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 higher-dimensional integrals. Set
v = (v1, ..., vd)
and
dv = (dv1 ... dvd),
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 Ut = U-1)
such that
UAU-1
is the diagonal matrix B
whose only
nonzero entries are
b11, ... , bdd
along the diagonal. Then
A = U-1BU
and
vtAv
= vt U-1B U v =
vtUtB U v =
wtB w
where w = Uv,
using Ut = U-1
and (Uv)t =
vtUt.
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 = (b1, ... , bd)
This is proven exactly like Proposition 2.
If we write this integral as Zb
then the integral of Proposition 2 is
Z0
and this proposition can be rewritten as
2. Facts from calculus and their d-dimensional analogues
.
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