# Need help deriving the energy of a signal

I am solving old exam problems in preparation of my exam in signals, and I'm having trouble with a question.

For a continuous aperiodic signal with the spectrum $X(\omega)=\exp(-\omega^2)$ I am to derive the energy of the signal. I have made it this far, in agreement with the solution manual:

$W = \frac{1}{2\pi}\int_{-\infty}^{\infty} \exp(-(\sqrt{2}\omega)^2)\,\mathrm{d}\omega$

The assignment then suggests substituting in the error function

$\mathrm{erf}(z) = \frac{1}{\sqrt{\pi}} \int_{-z}^z \exp(-y^2)\,\mathrm{d}y$

and exploiting that $\mathrm{erf}(z) \rightarrow 1$ towards infinity, which makes good sense, but the result in the solution manual before reducing the expression is

$W = \frac{1}{2\pi} \sqrt{\frac{\pi}{2}} \mathrm{erf}(\sqrt{2}\omega)\Big\rvert_{\omega \rightarrow \infty}$

I don't understand where the $\frac{1}{2}$ in the square root comes from, so I would appreciate if anyone could help me understand this.

The factor just comes from the substitution $y=\sqrt{2}\omega$ in the integral defining the error function:
\begin{align}E_x=\frac{1}{2\pi}\int_{-\infty}^{\infty}|X(\omega)|^2d\omega&=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-2\omega^2}d\omega\\&=\frac{1}{2\pi}\frac{1}{\sqrt{2}}\int_{-\infty}^{\infty}e^{-(\sqrt{2}\omega)^2}d(\sqrt{2}\omega)\\&=\frac{1}{2\pi}\frac{1}{\sqrt{2}}\int_{-\infty}^{\infty}e^{-y^2}dy\\&=\frac{1}{2\pi}\sqrt{\frac{\pi}{2}}\lim_{x\to\infty}\textrm{erf}(x)\\&=\frac{1}{2\sqrt{2\pi}}\end{align}