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

find U_n = ∫_0 ^(+∞) (((−1)^x )/(2^x^2 (x^2 +4n^2 )))dx (n from N and n≥1) study nature of the serie Σ 2^n^2 U_n

$${find}\:{U}_{{n}} =\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{x}} }{\mathrm{2}^{{x}^{\mathrm{2}} } \left({x}^{\mathrm{2}} \:+\mathrm{4}{n}^{\mathrm{2}} \right)}{dx}\:\:\:\:\:\left({n}\:{from}\:{N}\:{and}\:{n}\geqslant\mathrm{1}\right) \\ $$$${study}\:{nature}\:{of}\:{the}\:{serie}\:\:\Sigma\:\mathrm{2}^{{n}^{\mathrm{2}} } {U}_{{n}} \\ $$

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

we have U_n =∫_0 ^∞ ((2^(−x^2 ) (−1)^x )/((x^2 +4n^2 ))) =∫_0 ^∞ ((2^(−x^2 ) e^(iπx) )/((x^2 +4n^2 ))) =(1/2) ∫_(−∞) ^(+∞) ((2^(−x^2 ) e^(iπx) )/(x^2 +4n^2 )) let ϕ(z) =((2^(−z^2 ) e^(iπz) )/(z^2 +4n^2 )) ⇒ϕ(z) =((2^(−z^2 ) e^(iπz) )/((z−2ni)(z+2ni))) the poles of ϕ are 2ni and −2ni residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,2ni) Res(ϕ,2ni) =lim_(z→2ni) (z−2ni)ϕ(z) =((2^(−(2ni)^2 ) e^(iπ(2ni)) )/(4ni)) =((2^(4n^2 ) e^(−2nπ) )/(4ni)) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((2^(4n^2 ) e^(−2nπ) )/(4ni)) =(π/(2n)) 2^(4n^2 ) e^(−2nπ) ⇒ U_n =(π/(4n)) 2^(4n^2 ) e^(−2nπ)

$${we}\:{have}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}^{−{x}^{\mathrm{2}} } \left(−\mathrm{1}\right)^{{x}} }{\left({x}^{\mathrm{2}} \:+\mathrm{4}{n}^{\mathrm{2}} \right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}^{−{x}^{\mathrm{2}} } {e}^{{i}\pi{x}} }{\left({x}^{\mathrm{2}} \:+\mathrm{4}{n}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\frac{\mathrm{2}^{−{x}^{\mathrm{2}} } {e}^{{i}\pi{x}} }{{x}^{\mathrm{2}} \:+\mathrm{4}{n}^{\mathrm{2}} }\:\:{let}\:\varphi\left({z}\right)\:=\frac{\mathrm{2}^{−{z}^{\mathrm{2}} } {e}^{{i}\pi{z}} }{{z}^{\mathrm{2}} \:+\mathrm{4}{n}^{\mathrm{2}} }\:\Rightarrow\varphi\left({z}\right)\:=\frac{\mathrm{2}^{−{z}^{\mathrm{2}} } {e}^{{i}\pi{z}} }{\left({z}−\mathrm{2}{ni}\right)\left({z}+\mathrm{2}{ni}\right)} \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:\mathrm{2}{ni}\:{and}\:−\mathrm{2}{ni}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\mathrm{2}{ni}\right) \\ $$$${Res}\left(\varphi,\mathrm{2}{ni}\right)\:={lim}_{{z}\rightarrow\mathrm{2}{ni}} \:\:\left({z}−\mathrm{2}{ni}\right)\varphi\left({z}\right)\:=\frac{\mathrm{2}^{−\left(\mathrm{2}{ni}\right)^{\mathrm{2}} } {e}^{{i}\pi\left(\mathrm{2}{ni}\right)} }{\mathrm{4}{ni}} \\ $$$$=\frac{\mathrm{2}^{\mathrm{4}{n}^{\mathrm{2}} } {e}^{−\mathrm{2}{n}\pi} }{\mathrm{4}{ni}}\:\Rightarrow\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\mathrm{2}^{\mathrm{4}{n}^{\mathrm{2}} } {e}^{−\mathrm{2}{n}\pi} }{\mathrm{4}{ni}}\:=\frac{\pi}{\mathrm{2}{n}}\:\mathrm{2}^{\mathrm{4}{n}^{\mathrm{2}} } {e}^{−\mathrm{2}{n}\pi} \:\Rightarrow \\ $$$${U}_{{n}} =\frac{\pi}{\mathrm{4}{n}}\:\mathrm{2}^{\mathrm{4}{n}^{\mathrm{2}} } \:{e}^{−\mathrm{2}{n}\pi} \\ $$

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

2^n^2 U_n =(π/(4n)) 2^(5n^2 ) e^(−2nπ) =(π/(4n)) e^(5n^2 ln(2)−2nπ) =(π/(4n)) e^(n^2 (5ln(2)−((2π)/n))) lim_(n→+∞) 2^n^2 U_n =+∞ ⇒Σ 2^n^2 U_n diverges

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

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