Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

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

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.

share|cite|improve this answer
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

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.