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Question Number 60263 by maxmathsup by imad last updated on 19/May/19

$$let{U}_n={\int }_0^{\mathrm{\infty }}\frac{e^{-n\left[x^2\right]}}{x^2+3}dx$$$$1)calculate{U}_{n}intermsofn$$$$2)find{lim}_{n\to +\mathrm{\infty }}n{U}_n$$$$3)determinenatureoftheserie\mathrm{\Sigma }{U}_n$$

Commented by maxmathsup by imad last updated on 20/May/19

$$1)wehave2U_n={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{-n\left[x^2\right]}}{x^2+3}dxlet\phi \left(z\right)=\frac{e^{-n\left[z^2\right]}}{z^2+3}wehave$$$$\phi \left(z\right)=\frac{e^{-n\left[z^2\right]}}{(z-i\sqrt{3})(z+i\sqrt{3})}sothepolesof\phi are\stackrel{-}{+}i\sqrt{3}residustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i\sqrt{3}3$$$$Res(\phi ,i)={lim}_{z\to i\sqrt{3}}(z-i\sqrt{3})\phi \left(z\right)={lim}_{z\to i}\frac{e^{-n\left[z^2\right]}}{z+i\sqrt{3}}=\frac{e^{-n[-3]}}{2i\sqrt{3}}$$$$=\frac{e^{3n}}{2i\sqrt{3}}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{e^{3n}}{2i\sqrt{3}}=\frac{\pi }{\sqrt{3}}{e}^{3n}\Rightarrow U_n=\frac{\pi }{2\sqrt{3}}{e}^{3n}.$$$$2){lim}_{n\to +\mathrm{\infty }}{nU}_n=+\mathrm{\infty }$$$$3)itsclearthat\mathrm{\Sigma }{U}_{n}diverges..$$

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