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Question Number 132674 by aurpeyz last updated on 15/Feb/21

i n t e g r a l o f e^{- 2 {a x}^{2}}

Answered by Olaf last updated on 15/Feb/21

Let Ω_{a} (x) = \int_{0}^{x} e^{- 2 {a t}^{2}} d t

1) a = 0

Ω_{0} (x) = \int_{0}^{x} d t = x

2) a > 0

Let u = \sqrt{2 a} t

Ω_{a} (x) = \int_{0}^{\sqrt{2 a} x} e^{- u^{2}} . \frac{d u}{\sqrt{2 a}}

Ω_{a} (x) = \frac{\sqrt{π}}{2 \sqrt{2 a}} (\frac{2}{\sqrt{π}} \int_{0}^{\sqrt{2 a} x} e^{- u^{2}} d u)

Ω_{a} (x) = \sqrt{\frac{π}{8 a}} . \erf (\sqrt{2 a} x)

3) a < 0

Let u = i \sqrt{- 2 a} t

Ω_{a} (x) = \int_{0}^{i \sqrt{- 2 a} x} e^{- u^{2}} . \frac{d u}{i \sqrt{- 2 a}}

Ω_{a} (x) = \frac{\sqrt{π}}{2 i \sqrt{- 2 a}} (\frac{2}{\sqrt{π}} \int_{0}^{- i \sqrt{2 a} x} e^{- u^{2}} d u)

Ω_{a} (x) = \sqrt{- \frac{π}{8 a}} . \frac{\erf (i \sqrt{- 2 a} x)}{i}

Ω_{a} (x) = \sqrt{- \frac{π}{8 a}} . erfi (\sqrt{- 2 a} x)

Commented by aurpeyz last updated on 16/Feb/21

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