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

Then:

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

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?

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2  
You're proving that rational functions are integrable in closed form, not that polynomials are integrable. –  Robert Israel Apr 19 '12 at 22:53
2  
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
2  
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.

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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
2  
@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

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