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Question Number 36916 by maxmathsup by imad last updated on 07/Jun/18

let z=r e^(iθ) fins f(z) = ∫_(−∞) ^(+∞) z^(−x^2 ) dx

$${let}\:{z}={r}\:{e}^{{i}\theta} \:\:\:\:{fins}\:{f}\left({z}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:{z}^{−{x}^{\mathrm{2}} } {dx} \\ $$

Commented by math khazana by abdo last updated on 09/Jun/18

f(z) = ∫_(−∞) ^(+∞) (r e^(iθ) )^(−x^2 ) dx =∫_(−∞) ^(+∞) { e^(ln(r) +iθ) }^(−x^2 ) dx = ∫_(−∞) ^(+∞) e^(−(ln(r) +iθ)x^2 ) dx but changement (√(ln(r) +iθ))x =t give f(z) = (1/(√(ln(r) +iθ))) ∫_(−∞) ^(+∞) e^(−t^2 ) dt =((√π)/(√(ln(r)+iθ))) .

$${f}\left({z}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\left({r}\:{e}^{{i}\theta} \right)^{−{x}^{\mathrm{2}} } {dx} \\ $$$$=\int_{−\infty} ^{+\infty} \:\:\left\{\:{e}^{{ln}\left({r}\right)\:+{i}\theta} \right\}^{−{x}^{\mathrm{2}} } {dx} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:{e}^{−\left({ln}\left({r}\right)\:+{i}\theta\right){x}^{\mathrm{2}} \:} \:{dx}\:\:{but}\:{changement}\: \\ $$$$\sqrt{{ln}\left({r}\right)\:+{i}\theta}{x}\:={t}\:{give} \\ $$$${f}\left({z}\right)\:=\:\frac{\mathrm{1}}{\sqrt{{ln}\left({r}\right)\:+{i}\theta}}\:\int_{−\infty} ^{+\infty} \:\:\:{e}^{−{t}^{\mathrm{2}} } {dt} \\ $$$$=\frac{\sqrt{\pi}}{\sqrt{{ln}\left({r}\right)+{i}\theta}}\:. \\ $$

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