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Question Number 65455 by mathmax by abdo last updated on 30/Jul/19

$$find{U}_n={\int }_0^{+\mathrm{\infty }}\frac{{(-1)}^x}{2^{x^2}(x^2+4n^2)}dx(nfromNandn⩾1)$$$$studynatureoftheserie\mathrm{\Sigma }{2}^{n^2}{U}_n$$

Commented by mathmax by abdo last updated on 31/Jul/19

$$wehave{U}_n={\int }_0^{\mathrm{\infty }}\frac{2^{-x^2}{(-1)}^x}{(x^2+4n^2)}={\int }_0^{\mathrm{\infty }}\frac{2^{-x^2}{e}^{i\pi x}}{(x^2+4n^2)}$$$$=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{2^{-x^2}{e}^{i\pi x}}{x^2+4n^2}let\phi \left(z\right)=\frac{2^{-z^2}{e}^{i\pi z}}{z^2+4n^2}\Rightarrow \phi \left(z\right)=\frac{2^{-z^2}{e}^{i\pi z}}{(z-2ni)(z+2ni)}$$$$thepolesof\phi are2niand-2niresidustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,2ni)$$$$Res(\phi ,2ni)={lim}_{z\to 2ni}(z-2ni)\phi \left(z\right)=\frac{2^{-{\left(2ni\right)}^2}{e}^{i\pi \left(2ni\right)}}{4ni}$$$$=\frac{2^{4n^2}{e}^{-2n\pi }}{4ni}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{2^{4n^2}{e}^{-2n\pi }}{4ni}=\frac{\pi }{2n}{2}^{4n^2}{e}^{-2n\pi }\Rightarrow$$$$U_n=\frac{\pi }{4n}{2}^{4n^2}{e}^{-2n\pi }$$

Commented by mathmax by abdo last updated on 31/Jul/19

$$2^{n^2}{U}_n=\frac{\pi }{4n}{2}^{5n^2}{e}^{-2n\pi }=\frac{\pi }{4n}{e}^{5n^{2}ln\left(2\right)-2n\pi }=\frac{\pi }{4n}{e}^{n^2(5ln\left(2\right)-\frac{2\pi }{n})}$$$${lim}_{n\to +\mathrm{\infty }}{2}^{n^2}{U}_n=+\mathrm{\infty }\Rightarrow \mathrm{\Sigma }{2}^{n^2}{U}_{n}diverges$$

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