Gauss integral with sine. How to calculate this $ \int^{\infty}_0 e^{- \alpha x^2} \sin(\beta x) \,\mathrm{d} x $ ? I've tried to get a differential equation, but is seems not to be easily solvable.
 A: Using the fact that the value of the Gaussian integral is $\displaystyle\int_0^\infty e^{-x^2}dx=\sqrt\frac\pi2$ , and recalling Euler's famous formula $e^{ix}=\cos x+i\sin x$, discovered by Abraham de Moivre, the integral becomes:
$$I(a,b)=\int_0^\infty e^{-ax^2}\sin(bx)dx=\int_0^\infty e^{-ax^2}\Im\left(e^{ix}\right)dx=\Im\left(\int_0^\infty e^{-(ax^2-ibx)}dx\right)=$$
$$=\Im\bigg(\int_0^\infty \exp\bigg[-\bigg(x\sqrt a-i\frac b{2\sqrt a}\bigg)^2-\frac{b^2}{4a}\bigg]dx\bigg)=\Im\bigg(\int_{-ic}^{\infty-ic}e^{-(t^2+c^2)}\frac{dt}{\sqrt a}\bigg)=$$
$$=\frac{e^{-c^2}}{\sqrt a}\cdot\Im\bigg(\int_{-ic}^{\infty-ic}e^{-t^2}dt\bigg)=\frac{e^{-c^2}}{\sqrt a}\cdot\Im\bigg[\frac{\sqrt\pi}2\bigg(1+\text{Erf}(ic)\bigg)\bigg]=\sqrt\frac\pi a\cdot\frac{e^{-c^2}}2\cdot\Im\Big[\text{Erf}(ic)\Big],$$
where $c=-\displaystyle\frac b{2\sqrt a}$ , $t=x\sqrt a-ic$ , $\Im$ represents the imaginary part , and Erf is the error function.
A: I don't believe that this integral cannot solved by ODE approach.
See http://tw.knowledge.yahoo.com/question/article?qid=1712010766078 or http://tw.knowledge.yahoo.com/question/article?qid=1712010194980.
