ’\mathrm{1}} }{{n}^{\mathrm{2}} }{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}āˆ’{x}\right){log}\left({x}\right)}{{x}}{dx} \\ $$$$=\mathrm{2}{log}\left({x}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{\mathrm{3}} }āˆ’\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}āˆ’\mathrm{1}} }{{n}^{\mathrm{3}} }{dx}+\aleph \\ $$$$=āˆ’\frac{\pi^{\mathrm{4}} }{\mathrm{45}}+\aleph=āˆ’\frac{\pi^{\mathrm{4}} }{\mathrm{36}} \\ $$$$\aleph=āˆ’\frac{\pi^{\mathrm{4}} }{\mathrm{180}}\:\:\left({May}\:{be}\:{done}\:{previously}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 04/Apr/21
thank you so much mr payan...     yes you are right     š›—=āˆ«_0 ^( 1) ((ln(x).ln^2 (1āˆ’x))/x)dx=((āˆ’Ļ€^4 )/(180))      ā‡“ ā‡“ ā‡“
$${thank}\:{you}\:{so}\:{much}\:{mr}\:{payan}… \\ $$$$\:\:\:{yes}\:{you}\:{are}\:{right} \\ $$$$\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}\right).{ln}^{\mathrm{2}} \left(\mathrm{1}āˆ’{x}\right)}{{x}}{dx}=\frac{āˆ’\pi^{\mathrm{4}} }{\mathrm{180}} \\ $$$$\:\:\:\:\Downarrow\:\Downarrow\:\Downarrow \\ $$
Answered by mnjuly1970 last updated on 04/Apr/21
Answered by mnjuly1970 last updated on 04/Apr/21

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