Evaluating Functional integrals

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To compute a Gaussian functional integral of the form e^{-iS_{0}[\phi]+i(J,\phi)}, one can express it as a determinant involving constants a, b, and c. When J is non-zero, the functional determinant becomes a function of J(x), complicating the evaluation. The discussion highlights the importance of completing the square to simplify the integral, allowing the extraction of terms involving J. Additionally, the challenge of obtaining finite results from expressions like δ^nZ[J]/δJ(x1)δJ(x2)... is noted, with perturbative methods being a common approach. The evaluation of the functional determinant remains a central issue in the discussion.
Klaus_Hoffmann
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HOw can you compute a Gaussian functional integral?

i mean integral of the type e^{-iS_{0}[\phi]+i(J,\phi)}

if J=0 then i believe that we can describe the Functional integral as

\frac{c}{(Det(a\partial +b)} a,b,c constant

so Det(a\partial +b)}= exp^{-\zeta '(0)}

\zeta (s) = \sum_{1 \le n}\lambda_{n} ^{-s}

my problem comes when J(x) is different from 0 so we have a functional determinant which is also a function (functional ?? ) of J(x) , then how could yo evaluate it?, also a Fourier transform of a gaussian is again another Gaussian.

Another question..how can you once you have obtained the functional integral Z[J] expressions of the form

\frac{ \delta ^{n}Z[J]}{ \delta J(x1) \delta J(x2)...}

so you get finite results ??
 
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You can't except perturbatively. The perturbation expansion is explained in any good book on QFT, try Zee's QFT in a Nutshell for example.
 
Sorry I see I misread, the trick is to complete the square. You add J^2 - J^2 write the whole thing as (phi + J)^2 and pull the J^2 term out of the integral.
 
thank you everybody.. f-h hit the correct answer (thankx) however still we have the problem that how could you evaluate the functional Determinant.

Det( a\partial ^{2} +b) of the operator:

( a\partial ^{2} +b)(\phi +J(x))
 
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