- 983

- 173

$$ I = \int_1^{\infty} \frac{dx}{p(x)} $$

can be done in indefinite form for n = 0 or 1.

For n = 0, p(x) = 1, and the indefinite integral is x. It diverges for ##x \to \infty##.

For n = 1, p(x) = x + a

_{1}, and the indefinite integral is ##\log(x + a_1)##. It also diverges for ##x \to \infty##.

Thus, the integral ##I## is always divergent for ##n \leq 1##.

This leaves the case where ##n \geq 2##. This problem will be solved by bounding the integrand, and showing that this gives a finite integral. That will then make ##I## finite.

If p(x) has all-negative roots, then it must be uniformly positive or negative over x in the domain of integration of the problem's integral, ##[1,\infty)##. Let us choose a form for p(x) that will make it possible to show that ##I## is bounded from above. This form is p(x) = x

^{n}q(1/x), where

$$q(u) = 1 + a_{n-1} u + a_{n-2} u^2 + \cdots + a_0 u^n$$

As ##x \to \infty##, ##q(1/x) \to 1##, which is positive, thus, p(x) is uniformly positive over x in ##[1,\infty)##. That also makes q(1/x) uniformly positive over that interval. This also means that 1/p(x) is uniformly positive, thus making ##I## also positive, bounded from below by 0.

We must now show that 1/p(x) is bounded from above, and bounded from above in a way that makes ##I## finite. This is equivalent to showing that p(x) is bounded from below in this sort of fashion, and this can be done by bounding q(u) for u in ##(0,1]##. By the extreme value theorem, q(u) must have a minimum value for some value of u in that interval. Since q(u) is always positive in that interval, that means that its minimum value in that integral must also be positive. I will call that value Q. Thus, ##q(1/x) \geq Q## for all x in ##[1,\infty)##, giving ##p(x) = x^n q(1/x) \geq Q x^n## for all x in that interval. This makes ##1/p(x) \leq 1/(Q x^n)## over that interval, and thus,

$$ I = \int_1^{\infty} \frac{dx}{p(x)} \leq \int_1^{\infty} \frac{dx}{Q x^n} = \frac{1}{(n-1)Q} $$

Thus, ##I## is bounded from both above and below, and the integral always converges for ##n \geq 2##.

In summary, this integral always diverges for ##n \leq 1## and always converges for ##n \geq 2##.