Which conditions makes integrals of the form $\int_{-\infty}^\infty \left|\frac{\mathbb{P}(w^a)}{\mathbb{P}(w^b)} \right|dw < \infty$ converge?

Question

Which conditions must fulfill an integral to be convergent $\int_{-\infty}^\infty \left|\frac{\mathbb{P}(w^a)}{\mathbb{P}(w^b)} \right|dw < \infty$ with $\mathbb{P}(w^m)$ a polynomial with coefficients $\in \mathbb{C}$?

I am specially interested in the cases of ${a, b} \in \{0, 1, 2, 3\}$.

By experimenting in Wolfram-Alpha, I believe that this is happening: $$ \int\limits_{-\infty}^{\infty} \left| \frac{1}{w^2+ib} \right| dw < \infty,\,\,\,\forall b\in \mathbb{R} $$

But, $$ \int\limits_{-\infty}^{\infty} \left| \frac{1}{w^2+b} \right| dw \begin{cases} = \infty,\,\,\, b \leq 0 \\ < \infty,\,\,\, b > 0 \end{cases}$$

So the complex unit becomes harder to form an intuition related to this, I remember that the Laplace transform have this kind of issues also, but I don't know how to relate them.

I am working in a problem where I can use these kind of integral as upper bounds, and I don´t know how to analyze this can of problems about the convergence of integrals, so if you have any bibliography you can share I will be thankful.

Answer 1

Let $p(z)$ and $q(z)$ be non-zero polynomials with coefficients in $\mathbb{C}$. Let $r(z) = p(z)/q(z)$.

1. If $r(z)$ has a pole of order $k$ at some $x_0 \in \mathbb{R}$, then $r(z) \sim \frac{c}{(z-x_0)^k}$ as $z \to x_0$ for some constant $c \neq 0$. In particular, there exists $\delta > 0$ such that

$$ \left| \frac{r(z)}{c/(z-x_0)^k} - 1 \right| < \frac{1}{2} \qquad\text{whenever}\quad |z - x_0| < \delta. $$

This then implies that

$$ \left|r(z)\right| \geq \frac{|c|}{2|z - x_0|^k} \qquad\text{whenever} \quad |z - x_0| < \delta, $$

and so, $\int_{x_0 - \delta}^{x_0 + \delta} | r(x) | \, \mathrm{d}x = \infty$ by the comparison test.

2. Note that $r(z) \sim c z^{\deg p - \deg q}$ as $z \to \infty$, for some constant $c \neq 0$. Similarly as before, this implies that there exists $R > 0$ such that

$$ \frac{|c|}{2}|z|^{\deg p - \deg q} \leq \left|r(z)\right| \leq \frac{3|c|}{2}|z|^{\deg p - \deg q} \qquad\text{whenever}\quad |z| > R. $$

So by the comparison test, $\int_{|x|>R} | r(x) | \, \mathrm{d}x$ diverges if and only if $\int_{|x|>R} |x|^{\deg p - \deg q} \, \mathrm{d}x$ diverges, which occurs precisely when $\deg p - \deg q \geq -1$ by the p-test.

3. The two observations together show that $\int_{-\infty}^{\infty} | r(x) | \, \mathrm{d}x$ diverges if either $r(z)$ has a pole on $\mathbb{R}$ or $\deg q \leq (\deg p) + 1 $.

Now suppose $r(z)$ has no pole on $\mathbb{R}$ and $\deg q > (\deg p) + 1$. By writing $r(z) = p(z)/q(z)$ in lowest terms, we may assume that $q(z)$ has no zero on $\mathbb{R}$. Let $c$ and $R$ be as in Step 2. Then

• $\int_{|x|>R} | r(x) | \, \mathrm{d}x$ converges by Step 2.

• $\int_{|x|\leq R} | r(x) | \, \mathrm{d}x$ converges, since $r(x)$ is continuous on $[-R, R]$.

Therefore $\int_{-\infty}^{\infty} | r(x) | \, \mathrm{d}x$ converges.