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&=& 2 \int_{-\mu}^{\infty} \frac{(y+\mu)}{\sqrt(2\pi)\sigma}\exp\left(-\frac{y^2}{2\sigma^2}\right) \\  &=& 2 \int_{-\mu}^{0} \frac{(y+\mu)}{\sqrt(2\pi)\sigma}\exp\left(-\frac{y^2}{2\sigma^2}\right) + 2 \int_{0}^{\infty} \frac{(y+\mu)}{\sqrt(2\pi)\sigma}\exp\left(-\frac{y^2}{2\sigma^2}\right) \\  &=& 2 \int_{0}^{\infty} \frac{\sigma dt}{\sqrt{2\pi}}\exp(-t) \\  &=& \mu +  \sqrt{\frac{2}{\pi}} \sigma \end{eqnarray}  Therefore we should subtract this quantity from $|f|$ to obtain a noise debiased measure of f  \begin{equation}