Question Number 36917 by maxmathsup by imad last updated on 07/Jun/18
$${calculate}\:\int_{\mathrm{0}} ^{+\infty} \left(\mathrm{1}+{i}\right)^{−{x}^{\mathrm{2}} } {dx} \\ $$
Commented by math khazana by abdo last updated on 09/Jun/18
$${we}\:{have}\:\left(\mathrm{1}+{i}\right)=\sqrt{\mathrm{2}}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\left(\sqrt{\mathrm{2}}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)^{−{x}^{\mathrm{2}} } {dx} \\ $$$$=\:\int_{\mathrm{0}} ^{+\infty} \:\left\{{e}^{{ln}\left(\sqrt{\mathrm{2}}\right)+{i}\frac{\pi}{\mathrm{4}}} \right\}^{−{x}^{\mathrm{2}} } {dx} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \:\:{e}^{−\left\{{ln}\left(\sqrt{\mathrm{2}}\right)\:+{i}\frac{\pi}{\mathrm{4}}\right\}{x}^{\mathrm{2}} } {dx}\:\:{changement} \\ $$$$\sqrt{{ln}\left(\sqrt{\mathrm{2}}\right)+{i}\frac{\pi}{\mathrm{4}}}\:{x}\:={t}\:{give} \\ $$$${I}\:\:=\:\frac{\mathrm{1}}{\:\sqrt{{ln}\left(\sqrt{\mathrm{2}}\right)\:+{i}\frac{\pi}{\mathrm{4}}}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:{e}^{−{u}^{\mathrm{2}} } {du} \\ $$$${I}\:=\:\frac{\sqrt{\pi}}{\mathrm{2}\left\{\sqrt{\:{ln}\left(\sqrt{\mathrm{2}}\right)+{i}\frac{\pi}{\mathrm{4}}}\right\}}\:\:. \\ $$