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 the integral in polar coordinates. The general identity follows by change of variable from x to .
This fact generalizes to higherdimensional integrals. Set v = (v^{1}, ..., v^{d}) and dv = (dv^{1} ... dv^{d}), and let A be a symmetric d by d matrix.
We use the fact that a symmetric matrix A is diagonalizable: there exists an orthogonal matrix U (so U^{t} = U^{1}) such that UAU^{1} is the diagonal matrix B whose only nonzero entries are b_{11}, ... , b_{dd} along the diagonal. Then A = U^{1}BU and v^{t}Av = v^{t} U^{1}B U v = v^{t}U^{t}B U v = w^{t}B w where w = Uv, using U^{t} = U^{1} and (Uv)^{t} = v^{t}U^{t}. Since U is orthogonal detU = 1 and the change of variable from v to w does not change 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 with the 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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