Gaussian integral but different

Solution:

First, we can apply the sine double angle formula

$\frac{1}{2}\int_0^{\infty}\sin\left(2x\right)e^{-x^2}\ dx \quad\quad\quad\color{#3D99F6}{\sin(2x)=2\sin(x)\cos(x)}$

Then we change sine to its exponential form

$=\frac{1}{2}\int_0^{\infty}-i\frac{e^{2ix}-e^{-2ix}}{2}e^{-x^2}dx \quad\quad\quad\color{#3D99F6}{\sin\left(x\right)=\frac{e^{ix}-e^{-ix}}{2i}}$

Then let's apply linearity of the integral

$=-\frac{1}{4}i\left(\int_0^{\infty}e^{2ix-x^2}\ dx-\int_0^{\infty}e^{-2ix-x^2}\ dx\right)$

Then we can change these to more familiar Gaussian integral

$=-\frac{1}{4}i\left(\int_0^{\infty}e^{-i^2+2ix-x^2+i^2}\ dx-\int_0^{\infty}e^{-i^2-2ix-x^2+i^2}\ dx\right)$

$=-\frac{e^{-1}}{4}i\left(\int_0^{\infty}e^{-i^2+2ix-x^2}\ dx-\int_0^{\infty}e^{-i^2-2ix-x^2}\ dx\right)$

$=-\frac{e^{-1}}{4}i\left(\int_0^{\infty}e^{-\left(x-i\right)^2}\ dx-\int_0^{\infty}e^{-\left(x+i\right)^2}\ dx\right)$

We have to use now the error function to these two integrals:

$\int e^{-\left(x-i\right)^2}\ dx\quad\quad\quad\quad\color{#3D99F6}{Let\quad u=x-i \Rightarrow du=dx}$

$\Rightarrow\frac{\sqrt{\pi}}{2}\int_{ }^{ }\frac{2}{\sqrt{\pi}}e^{-u^2}\ du=\frac{\sqrt{\pi}}{2}erf\left(u\right)+C=\frac{\sqrt{\pi}}{2}erf\left(x-i\right)+C$

$\int e^{-\left(x+i\right)^2}\ dx\quad\quad\quad\quad\color{#3D99F6}{Let\quad u=x+i \Rightarrow du=dx}$

$\Rightarrow\frac{\sqrt{\pi}}{2}\int_{ }^{ }\frac{2}{\sqrt{\pi}}e^{-u^2}\ du=\frac{\sqrt{\pi}}{2}erf\left(u\right)+C=\frac{\sqrt{\pi}}{2}erf\left(x+i\right)+C$

Next, we just have to plug in limits

$\Rightarrow-\frac{e^{-1}}{4}i\left(\int_0^{\infty}e^{-\left(x-i\right)^2}\ dx-\int_0^{\infty}e^{-\left(x+i\right)^2}\ dx\right)=-\frac{e^{-1}}{4}i\cdot\lim_{a\rightarrow\infty}\left(\bigg/_{\!\!\!\!\!0}^a\frac{\sqrt{\pi}}{2}erf\left(x-i\right)-\frac{\sqrt{\pi}}{2}erf\left(x+i\right)\right)$

$=-\frac{e^{-1}\sqrt{\pi}}{8}i\cdot\lim_{a\rightarrow\infty}\left(\bigg/_{\!\!\!\!\!0}^aerf\left(x-i\right)-erf\left(x+i\right)\right)$

$=-\frac{e^{-1}\sqrt{\pi}}{8}i\cdot\lim_{a\rightarrow\infty}\left(erf\left(a-i\right)-erf\left(a+i\right)-\left(erf\left(0-i\right)-erf\left(0+i\right)\right)\right)$

Those parts with a limit are going to cancel each other and we are left with

$=-\frac{e^{-1}\sqrt{\pi }}{8}i\cdot \left(-\left(erf\left(-i\right)-erf\left(i\right)\right)\right)=-\frac{e^{-1}\sqrt{\pi }}{8}i\cdot \left(-\left(-erf\left(i\right)-erf\left(i\right)\right)\right)\quad\quad\quad\quad\color{#3D99F6}{erf(-x)=-erf(x)}$

Finally, we are going to get

$=-\frac{e^{-1}\sqrt{\pi}}{8}\cdot2erf\left(i\right)=-\frac{e^{-1}\sqrt{\pi}\ i\ erf\left(i\right)}{4}$

$\Rightarrow a\cdot b\cdot d\cdot c^2=1\cdot2\cdot4\cdot i^2=-8$