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I am trying to prove that a function $f(t)=\frac{p(t)}{q(t)}$ where $q$ and $p$ are polynomials over $\mathbb{R}$ can be integrated using only elementary functions. My idea is as follows:

Write $q(t)=(t-\alpha_1)^{j_1}\dots(t-\alpha_q)^{j_q}(t^2+\beta_1t+\gamma_1)^{k_1}\dots(t^2+\beta_rt+\gamma_r)^{k_r}$, where the quadratic terms are irreducible (over $\mathbb{R}$) since they correspond to the complex roots. We can then write: \begin{align*} f(t)=\frac{p(t)}{q(t)}=P(t)+\sum_{i=1}^q\sum_{n=1}^{j_i}\frac{A_{in}}{(x-\alpha_i)^n}+\sum_{i=1}^r\sum_{n=1}^{k_i}\frac{B_{in}t+\Gamma_{in}}{(t^2+\beta_it+\gamma_i)^n}. \end{align*} Where $A,B,\Gamma$ are constants and $P(t)$ a polynomial over $\mathbb{R}$. However now I run into problems, since $\frac{B_{in}t+\Gamma_{in}}{(t^2+\beta_it+\gamma_i)^n}$ doesn't integrate very nicely (except in some cases in $\arccos$ or $\arcsin$, but these are specific cases). How do I prove $\int f(t)dt$ can be expressed in elementary functions?

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When $B_{in}\neq 0$, that gives you a (multiple of a) derivative of $(t^2 + \beta_i t + \gamma_i)^{1-n}$, and something of the form $(t^2 + \beta_i t + \gamma_i)^{-n}$. By elementary transformations, all you have to do is find primitives of $\frac{1}{(1+t^2)^n}$. Substituting $t = \tan\varphi$ there looks promising. –  Daniel Fischer Oct 7 '13 at 12:25

1 Answer 1

$$\int\frac{B_{in}t+\Gamma_{in}}{(t^2+\beta_it+\gamma_i)^n}dt=\int\frac{\frac{B_{in}}{2}(2t+\Gamma_{in})+(B-\frac{B_{in}\beta_i}{2})}{(t^2+\beta_it+\gamma_i)^n}dt= \\=\frac{B_{in}}{2}\int\frac{2t+\Gamma_{in}}{(t^2+\beta_it+\gamma_i)^n}dt+\left(B-\frac{B_{in}\beta_t}{2}\right)\int\frac{dt}{(t^2+\beta_it+\gamma_i)^n}$$

The first integral is computed this way:

$$\int \frac{2t+\Gamma_{in}}{(t^2+\beta_it+\gamma_i)^n}dt=\{u=t^2+\beta_ix+\gamma_i, (2t+\beta_i)dt=du\}=\\=\int\frac{du}{u^k}$$

Let's denote the second integral as $I_n$.

$$I_n = \int\frac{dt}{(t^2+\beta_it+\gamma_i)^n}=\frac{dt}{\left(t+\frac{\beta_i}{2}\right)^2+\left(\gamma_i+\frac{\beta_i^2}{4}\right)}$$.

Redenote $u=t+\frac{\beta_i}{2}, \gamma_i + \frac{\beta_i^2}{4}=m^2$

$$I_n = \int\frac{du}{(u^2+m^2)^n}=\frac{1}{m^2}\int\frac{(u^2+m^2)-u^2}{(u^2+m^2)^n}=I_{n-1}-\frac{1}{m^2}\int\frac{u^2du}{(u^2+m^2)^k}$$

Apply шintegration by parts to the second integral to see that it's elementary function.

Since $I_n$ depends on $I_{n-1}$, find $I_1=\int\frac{du}{u^2+m^2}$ and you are done.

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There is a problem though. When you describe $I_n$ you say $(t^2+\beta_it+\gamma_i=(t+\frac{\beta_i^2}{2})+(\gamma_i+\frac{\beta_i^2}{4})$ which is wrong. You presumably meant: $(t+\frac{\beta_i}{2})^2+(\gamma_i-\frac{\beta_i^2}{4})$. However, the line afterwards you say $m=\gamma_i+\frac{\beta_i^2}{4}=m$ and then write that $I_n=\int\frac{dt}{(t^2+m^2)^n}$, but this not right, since $m^2\neq m$, and it doesn't help if I fix what I said above. –  matti0006 Oct 7 '13 at 14:32
    
And you run into problems if you try to do $(t^2+m)-t^2$, since $m$ can be complex, and if we don't square it we cannot be certain it will be real. –  matti0006 Oct 7 '13 at 14:51
    
Thank you, I miswrote some things indeed. –  Тимофей Ломоносов Oct 7 '13 at 15:33

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