e (n 'Tr th -} c (1.23) ° x We now use the result exp(—k2 /x)dx/ r exp(—k2 / x)dx k2 /(k1-1) (1.24) o A and obtain after a little algebra c0—A.=a+D (1.25) Substituting this result into equation (1.21) gives tz0 exp(--i 1 P(Wi) (1.26) xo Wi Note this result is independent of both c and in