## Finite-dimensional Feynman Diagrams

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### 2. Facts from calculus and their *d*-dimensional analogues

The basic fact from calculus that powers the whole discussion is:

The identity with *a* = 1 is proved by the trick of calculating the *square* of theintegral in polar coordinates. The general identity follows by change of variablefrom *x* to .

This fact generalizes to higher-dimensional integrals. Set **v** = (*v*^{1}, ..., *v*^{d})and *d***v** = (*dv*^{1} ... *dv*^{d}),and let *A* be a symmetric *d*by *d* matrix.

We use the fact that a symmetric matrix *A* isdiagonalizable: there exists an orthogonal matrix *U*(so *U*^{t} = *U*^{-1})such that *UAU*^{-1} is the diagonal matrix *B*whose onlynonzero entries are *b*_{11}, ... , *b*_{dd}along the diagonal. Then *A = U*^{-1}BUand **v**^{t}*A***v** = **v**^{t} *U*^{-1}*B U* **v** =**v**^{t}*U*^{t}*B U* **v** =**w**^{t}*B* **w** where **w** = *U***v**,using *U*^{t} = *U*^{-1} and (*U***v**)^{t} =**v**^{t}*U*^{t}. Since *U*is orthogonal det*U* = 1 and the change of variable from **v** to **w**does notchange the integral:

This follows from Proposition 1 by completion of the square in the exponent and a change of variables.

The generalization to *d*dimensions replaces *a*with *A*as before and *b*withthe vector **b** = (*b*^{1}, ... , *b*^{d})

This is proven exactly like Proposition 2.If we write this integral as *Z*_{b}then the integral of Proposition 2 is *Z*_{0}and this proposition can be rewritten as

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