Calculer ∫^( +∞) _( (1/α)) e^(−αt^2 +2t) dt


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Question Number 196145 by Erico last updated on 18/Aug/23

$Calculer {\int_{\frac{1}{α}}}^{+ \infty} e^{- α t^{2} + 2 t} d t$
	Answered by Mathspace last updated on 20/Aug/23
	$t=(1/α)+x ⇒ ∫_(1/α) ^∞ e^(−αt^2 +2t) dt=∫_0 ^∞ e^(−α((1/α)+x)^2 +2((1/α)+x)) dx =∫_0 ^∞ e^(−α{(1/α^2 )+((2x)/α)+x^2 }+(2/α)+2x) dx =∫_0 ^∞ e^(−(1/α)−2x−αx^2 +(2/α)+2x) dx =e^(1/α) ∫_0 ^∞ e^(−αx^2 ) dx ((√α)x=u) =e^(1/α) ∫_0 ^∞ e^(−u^2 ) (du/( (√α))) =(e^(1/α) /( (√α))).((√π)/2) ⇒ I=((√π)/(2(√α))) e^(1/α) (α>0)$
	$t = \frac{1}{α} + x \Rightarrow$ $\int_{\frac{1}{α}}^{\infty} e^{- α t^{2} + 2 t} d t = \int_{0}^{\infty} e^{- α {(\frac{1}{α} + x)}^{2} + 2 (\frac{1}{α} + x)} d x$ $= \int_{0}^{\infty} e^{- α {\frac{1}{α^{2}} + \frac{2 x}{α} + x^{2}} + \frac{2}{α} + 2 x} d x$ $= \int_{0}^{\infty} e^{- \frac{1}{α} - 2 x - α x^{2} + \frac{2}{α} + 2 x} d x$ $= e^{\frac{1}{α}} \int_{0}^{\infty} e^{- α x^{2}} d x (\sqrt{α} x = u)$ $= e^{\frac{1}{α}} \int_{0}^{\infty} e^{- u^{2}} \frac{d u}{\sqrt{α}}$ $= \frac{e^{\frac{1}{α}}}{\sqrt{α}} . \frac{\sqrt{π}}{2} \Rightarrow$ $I = \frac{\sqrt{π}}{2 \sqrt{α}} e^{\frac{1}{α}} (α > 0)$
	Commented by Mathspace last updated on 20/Aug/23

	$s o r r y I = \frac{\sqrt{π}}{2 \sqrt{α}} e^{- \frac{1}{α}}$
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