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Question Number 49344 by maxmathsup by imad last updated on 05/Dec/18

let α>0 calculate ∫_(−∞) ^(+∞) (1+αi)^(−x^2 ) dx .

$${let}\:\alpha>\mathrm{0}\:{calculate}\:\int_{−\infty} ^{+\infty} \:\left(\mathrm{1}+\alpha{i}\right)^{−{x}^{\mathrm{2}} } {dx}\:. \\ $$

Commented byAbdo msup. last updated on 06/Dec/18

let A =∫_(−∞) ^(+∞) (1+αi)^(−x^2 ) dx ⇒A =∫_(−∞) ^(+∞) e^(−x^2 ln(1+αi)) dx =_(x(√(ln(1+αi)))=t) ∫_(−∞) ^(+∞) e^(−t^2 ) (dt/(√(ln(1+αi)))) =((√π)/(√(ln(1+αi)))) but 1+αi =(√(1+α^2 ))((1/(√(1+α^2 ))) +i(α/(√(1+α^2 )))) =r e^(iθ) ⇒r =(1+α^2 )^(1/2) and tanθ =α ⇒θ =arctan(α) ⇒ ln(1+αi)=ln(r)+iθ =(1/2)ln(1+α^2 )+iarctan(α) ⇒ A = ((√π)/(√((1/2)ln(1+α^2 )+i arctan(α)))) leyZ=(1/2)ln(1+α^2 )+i arctan(α) we have ∣Z∣=(√((1/4)ln^2 (1+α^2 )+arctan^2 (α))) ⇒ Z =∣Z∣{((ln(1+α^2 ))/(2(√((1/4)ln^2 (1+α^2 )+arctan^2 (α))))) +i((arctan(α))/(√((1/4)ln^2 (1+α^2 )+arctan^2 (α))))} =r e^(iθ) ⇒r =∣Z∣ and tanθ = ((2arctan(α))/(ln(1+α^2 ))) ⇒ θ =arctan(2 ((arctan(α))/(ln(1+α^2 ))))...be continued...

$${let}\:{A}\:=\int_{−\infty} ^{+\infty} \left(\mathrm{1}+\alpha{i}\right)^{−{x}^{\mathrm{2}} } {dx}\:\Rightarrow{A}\:=\int_{−\infty} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+\alpha{i}\right)} {dx} \\ $$ $$=_{{x}\sqrt{{ln}\left(\mathrm{1}+\alpha{i}\right)}={t}} \:\:\:\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{t}^{\mathrm{2}} } \:\:\frac{{dt}}{\sqrt{{ln}\left(\mathrm{1}+\alpha{i}\right)}} \\ $$ $$=\frac{\sqrt{\pi}}{\sqrt{{ln}\left(\mathrm{1}+\alpha{i}\right)}}\:\:{but}\:\mathrm{1}+\alpha{i}\:=\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }}\:+{i}\frac{\alpha}{\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }}\right) \\ $$ $$={r}\:{e}^{{i}\theta} \:\Rightarrow{r}\:=\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:{and}\:\:{tan}\theta\:=\alpha\:\Rightarrow\theta\:={arctan}\left(\alpha\right)\:\Rightarrow \\ $$ $${ln}\left(\mathrm{1}+\alpha{i}\right)={ln}\left({r}\right)+{i}\theta\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)+{iarctan}\left(\alpha\right)\:\Rightarrow \\ $$ $${A}\:=\:\frac{\sqrt{\pi}}{\sqrt{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)+{i}\:{arctan}\left(\alpha\right)}}\:\:{leyZ}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)+{i}\:{arctan}\left(\alpha\right) \\ $$ $${we}\:{have}\:\mid{Z}\mid=\sqrt{\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{1}+\alpha^{\mathrm{2}} \right)+{arctan}^{\mathrm{2}} \left(\alpha\right)}\:\Rightarrow \\ $$ $${Z}\:=\mid{Z}\mid\left\{\frac{{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}{\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{1}+\alpha^{\mathrm{2}} \right)+{arctan}^{\mathrm{2}} \left(\alpha\right)}}\:+{i}\frac{{arctan}\left(\alpha\right)}{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{1}+\alpha^{\mathrm{2}} \right)+{arctan}^{\mathrm{2}} \left(\alpha\right)}}\right\} \\ $$ $$={r}\:{e}^{{i}\theta} \:\:\Rightarrow{r}\:=\mid{Z}\mid\:{and}\:{tan}\theta\:=\:\frac{\mathrm{2}{arctan}\left(\alpha\right)}{{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\:\Rightarrow \\ $$ $$\theta\:={arctan}\left(\mathrm{2}\:\frac{{arctan}\left(\alpha\right)}{{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\right)...{be}\:{continued}... \\ $$

Answered by Smail last updated on 06/Dec/18

(1+αi)^(−x^2 ) =e^(−x^2 ln(1+αi)) u=x(√(ln(1+αi))) dx=(du/(√(ln(1+αi)))) ∫_(−∞) ^∞ (1+αi)^(−x^2 ) dx=(1/(√(ln(1+αi))))∫_(−∞) ^∞ e^(−u^2 ) du =((√π)/(√(ln(1+αi))))

$$\left(\mathrm{1}+\alpha{i}\right)^{−{x}^{\mathrm{2}} } ={e}^{−{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+\alpha{i}\right)} \\ $$ $${u}={x}\sqrt{{ln}\left(\mathrm{1}+\alpha{i}\right)} \\ $$ $${dx}=\frac{{du}}{\sqrt{{ln}\left(\mathrm{1}+\alpha{i}\right)}} \\ $$ $$\int_{−\infty} ^{\infty} \left(\mathrm{1}+\alpha{i}\right)^{−{x}^{\mathrm{2}} } {dx}=\frac{\mathrm{1}}{\sqrt{{ln}\left(\mathrm{1}+\alpha{i}\right)}}\int_{−\infty} ^{\infty} {e}^{−{u}^{\mathrm{2}} } {du} \\ $$ $$=\frac{\sqrt{\pi}}{\sqrt{{ln}\left(\mathrm{1}+\alpha{i}\right)}} \\ $$

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