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Question Number 6582 by Temp last updated on 04/Jul/16

$$\int e^{-{ix}^2}dx=??$$

Answered by Yozzii last updated on 04/Jul/16

$$e^u=\underset{r=0}{\sum ^{\mathrm{\infty }}}\frac{u^r}{r!}foranyu\in \mathbb{C}.$$$$\Rightarrow e^{-{ix}^2}=\underset{r=0}{\sum ^{\mathrm{\infty }}}\frac{{(-{ix}^2)}^r}{r!}=\underset{r=0}{\sum ^{\mathrm{\infty }}}\frac{{(-i)}^r{x}^{2r}}{r!}$$$$\Rightarrow \int e^{-{ix}^2}dx=\underset{r=0}{\sum ^{\mathrm{\infty }}}\frac{{(-i)}^r{x}^{2r+1}}{r!(2r+1)}+c$$$$$$

Commented by Temp last updated on 04/Jul/16

$$\mathrm{I}\mathrm{just}\mathrm{posted}\mathrm{a}\mathrm{more}\mathrm{specific}\mathrm{question}$$

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