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

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

$$\mathrm{Calculer}\:\underset{\:\frac{\mathrm{1}}{\alpha}} {\int}^{\:+\infty} {e}^{−\alpha{t}^{\mathrm{2}} +\mathrm{2}{t}} {dt} \\ $$

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)

$${t}=\frac{\mathrm{1}}{\alpha}+{x}\:\Rightarrow \\ $$$$\int_{\frac{\mathrm{1}}{\alpha}} ^{\infty} \:{e}^{−\alpha{t}^{\mathrm{2}} +\mathrm{2}{t}} {dt}=\int_{\mathrm{0}} ^{\infty} {e}^{−\alpha\left(\frac{\mathrm{1}}{\alpha}+{x}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{\mathrm{1}}{\alpha}+{x}\right)} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{−\alpha\left\{\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }+\frac{\mathrm{2}{x}}{\alpha}+{x}^{\mathrm{2}} \right\}+\frac{\mathrm{2}}{\alpha}+\mathrm{2}{x}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\frac{\mathrm{1}}{\alpha}−\mathrm{2}{x}−\alpha{x}^{\mathrm{2}} +\frac{\mathrm{2}}{\alpha}+\mathrm{2}{x}} {dx} \\ $$$$={e}^{\frac{\mathrm{1}}{\alpha}} \:\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\alpha{x}^{\mathrm{2}} } {dx}\:\:\left(\sqrt{\alpha}{x}={u}\right) \\ $$$$={e}^{\frac{\mathrm{1}}{\alpha}} \int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}^{\mathrm{2}} } \frac{{du}}{\:\sqrt{\alpha}} \\ $$$$=\frac{{e}^{\frac{\mathrm{1}}{\alpha}} }{\:\sqrt{\alpha}}.\frac{\sqrt{\pi}}{\mathrm{2}}\:\Rightarrow \\ $$$${I}=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\alpha}}\:{e}^{\frac{\mathrm{1}}{\alpha}} \:\:\:\left(\alpha>\mathrm{0}\right) \\ $$

Commented by Mathspace last updated on 20/Aug/23

sorry I=((√π)/(2(√α))) e^(−(1/α))

$${sorry}\:{I}=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\alpha}}\:{e}^{−\frac{\mathrm{1}}{\alpha}} \\ $$

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