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
![Ī£_(n=1) ^ā ((Ļā²ā²(n))/n) =āā«_0 ^1 Ī£_(n=1) ^ā (1/n).((log^2 (x)x^(nā1) )/(1āx)) =āĪ£_(n=1) ^ā ā«_0 ^1 ((log^2 (x)(x^(nā1) /n))/(1āx))dx =ā«_0 ^1 ((log^2 (x)log(1āx))/(x(1āx)))dx =ā«_0 ^1 ((log^2 (x)log(1āx))/x)+ā«_0 ^1 ((log^2 (x)log(1āx))/(1āx))dx =[log^2 (x)Ī£_(n=1) ^ā ((āx^n )/n^2 )]+2ā«_0 ^1 log(x)Ī£_(n=1) ^ā (x^(nā1) /n^2 )dx+ā«_0 ^1 ((log^2 (1āx)log(x))/x)dx =2log(x)Ī£_(n=1) ^ā (x^n /n^3 )ā2Ī£_(n=1) ^ā ā«_0 ^1 (x^(nā1) /n^3 )dx+āµ =ā(Ļ^4 /(45))+āµ=ā(Ļ^4 /(36)) āµ=ā(Ļ^4 /(180)) (May be done previously)](Q137603.png)
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\psi''\left({n}\right)}{{n}} \\ $$$$=ā\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}.\frac{{log}^{\mathrm{2}} \left({x}\right){x}^{{n}ā\mathrm{1}} }{\mathrm{1}ā{x}} \\ $$$$=ā\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({x}\right)\frac{{x}^{{n}ā\mathrm{1}} }{{n}}}{\mathrm{1}ā{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({x}\right){log}\left(\mathrm{1}ā{x}\right)}{{x}\left(\mathrm{1}ā{x}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({x}\right){log}\left(\mathrm{1}ā{x}\right)}{{x}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({x}\right){log}\left(\mathrm{1}ā{x}\right)}{\mathrm{1}ā{x}}{dx} \\ $$$$=\left[{log}^{\mathrm{2}} \left({x}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{ā{x}^{{n}} }{{n}^{\mathrm{2}} }\right]+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left({x}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}ā\mathrm{1}} }{{n}^{\mathrm{2}} }{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}ā{x}\right){log}\left({x}\right)}{{x}}{dx} \\ $$$$=\mathrm{2}{log}\left({x}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{\mathrm{3}} }ā\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}ā\mathrm{1}} }{{n}^{\mathrm{3}} }{dx}+\aleph \\ $$$$=ā\frac{\pi^{\mathrm{4}} }{\mathrm{45}}+\aleph=ā\frac{\pi^{\mathrm{4}} }{\mathrm{36}} \\ $$$$\aleph=ā\frac{\pi^{\mathrm{4}} }{\mathrm{180}}\:\:\left({May}\:{be}\:{done}\:{previously}\right) \\ $$$$ \\ $$$$ \\ $$
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
