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Question Number 36919 by maxmathsup by imad last updated on 07/Jun/18

$$calculatef\left(\alpha \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{(1+\alpha i)}^{-x^2}dx.$$

Commented by abdo.msup.com last updated on 10/Jun/18

$$wehave1+\alpha i=\sqrt{1+{\alpha }^2}(\frac{1}{\sqrt{1+{\alpha }^2}}+\frac{\alpha }{\sqrt{1+{\alpha }^2}}i)$$$$=r{e}^{i\theta }\Rightarrow r=\sqrt{1+{\alpha }^2}andcos\theta =\frac{1}{\sqrt{1+{\alpha }^2}}$$$$sin\theta =\frac{\alpha }{\sqrt{1+{\alpha }^2}}\Rightarrow tan\theta =\alpha \Rightarrow \theta =arctan\left(\alpha \right)$$$$1+\alpha i=r{e}^{iarctan\left(\alpha \right)}\Rightarrow$$$$f\left(\alpha \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\left(r{e}^{iarctan\left(\alpha \right)}\right)}^{-x^2}dx$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\left(e^{ln\left(r\right)+iarctan\left(\alpha \right)}\right)}^{-x^2}dx$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-(ln{(\sqrt{1+{\alpha }^2}+iarctan\left(\alpha \right))}^{}{x}^2}dx$$$$changement\sqrt{ln(\sqrt{1+{\alpha }^2)}+iartan\left(\alpha \right)}x=t$$$$give$$$$f\left(\alpha \right)=\frac{1}{\sqrt{ln(\sqrt{1+{\alpha }^2}+iarctan\left(\alpha \right)}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-t^2}dt$$$$=\frac{\sqrt{\pi }}{\sqrt{ln(\sqrt{1+{\alpha }^2)}+iarctan\left(\alpha \right)}}.$$

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