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Question Number 146442 by qaz last updated on 13/Jul/21

$$\frac{\mathrm{d}}{\mathrm{dn}}{\mid }_{\mathrm{n}=1}{\mathrm{H}}_{\mathrm{n}}=?$$

Answered by mnjuly1970 last updated on 13/Jul/21

$$=\frac{d}{dn}\left({\int }_0^1\frac{1-x^n}{1-x}\right)={\int }_0^1\frac{-x^{n}ln\left(x\right)}{1-x}dx$$$$={\int }_0^1\frac{-xln\left(x\right)}{1-x}dx=-{\int }_0^1\frac{(1-x)ln(1-x)}{x}dx$$$$={li}_2\left(1\right)+{\int }_0^{1}ln(1-x)dx$$$$=\frac{{\pi }^2}{6}+{\int }_0^{1}ln\left(x\right)dx=-1+\frac{{\pi }^2}{6}....$$

Answered by mnjuly1970 last updated on 13/Jul/21

$$\psi (n+1):=-\gamma +{\mathrm{H}}_n$$$$\frac{1}{n}+\psi \left(n\right)=-\gamma +{\mathrm{H}}_n$$$$\frac{\mathrm{d}}{\mathrm{d}n}\left(\mathrm{H}n\right)\underset{}{\mid }=-1+{\psi }^{\prime }\left(1\right)$$$$=-1+\frac{{\pi }^2}{6}=-1+\zeta \left(2\right)...$$

Commented by qaz last updated on 13/Jul/21

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.$$$$\mathrm{I}\mathrm{think}\mathrm{it}\mathrm{is}\mathrm{possible}\mathrm{to}\mathrm{use}\mathrm{differential}\mathrm{chain}\mathrm{rule},$$$$\mathrm{same}\mathrm{like}\frac{\partial }{\partial \mathrm{x}}{\int }_{\mathrm{h}\left(\mathrm{x}\right)}^{\mathrm{g}\left(\mathrm{x}\right)}\mathrm{f}(\mathrm{x},\mathrm{t})\mathrm{dt}={\int }_{\mathrm{h}\left(\mathrm{x}\right)}^{\mathrm{g}\left(\mathrm{x}\right)}\frac{\partial }{\partial \mathrm{x}}\mathrm{f}(\mathrm{x},\mathrm{t})\mathrm{dt}+\mathrm{g}{\left(\mathrm{x}\right)}^{\prime }\mathrm{f}(\mathrm{x},\mathrm{g}\left(\mathrm{x}\right))-\mathrm{h}{\left(\mathrm{x}\right)}^{\prime }\mathrm{f}(\mathrm{x},\mathrm{h}\left(\mathrm{x}\right))$$$$,\mathrm{to}\mathrm{differential}\mathrm{to}\mathrm{a}\mathrm{seiris}\mathrm{summation}.....$$

Commented by mnjuly1970 last updated on 13/Jul/21

$$thankyousirqaz$$

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