# functional determinant question

why when we want to evaluate a functional determinant we use the expansion near $t=0$ for the partition function $\sum_{n}e^{-tE_{n}} \sim \sum_{n} a_{n} (t)t^{n}$ instead of just using the semiclassical partition function ?? $\iint _{\Gamma}dx dpe^{-tp^{2}-tV(x)}$ which is just easier to handle with ? fo rexamplelin almost all papers i have seen this asymptotic approximation for the partition function near $t=0$ although coefficients are hard to evaluate.

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The usual expansions are not exactly in t, but they do Taylor expand the exponential. The semiclassical expansion is also used, for example, to calculate instanton decays. The question isn't stated so clearly--- are you asking why Taylor expand rather than do semiclassical? –  Ron Maimon Aug 23 '12 at 6:03
i was asking why do not use semiclassical expansion instead of the taylor power series near $t=0$ that was all –  Jose Javier Garcia Aug 23 '12 at 9:10