Suppose

No division is necessary in this example, and the denominator is already factored, except that we should replace with . Then the classical partial fraction form is

I will sketch here an efficient and self-checking way to identify the by
hand, which is admittedly complicated in this example because of the
presence of the algebraic extensions **i** and . We take the first
term in the Laurent series of both sides about one root, say .
This is perfectly feasible to do by hand.
This gives, on equating leading terms

(okay, so I used Maple to evaluate the constant! An HP calculator could also have been used.) This identifies . We now take this known term onto the left hand side, subtract, and divide into the new numerator. If this division is not exact, then we have made an error (this is the self-checking part of the method). We have, here,

and on putting the left-hand side over a common denominator and
dividing out (which * must* be a factor) we have

where is here just some numerical constant that I suppressed so the expression would fit on one line, and we see one of the factors of now cancels. We then can identify by computing the leading term of the Laurent series about , as before.

Obviously (I love that word) this process can be repeated. I suggest that
as an exercise, you find (by hand) the coefficients ,
, and corresponding to the expansion about the
pole **z=-1**. Start from the original problem, and that way you won't
have any ugly 's to mess up your arithmetic. (I get the numbers
, , and .
I haven't checked the last one yet).

It is clear (another useful phrase) that this method is actually an
algorithm. It obviously terminates, because the degree of the
denominator is decreased by 1 each time. It is also clear that
radical extensions (**i** and ) make our life more difficult.
This is true in general, and we are going to explore ways of eliminating
them whenever we can.

One tempting method to eliminate them is to use floating-point arithmetic from the start. Once the denominator has been completely factored, this is not a bad option, but in your assignment you will explore the kinds of things that can go wrong if near-multiple roots are not detected on the factorization.

** Remark**. Our final answer is a sum of terms with complex
numbers in it. Is this answer, namely equation (4),
at all useful? Don't the complex logarithms cause trouble?

It turns out that the answer to that is yes, but not as much as you'd
think, and some of these problems do * not* go away with the
more sophisticated algorithms used later, or even the standard
real form. In particular, the issue of branch switching, and choosing
a form of your answer that is always continuous, is an open problem
at this time of writing.

For now, note that

which is perfectly correct. The imaginary part is just a constant,
which cancels. This will always happen, provided that a branch
point or branch cut is not * crossed* (here we were moving along
the branch cut without crossing it). Going through 0 gives us a
good indication that something is wrong: is clearly
not , whatever it is, and indeed this integral diverges.

But as was mentioned previously, knowing whether the branch cuts were crossed in evaluating is an open problem.

What do Computer Algebra systems do about the problem? They mostly ignore it for indefinite integrals, and Maple (like some other systems) tries to look for singularities in the range of integration for definite integrals. This is only a partial answer, though, as the location of the branch cuts is more difficult than finding the singularities: effectively the branch cuts are the algebraic curves described by , for .

Thu Nov 16 13:46:20 PST 1995