# 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.

• Do you know how to find the fourier transform of $e^{-\alpha x^2 }$? – Mhenni Benghorbal Oct 26 '13 at 13:20
• Write sin in terms of the expoential, complete the square of the exponent, and obtain the integral in terms of the error function erf. – Jas Ter Oct 26 '13 at 13:21
• I know that there is a simpler solution that in terms of erf(I mean elementary functions). – Alex Oct 26 '13 at 13:27
• One nice thing to notice is that if we call the integral $y(\alpha,\beta)$, $${\partial y \over \partial \alpha} = {\partial ^2y\over\partial \beta^2},$$ meaning that the integral is a solution to the heat equation with the condition $y(\alpha,0)=0$. Note that Wolfram Alpha evaluates $$y(1,1) = F\left(1\over2\right),$$ where $F(\cdot)$ is Dawson's Integral, $$F(x) = \mathrm{e}^{-x^2}\int_0^{x} \mathrm{e}^{y^2} \ \mathrm{d}y.$$ This is, I believe, a solution to the heat equation, and can be written in terms of the imaginary error function. – Bennett Gardiner Oct 26 '13 at 13:28
• Mhenni, yes, I do. But in that case it would have different limits of integral... – Alex Oct 26 '13 at 13:30

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.