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I_2 &=& \mu + 2 \int_{0}^{\infty} \frac{\sigma dt}{\sqrt{2\pi}}\exp(-t) \\  &=& \mu + \sqrt{\frac{2}{\pi}} \sigma  \end{eqnarray}  Now I_1 is:  \begin{eqnarray}  I_1 &=& \mu + 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}  F = \int dx dy \left(|f| - \sqrt{\frac{2}{\pi}}\sigma\right) P(f, \sigma)