Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is this a correct proof that all rational functions are integrable?

I could be horribly wrong, but here goes:

First, by trivial inspection, an $n$th degree polynomial is integrable: $$\begin{align} \int p(x) dx&=\int \sum_{i=0}^{n}\alpha_ix^i dx\\ &=\sum_{i=0}^{n}\alpha_i\int x^i dx\\ &=\sum_{i=0}^{n}\frac{\alpha_i}{i+1}x^{i+1}+C\\ \end{align} $$

Now, define: $$r(x)=\sum_{i=0}^{n}c_ix^i \text{ and } s(x)=\sum_{i=0}^{m}\gamma_ix^i$$ By the fundamental theorem of algebra, $$r(x)=c_n\prod_{i=1}^{n}(x-r_i) \text{ and } s(x)=\gamma_n\prod_{i=1}^{m}(x-\phi_i)$$ where $r_i$ and $\phi_i$ are the real (and complex, if they exist) roots of $r(x)$ and $s(x)$, respectively.

Now, proving that a fraction of polynomials is integrable: $$\begin{align} \int \frac{r(x)}{s(x)} dx&=\int \frac{c_n(x-r_1)\cdots(x-r_n)}{\gamma_n(x-\phi_1)\cdots(x-\phi_m)}dx\\ &=\frac{c_n}{\gamma_n}\int\frac{(x-r_1)\cdots(x-r_n)}{(x-\phi_1)\cdots(x-\phi_m)} dx \end{align}$$

Suppose: $$ \frac{(x-r_1)\cdots(x-r_n)}{(x-\phi_1)\cdots(x-\phi_m)}=\frac{K_1}{x-\phi_1}+\frac{K_2}{x-\phi_2}+\dots+\frac{K_{m-1}}{x-\phi_{m-1}}+\frac{K_m}{x-\phi_m} \text{ where } K_i \in \mathbb{C} $$


$$\begin{align} \int \frac{r(x)}{s(x)} dx&=\frac{c_n}{\gamma_n}\int \sum_{i=1}^{m}\frac{K_i}{x-\phi_i} dx\\ &=\frac{c_n}{\gamma_n}\sum_{i=1}^{m}K_i\int \frac{1}{x-\phi_i} dx\\ &=\frac{c_n}{\gamma_n}\sum_{i=1}^{m}K_i \log(x-\phi_i)+C \end{align}$$

I guess my issue here is: How do I know that the set of $K$'s exist? (By set of $K$'s, I mean exactly that elements of the set $\{K_1,K_2,\dots,K_{m-1},K_{m}\}$ exist.) And, how do I know that these elements are complex numbers rather than polynomials themselves?

It has been clarified by Robert Israel that my use of terminology is incorrect. What I am proving is that rational functions are integrable in closed form, not that polynomials are integrable.

It has also been clarified that this proof is valid if and only if $s(x)$ does not have repeated roots ($\phi_i=\phi_j \text{ for } i\neq j$) and that the degree of $s(x)$ is less than the degree of $r(x)$.

An attempt at a general proof:

Let $v(x)$ and $w(x)$ be two polynomials. ($\deg v>\deg w$)

Now, define the following rational function: $$\mathfrak{J}(x)=\frac{v(x)}{w(x)}$$

\begin{align} \int \mathfrak{J}(x)\,dx&=\int \frac{v(x)}{w(x)}\,dx\\ \frac{v(x)}{w(x)}&=q(x)+\frac{r(x)}{w(x)} \quad \deg r<\deg w\\ &=\int q(x)+\frac{r(x)}{w(x)}\,dx\\ &=\int q(x)\,dx+\int\frac{r(x)}{w(x)}\,dx\\ &=\mathfrak{a}(x)+\int\frac{r(x)}{w(x)}\,dx\\ r(x)&=\sum_{i=1}^{n}c_ix^i\\ \int\frac{r(x)}{w(x)}&=\int \frac{c_nx^n}{w(x)}+\dots+\int \frac{c_0}{w(x)}\\ \int \frac{c_ix^i}{w(x)}&=\int \frac{k_{1,i}}{(x-r_1)^{\alpha_1}}+\dots+\int\frac{k_{m,i}}{(x-r_m)^{\alpha_m}}\\ &=k_{1,i}\int \frac{1}{(x-r_1)^{\alpha_1}}+\dots+k_{m,i}\int\frac{1}{(x-r_m)^{\alpha_m}}\\ &=k_{1,i}p_{1,i}(x)+\dots+k_{m,i}p_{m,i}(x)\\ &=\sum_{j=1}^{m}k_{j,i}p_{j,i}(x)\\ \int\frac{r(x)}{w(x)}&=\sum_{j=1}^{m}k_{j,n}p_{j,n}(x)+\dots+\sum_{q=1}^{m}k_{q,0}p_{q,0}(x)\\ &=\sum_{\alpha=0}^{n}\sum_{\beta=1}^{m}k_{\beta,\alpha}p_{\beta,\alpha}(x)\\ \mathfrak{b}(x)&=\sum_{\alpha=0}^{n}\sum_{\beta=1}^{m}k_{\beta,\alpha}p_{\beta,\alpha}(x) \end{align}

$$\therefore \int\mathfrak{J}(x)\,dx=\mathfrak{a}(x)+\mathfrak{b}(x)$$

There is obviously a lot of details left out; but do I need them?

share|improve this question
You're proving that rational functions are integrable in closed form, not that polynomials are integrable. –  Robert Israel Apr 19 '12 at 22:53
Your argument only works if $s$ has distinct roots; partial fraction decomposition is more complicated in general. –  Qiaochu Yuan Apr 19 '12 at 22:54
@QiaochuYuan, by distinct roots, do you mean numerical roots of the form $a+bi$ where $a$ and $b$ are integers? And, could you give a more expansive proof? –  000 Apr 19 '12 at 22:56
Distinct roots means $\phi_i \ne \phi_j$ for $i \ne j$. –  Robert Israel Apr 19 '12 at 22:58
@Limitless: Why would "distinct roots" possibly mean that?! The only explanation I can think of is a complot by mathematicians with the sole purpose of not being understood by anyone! –  Mariano Suárez-Alvarez Apr 22 '12 at 4:00

1 Answer 1

up vote 3 down vote accepted

Your partial fraction decomposition is only valid if $n < m$ and there are no repeated roots in the denominator. In general the antiderivative will have a polynomial part and a rational part as well as the logarithmic part you are getting.

share|improve this answer
Ah. So it's not that I'm wrong, it's that I am not covering all the cases? –  000 Apr 19 '12 at 22:58
@Limitless You are generalizing too much, yes. You need to consider $\deg p < \deg q$, $\deg p = \deg q$ and $\deg p > \deg q$, I suppose. –  Pedro Tamaroff Apr 19 '12 at 22:59
@PeterT.off Very interesting! I will have to consider those cases and see if I can construct an entire proof! :) –  000 Apr 19 '12 at 23:01
@Limitless: The $\mathrm{deg}\,p > \mathrm{deg}\,q$ case can be reduced to fretting about the $\mathrm{deg}\,p < \mathrm{deg}\,q$ case, by separating the function out into a polynomial part and a "proper fraction" part (akin to mixed numbers)... –  J. M. Apr 20 '12 at 1:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.