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Question Number 125098 by mnjuly1970 last updated on 08/Dec/20

$$...nicecalculus...$$$$provethat::{Apery}^{\prime }sconstant$$$$\varphi ={\int }_0^1\left\{(4x^2+4^2{x}^{2^2}+4^3{x}^{2^3}+...)\frac{{ln}^2\left(x\right)}{x(1+x)}\right\}dx$$$$=2\zeta \left(3\right)-1$$

Commented by talminator2856791 last updated on 08/Dec/20

$$\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{answer}$$

Answered by Dwaipayan Shikari last updated on 08/Dec/20

$${\int }_0^1(4x^2+4^2{x}^{2^2}+4^3{x}^{2^3}+..)\frac{{log}^{2}x}{x(1+x)}dx$$$$={\int }_0^1\sum ^{\mathrm{\infty }}{4}^n{x}^{2^n}\frac{{log}^{2}x}{x(1+x)}dx$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^1\frac{2^{2n}{x}^{2^n}{log}^2\left(x\right)}{x(1+x)}dx$$$$=\underset{n⩾1}{\sum ^{\mathrm{\infty }}}{2}^{2n}({\int }_0^1{x}^{2^n-1}{log}^2\left(x\right)-{\int }_0^1\frac{x^{2^n}{log}^2\left(x\right)}{1+x}dx)$$$$=\underset{n⩾1}{\sum ^{\mathrm{\infty }}}{2}^{2n}({\int }_{-\mathrm{\infty }}^0{e}^{2^{n}t}{t}^2-{\int }_0^1\frac{x^{2^n}{log}^2\left(x\right)}{1+x}dx)Replacetas-t$$$$=\underset{n⩾1}{\sum ^{\mathrm{\infty }}}{2}^{2n}({\int }_0^{\mathrm{\infty }}{e}^{-2^{n}t}{t}^2-{\int }_0^1\frac{x^{2^n}{log}^2\left(x\right)}{1+x}dx)2^{n}t=u\Rightarrow 2^n=\frac{du}{dt}$$$$=\underset{n⩾1}{\sum ^{\mathrm{\infty }}}\frac{1}{2^n}\mathrm{\Gamma }\left(2\right)-\underset{n⩾1}{\sum ^{\mathrm{\infty }}}{2}^{2n}{\int }_0^1\frac{x^{2^n}{log}^2\left(x\right)}{1+x}dx$$$$=1-\underset{n⩾1}{\sum ^{\mathrm{\infty }}}{2}^{2n}\underset{k⩾0}{\sum ^{\mathrm{\infty }}}{(-1)}^k{\int }_0^1{x}^{2^n-k}{log}^2\left(x\right)dx$$$$=1-\underset{n⩾1}{\sum ^{\mathrm{\infty }}}{2}^{2n}\underset{k⩾0}{\sum ^{\mathrm{\infty }}}{(-1)}^k{\int }_{-\mathrm{\infty }}^0{e}^{t(2^n-k+1)}{t}^{2}dt$$$$=1-\underset{n⩾1}{\sum ^{\mathrm{\infty }}}{2}^{2n}\underset{k⩾0}{\sum ^{\mathrm{\infty }}}{(-1)}^k{\int }_0^{\mathrm{\infty }}{e}^{-t(2^n-k+1)}{t}^{2}dt$$$$=1-\underset{n⩾1}{\sum ^{\mathrm{\infty }}}{2}^{2n}\underset{k⩾0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k}{{(2^n-k+1)}^3}=1-\underset{k⩾0}{\sum ^{\mathrm{\infty }}}{(-1)}^k\underset{n⩾1}{\sum ^{\mathrm{\infty }}}\frac{2^{2n}}{{(2^n-k+1)}^3}...$$

Commented by talminator2856791 last updated on 08/Dec/20

$$\mathrm{is}\log \mathrm{the}\mathrm{same}\mathrm{as}\ln ?$$$$\mathrm{because}\mathrm{in}\mathrm{the}\mathrm{question}\mathrm{it}\mathrm{says}\ln \mathrm{but}\mathrm{you}\mathrm{say}\log$$

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