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Integral of a gaussian with random variance

Assuming: $$X(x,\mu)=\frac{1}{\sqrt{(2\pi)\sigma^2}} \exp[-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}]$$ the integral of $X(x,\mu)$ from $-\infty$ to $+\infty$ is: $$S=\int_{-\infty}^{+\infty}dx ...