Question Number 36915 by maxmathsup by imad last updated on 07/Jun/18
$${let}\:{z}\:={a}+{ib}\:\:\:{find}\:\:{f}\left({z}\right)\:=\:\int_{−\infty} ^{+\infty} \:{z}^{−{x}^{\mathrm{2}} } {dx} \\ $$
Commented by math khazana by abdo last updated on 02/Aug/18
$${we}\:{have}\:\mid{z}\mid=\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }\:\Rightarrow{z}\:=\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }\left(\:\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:+\frac{{b}}{\:\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}\right)\Rightarrow \\ $$$$={r}\:{e}^{{i}\theta} \:\Rightarrow\:{r}\:=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:{and}\:{tan}\theta=\frac{{b}}{{a}}\left({we}\:{take}\:{a}\neq{o}\right)\Rightarrow \\ $$$$\theta={arctan}\left(\frac{{b}}{{a}}\right)\:\Rightarrow{z}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{e}^{{i}\:{arctan}\left(\frac{{b}}{{a}}\right)} \:\Rightarrow \\ $$$${f}\left({z}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\left({re}^{{i}\:\theta} \right)^{−{x}^{\mathrm{2}} } {dx} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\left({e}^{{ln}\left({r}\right)+{i}\:\theta} \right)^{−{x}^{\mathrm{2}} } {dx} \\ $$$$\:\int_{−\infty} ^{+\infty} \:\:\:{e}^{\left.−\left({ln}\left({r}\right)+{i}\theta\right)\right){x}^{\mathrm{2}} } {dx}\:{changement} \\ $$$$\sqrt{{ln}\left({r}\right)+{i}\theta}{x}={u}\:{give} \\ $$$${f}\left({z}\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\:{e}^{−{u}^{\mathrm{2}} } \:\:\:\frac{{du}}{\:\sqrt{{ln}\left({r}\right)+{i}\theta}} \\ $$$$=\frac{\pi}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\:+{iarctan}\left(\frac{{b}}{{a}}\right)}}\:\:{wih}\:{z}={a}+{ib} \\ $$$${and}\:{a}\neq\mathrm{0} \\ $$