Functional integration: Difference between revisions

Revision as of 23:17, 24 April 2023

Revision as of 23:18, 24 April 2023

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<math>

\frac{\displaystyle\int f(a)f(b)\exp\left\lbrace-\frac{1}{2} \int_{\mathbb{R}^2} f(x) K(x,;y) f(y)\, dx\,dy\right\rbrace \mathcal{D}[f]}

{\displaystyle\int \exp\left\lbrace-\frac{1}{2} \int_{\mathbb{R}^2} f(x) K(x,;y) f(y) \,dx\,dy\right\rbrace \mathcal{D}[f]} =

K^{-1}(a;b)\,.

</math>