- #1
Kevin_spencer2
- 29
- 0
To evaluate the integral
\int_{-\infty}^{\infty}dt e^{xf(t)} whenever x is 'big' (tending to infinity) we use the saddle point expansion so:
\int_{-\infty}^{\infty}dt e^{xf(t)}\sim g(x)\sum_{n=0}^{\infty}a_{n}x^{-n}
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):
\sum_{n=0}^{\infty}a(n) = S will at least S exist in the sense of a 'Borel summable' series to calculate the integral for x=1,2,3,4,...
\int_{-\infty}^{\infty}dt e^{xf(t)} whenever x is 'big' (tending to infinity) we use the saddle point expansion so:
\int_{-\infty}^{\infty}dt e^{xf(t)}\sim g(x)\sum_{n=0}^{\infty}a_{n}x^{-n}
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):
\sum_{n=0}^{\infty}a(n) = S will at least S exist in the sense of a 'Borel summable' series to calculate the integral for x=1,2,3,4,...
