The solution given by Eq. (8.49) presents some new challenges for analysis, because of the positive exponential in the outer integrals. Let us begin by thinking of the integral over s and making the transformation of variables u = s//e so that the integral over s becomes Jo^ exp(—u2)pedu. Now if we think that the noise is small (e ^ 1) then when y gets away from 0 the upper limit is getting large. We recognize then that we are computing the normalization constant for a Gaussian distribution once again.

Exercise 8.11 (E)

Show that J1 exp(—u2)du = -y/p. Here is a hint: remember that

If we then approximate the integral over s in Eq. (8.49) by ^/Pe we can conclude that

Now the integral in Eq. (8.50) is something new for us, because of the positive exponent. Just looking at this integral suggest that the main contribution to it will come from the vicinity of L, because the integrand is largest there. We can make this more precise. First, let us make the change of variables v = y/ve so that the integral we have to consider is x x

I = J^/p"" exp(v2)dv. We integrate this by parts, much as we did in the expansion of the tail of the Gaussian distribution:

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