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Question Number 200444 by Calculusboy last updated on 18/Nov/23

Answered by Frix last updated on 19/Nov/23

$$\int {\mathrm{e}}^{-\mathrm{i}{x}^2}dx\stackrel{t={\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}x}{=}$$$$=\frac{\sqrt{2}(1-\mathrm{i})}{2}\int {\mathrm{e}}^{-t^2}dt=\frac{\sqrt{2\pi }(1-\mathrm{i})}{4}\int \frac{{2\mathrm{e}}^{t^3}}{\sqrt{\pi }}dt=$$$$=\frac{\sqrt{2\pi }(1-\mathrm{i})}{4}\mathrm{erf}t=$$$$=\frac{\sqrt{2\pi }(1-\mathrm{i})}{4}\mathrm{erf}\frac{\sqrt{2}(1+\mathrm{i})x}{2}+C$$$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-\mathrm{i}{x}^2}dx=\frac{\sqrt{2\pi }(1-\mathrm{i})}{4}$$

Commented by Calculusboy last updated on 19/Nov/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}$$

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