Question Number 49344 by maxmathsup by imad last updated on 05/Dec/18
$${let}\:\alpha>\mathrm{0}\:{calculate}\:\int_{−\infty} ^{+\infty} \:\left(\mathrm{1}+\alpha{i}\right)^{−{x}^{\mathrm{2}} } {dx}\:. \\ $$
Commented by Abdo msup. last updated on 06/Dec/18
$${let}\:{A}\:=\int_{−\infty} ^{+\infty} \left(\mathrm{1}+\alpha{i}\right)^{−{x}^{\mathrm{2}} } {dx}\:\Rightarrow{A}\:=\int_{−\infty} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+\alpha{i}\right)} {dx} \\ $$$$=_{{x}\sqrt{{ln}\left(\mathrm{1}+\alpha{i}\right)}={t}} \:\:\:\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{t}^{\mathrm{2}} } \:\:\frac{{dt}}{\:\sqrt{{ln}\left(\mathrm{1}+\alpha{i}\right)}} \\ $$$$=\frac{\sqrt{\pi}}{\:\sqrt{{ln}\left(\mathrm{1}+\alpha{i}\right)}}\:\:{but}\:\mathrm{1}+\alpha{i}\:=\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }}\:+{i}\frac{\alpha}{\:\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }}\right) \\ $$$$={r}\:{e}^{{i}\theta} \:\Rightarrow{r}\:=\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:{and}\:\:{tan}\theta\:=\alpha\:\Rightarrow\theta\:={arctan}\left(\alpha\right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+\alpha{i}\right)={ln}\left({r}\right)+{i}\theta\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)+{iarctan}\left(\alpha\right)\:\Rightarrow \\ $$$${A}\:=\:\frac{\sqrt{\pi}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)+{i}\:{arctan}\left(\alpha\right)}}\:\:{leyZ}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)+{i}\:{arctan}\left(\alpha\right) \\ $$$${we}\:{have}\:\mid{Z}\mid=\sqrt{\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{1}+\alpha^{\mathrm{2}} \right)+{arctan}^{\mathrm{2}} \left(\alpha\right)}\:\Rightarrow \\ $$$${Z}\:=\mid{Z}\mid\left\{\frac{{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}{\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{1}+\alpha^{\mathrm{2}} \right)+{arctan}^{\mathrm{2}} \left(\alpha\right)}}\:+{i}\frac{{arctan}\left(\alpha\right)}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{1}+\alpha^{\mathrm{2}} \right)+{arctan}^{\mathrm{2}} \left(\alpha\right)}}\right\} \\ $$$$={r}\:{e}^{{i}\theta} \:\:\Rightarrow{r}\:=\mid{Z}\mid\:{and}\:{tan}\theta\:=\:\frac{\mathrm{2}{arctan}\left(\alpha\right)}{{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$$\theta\:={arctan}\left(\mathrm{2}\:\frac{{arctan}\left(\alpha\right)}{{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\right)…{be}\:{continued}… \\ $$
Answered by Smail last updated on 06/Dec/18
$$\left(\mathrm{1}+\alpha{i}\right)^{−{x}^{\mathrm{2}} } ={e}^{−{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+\alpha{i}\right)} \\ $$$${u}={x}\sqrt{{ln}\left(\mathrm{1}+\alpha{i}\right)} \\ $$$${dx}=\frac{{du}}{\:\sqrt{{ln}\left(\mathrm{1}+\alpha{i}\right)}} \\ $$$$\int_{−\infty} ^{\infty} \left(\mathrm{1}+\alpha{i}\right)^{−{x}^{\mathrm{2}} } {dx}=\frac{\mathrm{1}}{\:\sqrt{{ln}\left(\mathrm{1}+\alpha{i}\right)}}\int_{−\infty} ^{\infty} {e}^{−{u}^{\mathrm{2}} } {du} \\ $$$$=\frac{\sqrt{\pi}}{\:\sqrt{{ln}\left(\mathrm{1}+\alpha{i}\right)}} \\ $$