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Question Number 147195 by mnjuly1970 last updated on 18/Jul/21

$$$$$$\mathrm{\Omega }:={\int }_0^1{\mathrm{Li}}_2\left(\sqrt{x}\right)dx=?$$$$$$

Answered by Olaf_Thorendsen last updated on 18/Jul/21

$$$$$$\mathrm{\Omega }={\int }_0^1{\mathrm{Li}}_2\left(\sqrt{x}\right)dx$$$$\mathrm{\Omega }={\int }_0^1\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{{\left(\sqrt{x}\right)}^k}{k^2}dx$$$$\mathrm{\Omega }={\int }_0^1\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{x^{\frac{k}{2}}}{k^2}dx$$$$\mathrm{\Omega }={\left[\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{x^{\frac{k}{2}+1}}{k^2(\frac{k}{2}+1)}\right]}_0^1$$$$\mathrm{\Omega }=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^2(\frac{k}{2}+1)}$$$$\mathrm{\Omega }=2\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^2(k+2)}$$$$\mathrm{\Omega }=\underset{k=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{k^2}-\frac{1}{2}.\frac{1}{k}+\frac{1}{2}.\frac{1}{k+2})$$$$\mathrm{\Omega }=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^2}-\frac{1}{2}\underset{k=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{k}-\frac{1}{k+2})$$$$\mathrm{\Omega }=\frac{{\pi }^2}{6}-\frac{1}{2}(1+\frac{1}{2})$$$$\mathrm{\Omega }=\frac{{\pi }^2}{6}-\frac{3}{4}$$

Commented by mnjuly1970 last updated on 19/Jul/21

$$mercey-thanksalotmaster$$$$olaf....$$

Answered by Kamel last updated on 18/Jul/21

$$\stackrel{t=\sqrt{x}}{=}2{\int }_0^1{tLi}_2\left(t\right)dt=\frac{{\pi }^2}{6}+{\int }_0^{1}tLn(1-t)dt$$$$=\frac{{\pi }^2}{6}-\frac{1}{2}{\int }_0^1(1+t)dt=\frac{{\pi }^2}{6}-\frac{1}{2}(1+\frac{1}{2})$$$$\therefore {\int }_0^1{Li}_2\left(\sqrt{x}\right)dx=\frac{{\pi }^2}{6}-\frac{3}{4}$$

Commented by mnjuly1970 last updated on 19/Jul/21

$$thankyousomuchmrkamel..$$

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