e-ix-2-dx-

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

\int e^{- {i x}^{2}} d x = ? ?

Answered by Yozzii last updated on 04/Jul/16

e^{u} = \underset{r = 0}{\sum^{\infty}} \frac{u^{r}}{r!} f o r a n y u \in C .

\Rightarrow e^{- {i x}^{2}} = \underset{r = 0}{\sum^{\infty}} \frac{{(- {i x}^{2})}^{r}}{r!} = \underset{r = 0}{\sum^{\infty}} \frac{{(- i)}^{r} x^{2 r}}{r!}

\Rightarrow \int e^{- {i x}^{2}} d x = \underset{r = 0}{\sum^{\infty}} \frac{{(- i)}^{r} x^{2 r + 1}}{r! (2 r + 1)} + c

Commented by Temp last updated on 04/Jul/16

I just posted a more specific question

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