Are there any shortcuts for computing the coefficients of partial fractions that are not covered by Cover Up Rule?

Question

I am referring to powers of linear factors higher than 1, and all quadratic factors. I was just wondering if there are any obscure techniques to solve the coefficients since in common practice I know there are none.

Answer 1

The cover-up method does work for higher powers of linear factors. When you find the coefficient of the most negative power, you subtract that term from the fraction, which leaves you with a fraction that contains that linear factor with a power in the denominator of one less than in the original one. So, you have a recursive method that will find all the coefficients in the partial fraction expansion.

However, this method amounts to performing a Laurent expansion around each pole, and you can then just as well simply perform that Laurent expansion in the mist convenient way, you don't need to proceed via the detailed cover-up formalism. In particular, it's not necessary to write down terms with undetermined coefficients and solve for those coefficients.

The main thing that can make the partial fraction expansion tedious is the presence of linear or quadratic factors to some power, even if you use the Laurent expansion method around each pole. But there exists a simple trick to get around this problem, which involves replacing the powers of the factors by 1 and inserting parameters in these factors. One can then obtain the desired partial fraction expansion by differentiating w.r.t. the inserted parameters or by performing a series expansion in these parameters.

The advantages of this method over the more traditional Laurent expansion method is that you now only need to work to some order in the expansion parameters so you don't need to work with the exact fraction as you can drop terms that are of higher order than you need in the final answer.

If the partial fraction expansion is needed to compute an integral, then you can compute the integral first and then perform the expansion. It can then happen that the degree of the numerator becomes equal or larger than that of the denominator when the powers in the denominator are reduced to one. In that case you don't need to perform a long division, you can work with the incomplete partial fraction expansion coming from the poles that will contribute to the partial fraction expansion of the original fraction.

What also saves work is that for real fractions with complex poles, you only need to consider one of the complex conjugate pairs. Let's consider a simple example of this method. Let's evaluate:

$$ \int\frac{x^2–3 x +7}{ \left(x^2–4x+6\right)^2}dx$$

We define: $$\begin{split}p(x) &= x^2 - 3x+7\\ q(x) &= x^2–4x+6\end{split}$$ So, we want to integrate the function: $$ S(x) = \frac{p(x)}{q(x)^2}$$ We then start with considering the rational function: $$ R(x,u) = \frac{p(x)}{q(x) - u}$$ The integral of $S(x)$ is then the coefficient of $u$ of the integral of $R(x,u)$

Since the degree of the numerator of $S(x)$ is smaller than the degree of the denominator, it suffices to consider only the sum of the two expansions around the two singularities of the denominator in $R(x,u)$, even though this is not sufficient for $R(x,u)$ itself. For the case at hand the difference is just a constant, but we can still save some work by integrating the incomplete partial fraction expansion of $R(x,u)$ and then extracting the result from the coefficient of $u$.

We start with factoring the polynomial in the denominator of $R(x,u)$. Solving $q(x) - u = 0$ yields: $$ x = 2\pm\sqrt{2-u}i$$ Expanding to first order in $u$ yields: $$ x = 2\pm\sqrt{2}i \mp\frac{\sqrt{2}}{4}i u+\mathcal{O}\left(u^2\right)$$

We’re now going to write down the contribution to the partial fraction expansion coming from the singularity at the root with the upper sign. Let’s denote the singular point from which we are going to consider the contribution to the partial fraction expansion of $R(x,u)$by $r$: $$ r = 2+\sqrt{2}i -\frac{\sqrt{2}}{4}i u$$

The contribution to the partial fraction expansion from $r$ is then: $$ A(x,u) =\frac{C}{x-r}$$ Where: $$C = \lim_{x\to r}(x-r)R(x,u) = \frac{p(r)}{q’(r)}$$ If we put $$r_0 =2+\sqrt{2}i $$ then we have to first order in $u$: $$p(r) = p\left(r_0\right) - \frac{\sqrt{2}}{4}i p’\left(r_0\right) u +\mathcal{O}\left(u^2\right)$$ We then have: $$ C = \frac{\frac{1}{2}-\frac{3\sqrt{2}}{4}i-\frac{1}{8}(1+2\sqrt{2}i)u}{1-\frac{u}{4}} +\mathcal{O}\left(u^2\right)$$ We then expand the denominator: $$ \frac{1}{1-\frac{u}{4}} = \left(1+\frac{u}{4}+\mathcal{O}\left(u^2\right)\right)$$ Multiplying with the numerator and dropping terms that are of order 2 or higher in $u$, yields: $$ C = \frac{1}{2}-\frac{3\sqrt{2}}{4}i - \frac{7\sqrt{2}}{16}i u +\mathcal{O}\left(u^2\right)$$ The integral of $A(x,u)$ is: $$ B(x,u) = C \log(x-r)$$ We then expand the logarithm: $$\log(x-r)= \log\left(x-r_0\right) + \log\left(1 - \frac{ \frac{\sqrt{2}}{4}i}{x-r_0}u\right)=\log\left(x-r_0\right) - \frac{ \frac{\sqrt{2}}{4}i}{x-r_0}u +\mathcal{O}\left(u^2\right)$$

Multiplying with $C$ and extracting the coefficient of $u$ then yields a logarithmic term and a rational term. If we add the complex conjugate to the logarithmic term then this becomes: $$\frac{7\sqrt{2}}{16}i\log\left(\frac{x-2+\sqrt{2} i}{x-2–\sqrt{2} i}\right)$$ The argument of the logarithm is the ratio of a complex number and its complex conjugate, this has a modulus of 1. The argument of this complex number is twice the argument of the numerator, we thus have: $$ \frac{x-2+\sqrt{2} i}{x-2–\sqrt{2} i} = \exp\left[2i\arctan\left(\frac{\sqrt{2}}{x-2}\right)\right] $$ The term involving the logarithm can thus be written as: $$ \frac{7\sqrt{2}}{8}\arctan\left(\frac{x-2}{\sqrt{2}}\right)\tag{1}$$ where we’ve used that $$\arctan\left(\frac{1}{p}\right) = \frac{\pi}{2}-\arctan(p)$$ and we’ve dropped the constant term.

The term involving the rational function is: $$\frac{1}{8}\frac{3+\sqrt{2}i}{x-2-\sqrt{2} i} $$ We have to add the complex conjugate of this term, which amounts to taking twice the real part. Multiplying numerator and denominator by the complex conjugate of the denominator, dropping imaginary terms and multiplying the result by 2 yields: $$\frac{1}{4}\frac{3x-8}{x^2–4x+6} $$

The integral is thus given by the sum this term and the logaritmic term (1): $$ \int\frac{x^2–3 x +7}{ \left(x^2–4x+6\right)^2}dx =\frac{7\sqrt{2}}{8}\arctan\left(\frac{x-2}{\sqrt{2}}\right) + \frac{3x-8}{4\left(x^2–4x+6\right)} + c $$