Question Number 132674 by aurpeyz last updated on 15/Feb/21
$${integral}\:{of}\:{e}^{−\mathrm{2}{ax}^{\mathrm{2}} } \\ $$
Answered by Olaf last updated on 15/Feb/21
$$\mathrm{Let}\:\Omega_{{a}} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{{x}} {e}^{−\mathrm{2}{at}^{\mathrm{2}} } {dt} \\ $$$$\left.\mathrm{1}\right)\:{a}\:=\:\mathrm{0} \\ $$$$\Omega_{\mathrm{0}} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{{x}} {dt}\:=\:{x} \\ $$$$\left.\mathrm{2}\right)\:{a}\:>\:\mathrm{0} \\ $$$$\mathrm{Let}\:{u}\:=\:\sqrt{\mathrm{2}{a}}{t} \\ $$$$\Omega_{{a}} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}{a}}{x}} {e}^{−{u}^{\mathrm{2}} } .\frac{{du}}{\:\sqrt{\mathrm{2}{a}}} \\ $$$$\Omega_{{a}} \left({x}\right)\:=\:\frac{\sqrt{\pi}}{\:\mathrm{2}\sqrt{\mathrm{2}{a}}}\left(\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}{a}}{x}} {e}^{−{u}^{\mathrm{2}} } {du}\right) \\ $$$$\Omega_{{a}} \left({x}\right)\:=\:\sqrt{\frac{\pi}{\mathrm{8}{a}}}.\mathrm{erf}\left(\sqrt{\mathrm{2}{a}}{x}\right) \\ $$$$\left.\mathrm{3}\right)\:{a}\:<\:\mathrm{0} \\ $$$$\mathrm{Let}\:{u}\:=\:{i}\sqrt{−\mathrm{2}{a}}{t} \\ $$$$\Omega_{{a}} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{{i}\sqrt{−\mathrm{2}{a}}{x}} {e}^{−{u}^{\mathrm{2}} } .\frac{{du}}{\:{i}\sqrt{−\mathrm{2}{a}}} \\ $$$$\Omega_{{a}} \left({x}\right)\:=\:\frac{\sqrt{\pi}}{\:\mathrm{2}{i}\sqrt{−\mathrm{2}{a}}}\left(\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{−{i}\sqrt{\mathrm{2}{a}}{x}} {e}^{−{u}^{\mathrm{2}} } {du}\right) \\ $$$$\Omega_{{a}} \left({x}\right)\:=\:\sqrt{−\frac{\pi}{\mathrm{8}{a}}}.\frac{\mathrm{erf}\left({i}\sqrt{−\mathrm{2}{a}}{x}\right)}{{i}} \\ $$$$\Omega_{{a}} \left({x}\right)\:=\:\sqrt{−\frac{\pi}{\mathrm{8}{a}}}.\mathrm{erfi}\left(\sqrt{−\mathrm{2a}}{x}\right) \\ $$
Commented by aurpeyz last updated on 16/Feb/21
$${arrrrggggg} \\ $$