For what values of $p$ and $q$ does the integral converge? $$\int \limits _{-\infty}^{\infty} \frac{1}{|x|^p(1+|x|^q)}dx$$
Thus far, I have set $p=0$ and $q=0$ separately and tried to determine for which values of $q$ and $p$ the integral converges, respectively.
When $p=0$, I noticed that $q$ cannot be an odd integer, as it would allow for the denominator to equal 0, causing the function to explode. Even integers do seem to allow the integral to converge, but I'm not sure what to do about all other real numbers. 
When $q=0$, I noticed that the integral never converges, as $p$ would need to be greater than $1$ and less than $1$ for different ranges of the integral.
I am not sure where to go from here. What are some general methods to tackle such a problem?
 A: Hint 1:
$$\int \limits _{-\infty}^{\infty} \frac{1}{|x|^p(1+|x|^q)}dx =2 \int \limits _{0}^{\infty} \frac{1}{|x|^p(1+|x|^q)}dx =2 \int \limits _{0}^{1} \frac{1}{|x|^p(1+|x|^q)}dx+2 \int \limits _{1}^{\infty} \frac{1}{|x|^p(1+|x|^q)}dx$$
Therefore, by positivity, your integral is convergent if and only if 
$$\int \limits _{1}^{\infty} \frac{1}{|x|^p(1+|x|^q)}dx$$ and
$$\int \limits _{0}^{1} \frac{1}{|x|^p(1+|x|^q)}dx$$
are convergent.
Hint 2 You can limit compare 
$$\int \limits _{1}^{\infty} \frac{1}{|x|^p(1+|x|^q)}dx$$
with
$$\int \limits _{1}^{\infty} \frac{1}{|x|^{p+q}}dx$$
Hint 3 You can limit compare
$$\int \limits _{0}^{1} \frac{1}{|x|^p(1+|x|^q)}dx$$
with
$$\int \limits _{0}^{1} \frac{1}{|x|^p}dx$$
A: By parity, our integral equals:
$$ I(p,q) = 2\int_{0}^{+\infty}\frac{dx}{x^p(1+x^q)} \tag{1}$$
where the integrand function behaves like $x^{-p}$ in a right neighbourhood of zero and like $x^{-(p+q)}$ in a left neighbourhood of $+\infty$. In order to ensure integrability, we must have $\color{red}{p<1}$ and $\color{red}{p+q>1}$.
Under such assumptions, by using the Laplace (inverse) transform or other tools we have:

$$ I(p,q) = \color{red}{\frac{2\pi}{q \sin\left(\pi\cdot\frac{1-p}{q}\right)}}.\tag{2}$$

A: Hint: The integral converges if and only if both $\int_0^1$ and $\int_1^\infty$ converge.
Now $\int_0^1$ converges if and only if $\int_0^1 x^{-p}$ converges, because the other factor is bounded and bounded away from zero on $(0,1)$.
On the other hand, if $x>1$ then $$\frac1{1+x^q}\sim x^{-q},$$where $A\sim B$ means there exist constants $c$, $C$ with $cA\le B\le CA$. So $\int_1^\infty$ converges if and only if $\int_1^\infty x^{-(p+q)}$ converges.
A: Because we have :
$$\dfrac{1}{x^p(1+x^q)}\sim\dfrac{1}{x^{\min\{p,p+q\}}},x\to0^+\qquad\dfrac{1}{x^p(1+x^q)}\sim\dfrac{1}{x^{\max\{p,p+q\}}},x\to+\infty,$$
and
$$\int_{0}^{+\infty}\frac{dx}{x^p(1+x^q)}=\int_{0}^{1}\frac{dx}{x^p(1+x^q)}+\int_{1}^{+\infty}\frac{dx}{x^p(1+x^q)}.$$
so when $\min\{p,p+q\}<1$ and $\max\{p,p+q\}>1$, the  integral is convergent, and otherwise  integral is not convergent.
