Question Number 178563 by zaheen last updated on 18/Oct/22
$$\int{e}^{{x}^{\mathrm{2}} } {dx}=? \\ $$
Answered by Acem last updated on 18/Oct/22
$$ \\ $$$${We}\:{have}\:\int{e}^{−{x}^{\mathrm{2}} } {dx}=\:\frac{\sqrt{\pi}}{\mathrm{2}}\:{erf}\left({x}\right)+\:{c} \\ $$$${e}^{{x}^{\mathrm{2}} } =\:{e}^{−\left({ix}\right)^{\mathrm{2}} } \:{then}\:{supposed}\:{that}\:{u}=\:{ix}\Rightarrow\:{du}={idx} \\ $$$$\int{e}^{{x}^{\mathrm{2}} } {dx}=\:\frac{\mathrm{1}}{{i}}\int{e}^{−{u}^{\mathrm{2}} } {du}=\:−{i}.\frac{\sqrt{\pi}}{\mathrm{2}}\:{erf}\left({u}\right)+\:{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{−{i}\:\sqrt{\pi}}{\mathrm{2}}\:{erf}\left({ix}\right)\:+{c} \\ $$
Answered by Spillover last updated on 18/Oct/22
$$\int\mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \mathrm{dx}=\int\mathrm{e}^{\left.−\left(−\mathrm{x}^{\mathrm{2}} \right)\right)} \mathrm{dx} \\ $$$$\mathrm{u}=\mathrm{ix}\:\:\:\:\mathrm{x}=\frac{\mathrm{u}}{\mathrm{i}} \\ $$$$\mathrm{dx}=−\mathrm{idu} \\ $$$$\int\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } −\mathrm{idu} \\ $$$$−\mathrm{i}\int\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} \mathrm{du}} \\ $$$$−\mathrm{i}\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erf}\left(\mathrm{ix}\right)+\mathrm{D} \\ $$$$\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erfi}\left(\mathrm{x}\right)+\mathrm{D} \\ $$