Question Number 200444 by Calculusboy last updated on 18/Nov/23
Answered by Frix last updated on 19/Nov/23
$$\int\mathrm{e}^{−\mathrm{i}{x}^{\mathrm{2}} } {dx}\:\overset{{t}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} {x}} {=} \\ $$$$=\frac{\sqrt{\mathrm{2}}\left(\mathrm{1}−\mathrm{i}\right)}{\mathrm{2}}\int\mathrm{e}^{−{t}^{\mathrm{2}} } {dt}=\frac{\sqrt{\mathrm{2}\pi}\left(\mathrm{1}−\mathrm{i}\right)}{\mathrm{4}}\int\frac{\mathrm{2e}^{{t}^{\mathrm{3}} } }{\:\sqrt{\pi}}{dt}= \\ $$$$=\frac{\sqrt{\mathrm{2}\pi}\left(\mathrm{1}−\mathrm{i}\right)}{\mathrm{4}}\mathrm{erf}\:{t}\:= \\ $$$$=\frac{\sqrt{\mathrm{2}\pi}\left(\mathrm{1}−\mathrm{i}\right)}{\mathrm{4}}\mathrm{erf}\:\frac{\sqrt{\mathrm{2}}\left(\mathrm{1}+\mathrm{i}\right){x}}{\mathrm{2}}\:+{C} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−\mathrm{i}{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\mathrm{2}\pi}\left(\mathrm{1}−\mathrm{i}\right)}{\mathrm{4}} \\ $$
Commented by Calculusboy last updated on 19/Nov/23
$$\boldsymbol{{thanks}} \\ $$