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Question Number 47119 by math solver by abdo. last updated on 05/Nov/18

let u_n =∫_(−∞) ^∞ e^(−nx^2 +x) dx 1)calculate u_n 2)find Σ_n u_n

$${let}\:{u}_{{n}} =\int_{−\infty} ^{\infty} \:{e}^{−{nx}^{\mathrm{2}} +{x}} {dx} \\ $$$$\left.\mathrm{1}\right){calculate}\:{u}_{{n}} \\ $$$$\left.\mathrm{2}\right){find}\:\sum_{{n}} \:{u}_{{n}} \\ $$

Commented by maxmathsup by imad last updated on 05/Nov/18

1) we have u_n =∫_(−∞) ^(+∞) e^(−{((√n)x)^2 −2 x (1/(2(√n))) +(1/(4n))−(1/(4n))}) dx =e^(1/(4n)) ∫_(−∞) ^(+∞) e^(−((√n)x−(1/((√n) )))^2 ) dx =_((√n)x−(1/(√n))=t) e^(1/(4n)) ∫_(−∞) ^(+∞) e^(−t^2 ) (dt/(√n)) =(e^(1/(4n)) /(√n)) (√π) ⇒ u_n =((√π)/(√n)) e^(1/(4n)) .

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{u}_{{n}} =\int_{−\infty} ^{+\infty} \:{e}^{−\left\{\left(\sqrt{{n}}{x}\right)^{\mathrm{2}} \:\:−\mathrm{2}\:{x}\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{n}}}\:+\frac{\mathrm{1}}{\mathrm{4}{n}}−\frac{\mathrm{1}}{\mathrm{4}{n}}\right\}} {dx}\: \\ $$$$={e}^{\frac{\mathrm{1}}{\mathrm{4}{n}}} \:\int_{−\infty} ^{+\infty} \:\:{e}^{−\left(\sqrt{{n}}{x}−\frac{\mathrm{1}}{\sqrt{{n}}\:}\right)^{\mathrm{2}} } {dx}\:=_{\sqrt{{n}}{x}−\frac{\mathrm{1}}{\sqrt{{n}}}={t}} {e}^{\frac{\mathrm{1}}{\mathrm{4}{n}}} \:\int_{−\infty} ^{+\infty} \:{e}^{−{t}^{\mathrm{2}} } \:\frac{{dt}}{\sqrt{{n}}} \\ $$$$=\frac{{e}^{\frac{\mathrm{1}}{\mathrm{4}{n}}} }{\sqrt{{n}}}\:\sqrt{\pi}\:\Rightarrow\:{u}_{{n}} =\frac{\sqrt{\pi}}{\sqrt{{n}}}\:{e}^{\frac{\mathrm{1}}{\mathrm{4}{n}}} \:. \\ $$

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