IS-THIS-RIGHT-I-0-e-ix-2-dx-I-2-0-e-ix-2-dx-0-e-iy-2-dy-0-0-e-iy-2-dye-ix-2-dx-0-0-e-i-x-2-y-2-dydx-dydx-dA-rdrd-I-2-0-pi-2-0-

Question Number 193316 by gatocomcirrose last updated on 10/Jun/23

$$ \\ $$$$\mathrm{IS}\:\mathrm{THIS}\:\mathrm{RIGHT}? \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{ix}^{\mathrm{2}} } \mathrm{dx} \\ $$$$\mathrm{I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{ix}^{\mathrm{2}} } \mathrm{dx}\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{iy}^{\mathrm{2}} } \mathrm{dy} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{iy}^{\mathrm{2}} } \mathrm{dye}^{−\mathrm{ix}^{\mathrm{2}} } \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{i}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)} \mathrm{dydx} \\ $$$$\mathrm{dydx}=\mathrm{dA}=\mathrm{rdrd}\theta \\ $$$$\mathrm{I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{ir}^{\mathrm{2}} } \mathrm{rdrd}\theta \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{ir}^{\mathrm{2}} } \mathrm{rdr};\:\mathrm{u}=−\mathrm{ir}^{\mathrm{2}} \Rightarrow\mathrm{du}=−\mathrm{2irdr} \\ $$$$−\frac{\mathrm{i}}{\mathrm{2}}\int_{−\infty} ^{\mathrm{0}} \mathrm{e}^{\mathrm{u}} \mathrm{du}=−\frac{\mathrm{i}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\frac{\mathrm{i}}{\mathrm{2}}\mathrm{d}\theta=−\frac{\mathrm{i}\pi}{\mathrm{4}}\Rightarrow\mathrm{I}=\sqrt{\frac{−\mathrm{i}\pi}{\mathrm{4}}} \\ $$$$\mathrm{I}=\frac{\mathrm{i}}{\mathrm{2}}\sqrt{\mathrm{e}^{\mathrm{i}\pi/\mathrm{2}} \pi}=\frac{\mathrm{ie}^{\mathrm{i}\pi/\mathrm{4}} }{\mathrm{2}}\sqrt{\pi} \\ $$$$\mathrm{I}=\frac{\mathrm{i}\sqrt{\pi}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{i}\right)\right)=\frac{\mathrm{i}\sqrt{\mathrm{2}\pi}}{\mathrm{4}}\left(\mathrm{1}+\mathrm{i}\right) \\ $$

Commented by aba last updated on 10/Jun/23

$$\mathrm{wrong}\:\mathrm{I}=−\frac{\mathrm{i}\sqrt{\mathrm{2}\pi}}{\mathrm{4}}\left(\mathrm{i}+\mathrm{1}\right) \\ $$

Answered by aba last updated on 10/Jun/23

$$\mathrm{I}=\sqrt{\frac{−\mathrm{i}\pi}{\mathrm{4}}}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{e}^{−\mathrm{i}\frac{\pi}{\mathrm{2}}} \pi}=\frac{\mathrm{e}^{−\mathrm{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{2}}\sqrt{\pi}=\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{4}}\left(\mathrm{1}−\mathrm{i}\right)=−\frac{\mathrm{i}\sqrt{\mathrm{2}\pi}}{\mathrm{4}}\left(\mathrm{i}+\mathrm{1}\right)=\sqrt{\frac{\pi}{\mathrm{2}}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{i}}{\mathrm{2}}\right)\:\checkmark \\ $$

Answered by Mathspace last updated on 15/Jun/23

$${I}=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ix}^{\mathrm{2}} } {dx}\:\:{we}\:{do}\:\sqrt{{i}}{x}={t}\:\Rightarrow \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{2}} } \frac{{dt}}{\:\sqrt{{i}}}\:\:{or}\:{i}={e}^{\frac{{i}\pi}{\mathrm{2}}} \Rightarrow \\ $$$$\sqrt{{i}}={e}^{\frac{{i}\pi}{\mathrm{4}}} \Rightarrow{I}={e}^{−\frac{{i}\pi}{\mathrm{4}}} .\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{2}} } {dt} \\ $$$$=\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{i}\right)×\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\mathrm{1}−{i}\right) \\ $$

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