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&=& 2 \int_{0}^{\infty} \frac{\sigma dt}{\sqrt{2\pi}}\exp(-t) \\  &=& \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 (|f| - \sqrt{\frac{2}{\pi}}\sigma) P(f, \sigma)  \end{equation}