# improper integral equal to 1

For which constants $a$ and $b$ will

$$\lim_{p\to\infty}\int_{-p}^p\frac{x^3+ax^2+bx}{x^2+x+1}dx=1$$

Now I set the following:

$$\lim_{p\to\infty}\int_{-p}^p\frac{x^3+ax^2+bx}{x^2+x+1}dx=\lim_{p\to\infty}\int_{-p}^0\frac{x^3+ax^2+bx}{x^2+x+1}dx+\lim_{p\to\infty}\int_{0}^p\frac{x^3+ax^2+bx}{x^2+x+1}dx$$

Could I assume that both limits should be equal $\frac{1}{2}$ then try to integrate and try to find the constants? I don't get far by tring that. What else could I try? Thanks!

-

You want to keep $-p$ and $p$ together a little longer. If you separate, you get an expression of shape "$\infty -\infty$," which is not at all helpful.

My approach would involve taking the standard first step for such a problem, which is to divide $x^3+ax^2+bx$ by $x^2+x+1$.

The first term in the quotient is $x$, no problem, nice cancellation when you integrate from $-p$ to $p$.

The second term is more interesting, it is $a-1$. This one is very troublesome unless $\dots$.

-
the situation is simpliefied if I try to choose $a=1$. then I arrive at simplified expression $\lim_{p\to\infty}\int_{-p}^p\frac{x(b-1)}{x^2+x+1}dx$ am I moving to right direction? – Sarunas Jan 22 '13 at 22:02
Yes. You need to prove that $a\ne 1$ is bad. Basically that is because when you integrate you get $2p(a-1)$, and the remaining stuff cannot save you from having the limit be infinite. So now you are at a concrete question, how to choose $b$. This is an integration problem, which if necessary can be brute-forced, completing the square. – André Nicolas Jan 22 '13 at 22:15
the last steps are needed. I calculated the integral which is equal to $(b-1)(\frac{1}{2}log(x^2+x+1) - \frac{1}{\sqrt{3}}arctan(\frac{2x+1}{\sqrt{3}}))$ how do I proceed in finding constant $b$? – Sarunas Jan 23 '13 at 10:15
I haven't checked your integral. But if correct, plug in $p$, $-p$, subtract. The $\log$ term is $\log\frac{p^2+p+1}{p^2-p+1}$. As $p$ goes to infinity, this approaches $0$. For the $\arctan$ stuff, plug in, go to $\infty$. Get $\frac{1}{\sqrt{3}}\pi$ (from $\pi/2-(-\pi/2)$. So to make integral $1$, want $b-1=\frac{\sqrt{3}}{\pi}$. This is probably slightly wrong, when I have time I will check with paper, pen. – André Nicolas Jan 23 '13 at 12:52
you made only small mistake but I got the point. thanks a lot for your help! – Sarunas Jan 23 '13 at 13:20

Hints:

• First: choose $a$ such that the limit does in fact converge (there is only a single option there). Note that for the convergence it is important that the expansion for large $x$ (apart from terms which decay faster than $x^{-1}$) only involves odd powers as the odd powers get cancelled by the symmetric limit.

• Second: calculate the integral as a function of $b$

-