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Question Number 94340 by mathmax by abdo last updated on 20/May/20

$$1)\mathrm{calculate}{\mathrm{U}}_{\mathrm{n}}={\int }_0^1\ln \left(\mathrm{x}\right)\ln (1-\frac{\mathrm{x}}{\mathrm{n}})\mathrm{dx}(\mathrm{n}>0)$$ $$2)\mathrm{find}\mathrm{nature}\mathrm{of}\mathrm{\Sigma }{\mathrm{U}}_{\mathrm{n}}\mathrm{and}\mathrm{\Sigma }{\mathrm{nU}}_{\mathrm{n}}$$

Answered by abdomathmax last updated on 19/May/20

$$\mathrm{we}\mathrm{have}\sin \left(\mathrm{tx}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{(2\mathrm{n}+1)!}{\left(\mathrm{tx}\right)}^{2\mathrm{n}+1}\Rightarrow$$ $$\mathrm{f}\left(\mathrm{x}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2}\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{(2\mathrm{n}+1)!}{\mathrm{x}}^{2\mathrm{n}+1}{\mathrm{t}}^{2\mathrm{n}+1}\right)\mathrm{dt}$$ $$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}+1}}{(2\mathrm{n}+1)!}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2}{\mathrm{t}}^{2\mathrm{n}+1}\mathrm{dt}$$ $${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2}{\mathrm{t}}^{2\mathrm{n}+1}\mathrm{dt}{=}_{\mathrm{t}=\sqrt{\mathrm{u}}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{u}}{\left(\mathrm{u}\right)}^{\frac{2\mathrm{n}+1}{2}}\frac{\mathrm{du}}{2\sqrt{\mathrm{u}}}$$ $$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{u}}{\mathrm{u}}^{\mathrm{n}}\mathrm{du}\mathrm{we}\mathrm{know}\mathrm{\Gamma }\left(\mathrm{x}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{x}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\Rightarrow$$ $${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2}{\mathrm{t}}^{2\mathrm{n}+1}\mathrm{dt}=\frac{1}{2}\mathrm{\Gamma }(\mathrm{n}+1)=\frac{\mathrm{n}!}{2}\Rightarrow$$ $$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}\mathrm{n}!}{(2\mathrm{n}+1)!}{\mathrm{x}}^{2\mathrm{n}+1}$$ $$$$

Commented bymathmax by abdo last updated on 20/May/20

$$\mathrm{the}\mathrm{Q}\mathrm{here}\mathrm{is}\mathrm{developp}\mathrm{at}\mathrm{integr}\mathrm{serie}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2}\sin \left(\mathrm{tx}\right)\mathrm{dt}$$

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