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?
Best Answer
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.