User:Crdrost/Subpage

So we have the Gaussian integral, $$I = \int_{-\infty}^{\infty} dx~e^{-x^2/2},$$ which we can solve by a really slick trick which involves calculating its square,


 * $$I^2 = \int_{-\infty}^{\infty} dx~\int_{-\infty}^{\infty} dy~e^{-x^2/2 - y^2/2 }.$$

The trick is to use polar coordinates, where $$dx~dy$$ becomes $$r~dr~d\theta$$ and to notice that $$r~dr=d(r^2/2)$$, hence we find,
 * $$I^2 = \int_{0}^{2\pi} d\theta~\int_{0}^{\infty} d\left(\frac{r^2}2\right)~e^{-r^2/2} = 2\pi.$$

Therefore $$I=\sqrt{2\pi}.$$

We can also generalize to $$e^{-ax^2/2}$$ for $$a>0$$ and then we find the substitution,


 * $$ \int_{-\infty}^{\infty} dx~e^{-ax^2/2} = \frac{1}{\sqrt a}~\int_{-\infty}^{\infty} d(x \sqrt{a})~e^{-(x\sqrt{a})^2/2} = \sqrt{\frac{2\pi}a}. $$

And then we want to generalize to $$e^{-ax^2/2 + J~x}$$ and we can do that by completing the square; note that $$-\frac12 ~a~ (x - J/a)^2 = -\frac12 a x^2 + J x - \frac12 ~a^{-1} ~J^2,$$ so by multiplying by $$1=e^{J^2/(2a)}~e^{-J^2/(2a)}$$ one of those gives us the rightmost term in the expansion and the other lives out front:


 * $$ \int_{-\infty}^{\infty} dx~e^{-ax^2/2 + Jx} = \frac{1}{\sqrt a}~e^{J^2/(2a)}\int_{-\infty}^{\infty} d(x \sqrt{a})~e^{-(x\sqrt{a} -J/\sqrt{a})^2/2} = \sqrt{\frac{2\pi}a}~e^{J^2/(2a)}. $$

Finally we can deal with some matrix generalizations. Supposing that $$a\mapsto A_{ij}$$ and $$J\mapsto J_i$$ while $$x \mapsto x_i$$ we have the following integral in general:


 * $$ f(\mathbf A, \mathbf J) = \int_{-\infty}^{\infty} dx_1 \int_{-\infty}^{\infty} dx_2 \dots \int_{-\infty}^{\infty} dx_n ~ \exp\left(\sum_{ij} A_{ij} x_i x_j + \sum_{i} J_i x_i\right) = \oint d^n x~\exp(\mathbf x^T ~ \mathbf A ~\mathbf x + \mathbf J^T ~\mathbf x).$$

Supposing that $$\mathbf A = \mathbf P^{-1} \mathbf \Lambda \mathbf P $$ is diagonalizable we have,


 * $$ f(\mathbf A, \mathbf J) = \oint d^n x~\exp(\mathbf x^T ~\mathbf P^{-1}\mathbf P ~\mathbf A~ \mathbf P^{-1}\mathbf P~\mathbf x + \mathbf J^T \mathbf P^{-1}\mathbf P ~\mathbf x).$$

So without loss of generality we could say $$\det \mathbf P = 1,$$ and then \mathbf y = \mathbf P \mathbf x$ is a very useful substitution. This leads to a complicated formula that I don't know what to do with!