Question Number 36919 by maxmathsup by imad last updated on 07/Jun/18
$${calculate}\:{f}\left(\alpha\right)=\:\int_{−\infty} ^{+\infty} \:\left(\mathrm{1}+\alpha{i}\right)^{−{x}^{\mathrm{2}} } {dx}\:. \\ $$
Commented by abdo.msup.com last updated on 10/Jun/18
$${we}\:{have}\:\mathrm{1}+\alpha{i}\:=\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }}\:+\frac{\alpha}{\:\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }}{i}\right) \\ $$$$={r}\:{e}^{{i}\theta} \:\:\Rightarrow\:{r}\:=\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }\:\:\:{and}\:\:{cos}\theta\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }} \\ $$$${sin}\theta\:=\:\frac{\alpha}{\:\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }}\:\Rightarrow{tan}\theta\:=\alpha\:\Rightarrow\theta={arctan}\left(\alpha\right) \\ $$$$\mathrm{1}+\alpha{i}\:={r}\:{e}^{{i}\:{arctan}\left(\alpha\right)} \:\Rightarrow \\ $$$${f}\left(\alpha\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\left({r}\:{e}^{{iarctan}\left(\alpha\right)} \right)^{−{x}^{\mathrm{2}} } {dx} \\ $$$$=\int_{−\infty} ^{+\infty} \:\left({e}^{{ln}\left({r}\right)+{iarctan}\left(\alpha\right)} \right)^{−{x}^{\mathrm{2}} } {dx} \\ $$$$=\int_{−\infty} ^{+\infty} \:\:{e}^{−\left({ln}\left(\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} \:}+{i}\:{arctan}\left(\alpha\right)\right)^{} {x}^{\mathrm{2}} \right.} {dx}\: \\ $$$${changement}\:\sqrt{{ln}\left(\sqrt{\left.\mathrm{1}+\alpha^{\mathrm{2}} \right)}+{i}\:{artan}\left(\alpha\right)\right.}{x}={t} \\ $$$${give} \\ $$$${f}\left(\alpha\right)=\:\frac{\mathrm{1}}{\:\sqrt{{ln}\left(\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }+{i}\:{arctan}\left(\alpha\right)\right.}}\:\int_{−\infty} ^{+\infty} \:{e}^{−{t}^{\mathrm{2}} } {dt} \\ $$$$=\:\:\frac{\sqrt{\pi}}{\:\sqrt{{ln}\left(\sqrt{\left.\mathrm{1}+\alpha^{\mathrm{2}} \right)}+{i}\:{arctan}\left(\alpha\right)\right.}}. \\ $$