Kevin_spencer2
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To evaluate the integral
[tex]\int_{-\infty}^{\infty}dt e^{xf(t)}[/tex] whenever x is 'big' (tending to infinity) we use the saddle point expansion so:
[tex]\int_{-\infty}^{\infty}dt e^{xf(t)}\sim g(x)\sum_{n=0}^{\infty}a_{n}x^{-n}[/tex]
Of course the expansion above is just valid for x---> infinite, but what would happen if i put x=1 and hence i must find the sum for the a(n):
[tex]\sum_{n=0}^{\infty}a(n) = S[/tex] will at least S exist in the sense of a 'Borel summable' series to calculate the integral for x=1,2,3,4,...
[tex]\int_{-\infty}^{\infty}dt e^{xf(t)}[/tex] whenever x is 'big' (tending to infinity) we use the saddle point expansion so:
[tex]\int_{-\infty}^{\infty}dt e^{xf(t)}\sim g(x)\sum_{n=0}^{\infty}a_{n}x^{-n}[/tex]
Of course the expansion above is just valid for x---> infinite, but what would happen if i put x=1 and hence i must find the sum for the a(n):
[tex]\sum_{n=0}^{\infty}a(n) = S[/tex] will at least S exist in the sense of a 'Borel summable' series to calculate the integral for x=1,2,3,4,...