here is my attempt.is it correct? $$I = \int e^{-x^2}dx$$ let $u = e^{-x^2}$ and $v=x$ then : $$I = xe^{-x^2} - \int -2x^2e^{-x^2} dx ; J = \int -x^2e^{-x^2}dx$$ here is where I'm in doubt am I alowed to do this? let $x^2 = i\theta \Rightarrow x = \sqrt{i\theta}$ $$dx = \frac{\sqrt{i}}{2\sqrt{\theta}}$$ $$J = -\int \frac{\sqrt{i}}{2\sqrt{\theta}} i\theta e^{-i\theta}d\theta$$ $$ = $$ $$- \frac{i\sqrt{i}}{2}\int\sqrt{\theta}e^{-i\theta} d\theta = - \frac{i\sqrt{i}}{2} K$$ $$K = \int\sqrt{\theta}e^{-i\theta} d\theta = \int\sqrt{\theta}(\cos\theta - i\sin\theta) d\theta$$ with $$\Lambda (\theta)= \int\sqrt{\theta}\cos\theta d\theta = \frac{\sqrt{-i\theta} \Gamma(3/2, -i \theta) + \sqrt{i \theta} \Gamma(3/2, i \theta)}{2 \sqrt{\theta}} + C$$ And : $$\int\sqrt{\theta} (- i\sin\theta) d\theta = \Lambda (-\theta)$$ we have: $$K = \int\sqrt{\theta}(\cos\theta - i\sin\theta) d\theta = \Lambda (\theta) + i\Lambda (-\theta)$$ then: $$J = - \frac{i\sqrt{i}}{2} K = - \frac{i\sqrt{i}}{2}\Lambda (\theta) + \frac{\sqrt{i}}{2}\Lambda (-\theta)$$ we finally have : $$I = \int e^{-x^2}dx = xe^{-x^2} - 2J = xe^{-x^2} +i\sqrt{i}\Lambda (\theta) - \sqrt{i}\Lambda (-\theta) + \kappa$$
for $x^2 = i\theta$
by the way I am in high school so please refrain from technical talk!
