Editing Functional integration (section)

==Examples==
Most functional integrals are actually infinite, but often the limit of the [[quotient]] of two related functional integrals can still be finite. The functional integrals that can be evaluated exactly usually start with the following [[Gaussian integral]]:

:<math>
\frac{\displaystyle\int \exp\left\lbrace-\frac{1}{2} \int_{\mathbb{R}}\left[\int_{\mathbb{R}} f(x)  K(x;y) f(y)\,dy +  J(x)  f(x)\right]dx\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]} =
 \exp\left\lbrace\frac{1}{2}\int_{\mathbb{R}^2} J(x) \cdot K^{-1}(x;y) \cdot J(y) \,dx\,dy\right\rbrace\,,
</math>

in which <math>
K(x;y)=K(y;x)
</math>. By functionally differentiating this with respect to ''J''(''x'') and then setting  to 0 this becomes an exponential multiplied by a monomial in ''f''. To see this, let's use the following notation:

<math>
 G[f,J]=-\frac{1}{2} \int_{\mathbb{R}}\left[\int_{\mathbb{R}} f(x)  K(x;y) f(y)\,dy +  J(x)  f(x)\right]dx\, \quad,\quad W[J]=\int \exp\lbrace G[f,J]\rbrace\mathcal{D}[f]\;. 
</math>

With this notation the first equation can be written as:

<math>
\dfrac{W[J]}{W[0]}=\exp\left\lbrace\frac{1}{2}\int_{\mathbb{R}^2} J(x)  K^{-1}(x;y)  J(y) \,dx\,dy\right\rbrace.
</math>

Now, taking functional derivatives to the definition of <math>
W[J]
</math> and then evaluating in <math>
J=0
</math>, one obtains:

<math>
\dfrac{\delta }{\delta J(a)}W[J]\Bigg|_{J=0}=\int f(a)\exp\lbrace G[f,0]\rbrace\mathcal{D}[f]\;,
</math>

<math>
\dfrac{\delta^2 W[J]}{\delta J(a)\delta J(b)}\Bigg|_{J=0}=\int f(a)f(b)\exp\lbrace G[f,0]\rbrace\mathcal{D}[f]\;,
</math>

<math>
\qquad\qquad\qquad\qquad\vdots
</math>

which is the result anticipated. More over, by using the first equation one arrives to the useful result:

:<math>
    \dfrac{\delta^2}{\delta J(a)\delta J(b)}\left(\dfrac{W[J]}{W[0]}\right)\Bigg|_{J=0}=
 K^{-1}(a; b)\;;
</math>

Putting these results together and backing to the original notation we have:

<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>

Another useful integral is the functional [[delta function]]:

:<math>
\int \exp\left\lbrace \int_{\mathbb{R}} f(x) g(x)dx\right\rbrace \mathcal{D}[f] = \delta[g] = \prod_x\delta\big(g(x)\big),
</math>

which is useful to specify constraints. Functional integrals can also be done over [[Grassmann number|Grassmann-valued]] functions <math>\psi(x)</math>, where <math>\psi(x) \psi(y) = -\psi(y) \psi(x)</math>, which is useful in quantum electrodynamics for calculations involving [[fermions]].