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


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

$l e t u_{n} = \int_{- \infty}^{\infty} e^{- {n x}^{2} + x} d x$ $1) c a l c u l a t e u_{n}$ $2) f i n d \sum_{n} u_{n}$
Commented by maxmathsup by imad last updated on 05/Nov/18
$1) we have u_n =∫_(−∞) ^(+∞) e^(−{((√n)x)^2 −2 x (1/(2(√n))) +(1/(4n))−(1/(4n))}) dx =e^(1/(4n)) ∫_(−∞) ^(+∞) e^(−((√n)x−(1/((√n) )))^2 ) dx =_((√n)x−(1/(√n))=t) e^(1/(4n)) ∫_(−∞) ^(+∞) e^(−t^2 ) (dt/(√n)) =(e^(1/(4n)) /(√n)) (√π) ⇒ u_n =((√π)/(√n)) e^(1/(4n)) .$
$1) w e h a v e u_{n} = \int_{- \infty}^{+ \infty} e^{- {{(\sqrt{n} x)}^{2} - 2 x \frac{1}{2 \sqrt{n}} + \frac{1}{4 n} - \frac{1}{4 n}}} d x$ $= e^{\frac{1}{4 n}} \int_{- \infty}^{+ \infty} e^{- {(\sqrt{n} x - \frac{1}{\sqrt{n}})}^{2}} d x =_{\sqrt{n} x - \frac{1}{\sqrt{n}} = t} e^{\frac{1}{4 n}} \int_{- \infty}^{+ \infty} e^{- t^{2}} \frac{d t}{\sqrt{n}}$ $= \frac{e^{\frac{1}{4 n}}}{\sqrt{n}} \sqrt{π} \Rightarrow u_{n} = \frac{\sqrt{π}}{\sqrt{n}} e^{\frac{1}{4 n}} .$
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