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e-ix-2-dx-




Question Number 6582 by Temp last updated on 04/Jul/16
∫e^(−ix^2 ) dx=??
$$\int{e}^{−{ix}^{\mathrm{2}} } {dx}=?? \\ $$
Answered by Yozzii last updated on 04/Jul/16
e^u =Σ_(r=0) ^∞ (u^r /(r!)) for any u∈C.  ⇒e^(−ix^2 ) =Σ_(r=0) ^∞ (((−ix^2 )^r )/(r!))=Σ_(r=0) ^∞ (((−i)^r x^(2r) )/(r!))  ⇒∫e^(−ix^2 ) dx=Σ_(r=0) ^∞ (((−i)^r x^(2r+1) )/(r!(2r+1)))+c
$${e}^{{u}} =\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{u}^{{r}} }{{r}!}\:{for}\:{any}\:{u}\in\mathbb{C}. \\ $$$$\Rightarrow{e}^{−{ix}^{\mathrm{2}} } =\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−{ix}^{\mathrm{2}} \right)^{{r}} }{{r}!}=\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−{i}\right)^{{r}} {x}^{\mathrm{2}{r}} }{{r}!} \\ $$$$\Rightarrow\int{e}^{−{ix}^{\mathrm{2}} } {dx}=\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−{i}\right)^{{r}} {x}^{\mathrm{2}{r}+\mathrm{1}} }{{r}!\left(\mathrm{2}{r}+\mathrm{1}\right)}+{c} \\ $$$$ \\ $$
Commented by Temp last updated on 04/Jul/16
I just posted a more specific question
$$\mathrm{I}\:\mathrm{just}\:\mathrm{posted}\:\mathrm{a}\:\mathrm{more}\:\mathrm{specific}\:\mathrm{question} \\ $$

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