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Question Number 189091 by mnjuly1970 last updated on 12/Mar/23

$$$$$$1:\mathrm{\Omega }=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{\mathrm{H}}_n}{n^2}=?$$$$2:\eta (-1)=?$$$$$$$$$$

Answered by qaz last updated on 15/Mar/23

$$A=\mathrm{\Sigma }{(-1)}^n{H}_n{\int }_0^{\mathrm{\infty }}{xe}^{-nx}dx$$$$=-{\int }_0^{\mathrm{\infty }}\frac{xln(1+e^{-x})}{1+e^{-x}}dx={\int }_0^1\frac{lntln(1+t)}{t(1+t)}dt$$$$={\int }_0^1(\frac{lntln(1+t)}{t}-\frac{lntln(1+t)}{1+t})dt$$$$={\int }_0^{1}lnt{\left({Li}_2(-t)\right)}^{\prime }dt-{\int }_0^1\frac{(ln\frac{t}{1+t}+ln(1+t))ln(1+t)}{1+t}dt$$$$=-{\int }_0^1\frac{{Li}_2(-t)}{t}dt-\frac{1}{3}{ln}^{3}2-{\int }_0^1\frac{ln(1-\frac{1}{1+t})ln(1+t)}{1+t}dt$$$$={Li}_3(-1)-\frac{1}{3}{ln}^{3}2-{\int }_0^1{\left({Li}_2\left(\frac{1}{1+t}\right)\right)}^{\prime }ln(1+t)dt$$$$={Li}_3(-1)-\frac{1}{3}{ln}^{3}2-{Li}_2\left(\frac{1}{2}\right)ln2+{\int }_0^1\frac{{Li}_2\left(\frac{1}{1+t}\right)}{1+t}dt$$$$={Li}_3(-1)-\frac{1}{3}{ln}^{3}2-{Li}_2\left(\frac{1}{2}\right)ln2-{\int }_0^1{\left({Li}_3\left(\frac{1}{1+t}\right)\right)}^{\prime }dt$$$$={Li}_3(-1)-\frac{1}{3}{ln}^{3}2-{Li}_2\left(\frac{1}{2}\right)ln2-{Li}_3\left(\frac{1}{2}\right)+{Li}_3\left(1\right)$$$$={Li}_3(-1)-\frac{1}{3}{ln}^{3}2-(\frac{{\pi }^2}{12}-\frac{1}{2}{ln}^{2}2)ln2-[\frac{7}{8}{Li}_3\left(1\right)-\frac{{\pi }^2}{12}ln2+\frac{1}{6}{ln}^{3}2]+{Li}_3\left(1\right)$$$$={Li}_3(-1)+\frac{1}{8}{Li}_3\left(1\right)=(2^{1-3}-1)\zeta \left(3\right)+\frac{1}{8}\zeta \left(3\right)=-\frac{5}{8}\zeta \left(3\right)$$$$...$$

Commented by senestro last updated on 02/Apr/23

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