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.


1 Answer 1


The factor just comes from the substitution $y=\sqrt{2}\omega$ in the integral defining the error function:



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