Assume that you have an integral of the form , where
are polynomials. It is well-known that the best way to solve this is to do a partial fraction decomposition of
, which allows you to find an antiderivative of
; this can then be used to evaluate
. Partial fraction decomposition of
allows you to write
as a sum of a polynomial with a linear combination of terms of the form
, where the
range over the (complex) roots of
and
lies between 1 and the multiplicity of the root
of
. Now,
whence finding an antiderivative of this is easy.
In case one has , one has that
is a root of
as well, and one can combine the terms
and
to a term
, multiplied by a polynomial in
of degree
. This allows to give a partial fraction decomposition of
purely in term of real polynomials. Note that
is the minimal polynomial of
over
.
More generally, one can consider polynomials over an arbitrary field . Now
will in general not split into linear (or at most quadratic) factors over
. But nonetheless, one can still do a partial fraction decomposition. Moreover, one can continue this to another level of abstractness by working with arbitrary principal ideal domains.
The main ingredient which we need is the Chinese Remainder Theorem:
Let be a principal ideal domain and let
. Write
, where
are pairwise coprime elements. Then
is an isomorphism. Moreover, there exist such that
divides
, that the inverse of the above isomorphism is given by
In case is Euclidean, one can choose the
such that
.
Consider ,
. We first show that
. For that, note that
Now the are pairwise coprime, whence this is equivalent to
dividing
. Hence,
. Therefore,
induces a monomorphism
. We have to show that
is surjective, or alternatively, that it admits an inverse.
Let . We have that
is coprime to
; hence, we have a Bézout identity, i.e. we can write
with
; in case
is Euclidean, we can also assume that
,
(see my post on the Extended Euclidean Algorithm). But this means that
is zero in
for all
,
in
. Set
; therefore,
, whence
, as we had to show.
We need another simple lemma, which works for a certain class of Euclidean rings such as with
for some fixed
, and
with
. The two properties we need are the following:
We will say that an Euclidean domain is admissible if
is multiplicative and for all
,
one can write
with
such that
and
.
Let be an admissible Euclidean domain and
. Any
can be written in the form
with
,
, where
.
We show this by induction on . For
,
, whence
. Hence,
is of the required form. Now, assume that
. Write
with
,
and
. Hence,
and
, whence by induction, we can write
with suitable
,
. If we set
, we obtain
with
as required.
Now partial fraction decomposition is a direct corollary of these results:
Let be a principal ideal domain with field of fractions
, and let
be two elements. Assume that
, where
,
are pairwise coprime polynomials and
. Then there exist elements
such that
In case is Euclidean with a multiplicative degree function, we can assume that
for
,
.
As an example, we want to consider the Partial Fraction Decomposition for , i.e. for
,
,
. Now
and
, whence
is the partial fraction decomposition of in
.
We can assume that by replacing
by
.
By the Chinese Remainder Theorem, the map ,
is an isomorphism. The inverse is given by
for some
; in case
is Euclidean,
can be chosen such that
. Therefore, there exist
(where in case
is Euclidean,
) and
such that
Dividing by , we get
This shows the claim in case is not Euclidean.
In case is Euclidean, we can write
with
; by adding
to
, we can therefore assume that
satisfies
. By the previous lemma, we can now write
with
and
; as
, we get
. Therefore, we obtain
with
, which gives
and by reindexing the second sum we obtain the formula from the statement of the theorem.
Now assume that we have a rational function written as
where are pairwise coprime,
with
. Then
In case is prime, this can be considered an element in the localization
of
at the prime ideal
. Let
be the completion of
. Note that both
and
have the residue field
.
In the special case that , we see that
. Hence, since
is a prime element in
, we obtain an isomorphism
with
. In
, the element
corresponds to
In particular, we can use the Hasse derivative of
respectively
(the definition can be translated to power series rings as well) to obtain
In the case , this is Heaviside's formula.
Comments.