What is the antiderivative of one over an arbitrary polynomial?

Question

Suppose we have an arbitrary polynomial as follows. $$p(x):=\sum_{i=0}^n a_nx^n$$ And we want to find the antiderivative of $(p(x))^{-1}$ with respect to $x$. $$\int\frac{dx}{p(x)}$$ Messing around with an online CAS suggests that this would be equal to $$C+\sum_{z\in\{r\mid p(r)=0\}}\frac{\ln |x-Z|}{\sum^n_{i=1}ia_iz^{i-1}}$$ Which sort of makes sense to me, and I think I can visualize that, but my question is, how do you derive this result and how true is it? Do you have to employ a form of partial fraction decomposition? The website also says that this only applies to cases where each root is distinct. What happens in the case of repeated roots? Can I extend this to a power series?

Answer 1

The formula follows from a partial fraction decomposition when the polynomial has no repeated roots.

Note that the denominator in your formula, $\sum_{i=1}^nia_iz^{i-1}$, is just the derivative of $p(x)$ evaluated at the root, $p'(z)$. Since $p(x)$ and $p'(x)$ have a common root if and only if $p(x)$ has multiple roots, then the formula would require you to divide by $0$ if you had any multiple roots. So the formula cannot work for polynomials with multiple roots, at least as is.

Suppose first that $p(x)$ is monic and has pairwise distinct roots, $z_1,\ldots,z_n$, and the partial fraction decomposition is $$\frac{1}{p(x)} = \frac{A_1}{x-z_1}+\cdots + \frac{A_n}{x-z_n}.$$ If we do the operation on the right hand side, we get $$\frac{1}{p(x)} = \frac{\sum_{j=1}^n A_j\prod_{i\neq j}(x-z_i)}{p(x)}.$$ so that means that $$\sum_{j=1}^n A_j\prod_{i\neq j}(x-z_i) = 1.$$

Using the Heaviside cover-up method, we see by plugging in $z_t$ that $$A_t = \frac{1}{\prod_{i\neq t}(z_t-z_i)}.$$ But notice as well that $$p'(x) = \sum_{j=1}^n\left(\prod_{i\neq j}(x-z_i)\right),$$ so that $$p'(z_t) = \prod_{i\neq t}(z_t-z_i).$$ That is, $A_t = \frac{1}{p'(z_t)}$. Therefore, we can write the partial fraction decomposition as: $$\frac{1}{p(x)} = \frac{1}{p'(z_1)(x-z_1)} + \cdots + \frac{1}{p'(z_n)(x-z_n)}.$$ Integrating, we get $$\begin{align*} \int\frac{1}{p(x)}\,dx &= \int\left(\sum_{i=1}^n \frac{1}{p'(z_i)(x-z_i)}\right)\,dx\\ &= \sum_{i=1}^n\left(\int\frac{1}{p'(z_i)(x-z_i)}\,dx\right)\\ &= \sum_{i=1}^n\left(\frac{\ln|x-z_i|}{p'(z_i)}\right) + C, \end{align*}$$ which is exactly your formula.

If $p(x)$ is not monic, but has leading coefficient $a_n\neq 0$, then letting $P(x) = \frac{1}{a_n}p(x)$, then $P(x)$ is monic with the same roots as $p(x)$. So if $z_1,\ldots,z_n$ are the roots, then $$\begin{align*} \int \frac{1}{p(x)}\,dx &= \int \frac{1}{a_nP(x)}\,dx = \frac{1}{a_n}\int \frac{1}{P(x)}\,dx\\ &=\frac{1}{a_n}\left( \sum_{i=1}^n \frac{\ln|x-z_i|}{P'(z_i)}\right)+C\\ &= \frac{1}{a_n}\left(\sum_{i=1}^n\frac{\ln|x-z_i|}{\frac{1}{a_n}p'(z_i)}\right)+C\\ &= \frac{1}{a_n}\left(\sum_{i=1}^n\frac{a_n\ln|x-z_i|}{p'(z_i)}\right)+C\\ &= \left(\sum_{i=1}^n\frac{\ln|x-z_i|}{p'(z_i)}\right) + C. \end{align*}$$ So you get the same formula if $p(x)$ is not monic.

Answer 2

Well, if you have a polynomial of this form you can write it as a product of factors (although the roots will probably be mostly complex): $$p_n(x)=\sum_{k=1}^na_kx^k=C_1\prod_{k=1}^n(x-\alpha_k)$$ now if you have: $$\frac{1}{p_n(x)}=C_2\prod_{k=1}^n\frac{1}{(x-\alpha_k)}=C_2\sum_{k=1}^n\frac{b_k}{x-\alpha_k}$$ where each term can then be integrated individually, I think the hard part would be determining $\alpha_k,b_k$ from $a_k$