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Question Number 142429 by Mathspace last updated on 31/May/21

$$calculate{U}_n={\int }_0^{\mathrm{\infty }}\frac{{log}^{n}x}{1+x^n}dx$$$$findnatureoftheserie\mathrm{\Sigma }{U}_n$$

Answered by mathmax by abdo last updated on 12/Jun/21

$${\mathrm{U}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}\frac{{\log }^{\mathrm{n}}\left(\mathrm{x}\right)}{1+{\mathrm{x}}^{\mathrm{n}}}\mathrm{dx}{=}_{\mathrm{x}={\mathrm{t}}^{\frac{1}{\mathrm{n}}}}{\int }_0^{\mathrm{\infty }}\frac{{\log }^{\mathrm{n}}\left({\mathrm{t}}^{\frac{1}{\mathrm{n}}}\right)}{1+\mathrm{t}}\frac{1}{\mathrm{n}}{\mathrm{t}}^{\frac{1}{\mathrm{n}}-1}\mathrm{dt}$$$$=\frac{1}{{\mathrm{n}}^2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{n}}-1}}{1+\mathrm{t}}{\log }^{\mathrm{n}}\left(\mathrm{t}\right)\mathrm{dt}\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}}{1+\mathrm{t}}\mathrm{dt}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{(\mathrm{a}-1)\mathrm{logt}}}{1+\mathrm{t}}\mathrm{dt}\Rightarrow {\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\partial }^{\mathrm{n}}}{\partial {\mathrm{a}}^{\mathrm{n}}}\left(\frac{{\mathrm{e}}^{(\mathrm{a}-1)\mathrm{logt}}}{1+\mathrm{t}}\right)\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}{\log }^{\mathrm{n}}\left(\mathrm{t}\right)}{1+\mathrm{t}}\mathrm{dt}\Rightarrow {\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\frac{1}{\mathrm{n}}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{n}}-1}{\log }^{\mathrm{n}}\left(\mathrm{t}\right)}{1+\mathrm{t}}\mathrm{dt}\Rightarrow$$$${\mathrm{U}}_{\mathrm{n}}=\frac{1}{{\mathrm{n}}^2}{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\frac{1}{\mathrm{n}}\right)\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{\sin \left(\pi \mathrm{a}\right)}\Rightarrow$$$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{a}\right)=\pi {\left(\frac{1}{\sin \left(\pi \mathrm{a}\right)}\right)}^{\left(\mathrm{n}\right)}=\pi {\left(\frac{2\mathrm{i}}{{\mathrm{e}}^{\mathrm{i}\pi \mathrm{a}}-{\mathrm{e}}^{-\mathrm{i}\pi \mathrm{a}}}\right)}^{\left(\mathrm{n}\right)}$$$$=2\mathrm{i}\pi {\left(\frac{1}{{\mathrm{e}}^{\mathrm{i}\pi \mathrm{a}}-{\mathrm{e}}^{-\mathrm{i}\pi \mathrm{a}}}\right)}^{\left(\mathrm{n}\right)}.....\mathrm{be}\mathrm{continued}....$$

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