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I(\mu, \sigma) = \int_{0}^{\infty} ... + \int_{-\mu}^{0} = I_1 + I_2  \end{equation}  Let us consider $I_1$ first.  \begin{equation}  I(\mu, \sigma) = \int_{0}^{\infty} df' (f'+\mu) \frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{f'^2}{2\sigma^2} \right)  \end{equation}  \begin{equation}  I(\mu, \sigma) = \frac{\mu}{2} + \int_{0}^{\infty} df' (f'+\mu) \frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{f'^2}{2\sigma^2} \right)  \end{equation}  \begin{equation}  \,\, = 2 \int_{-\mu}^{\infty} \frac{(y+\mu)}{\sqrt{2\pi}\sigma}\exp\left(-\frac{y^2}{2\sigma^2}\right)dy \\