For which values of $p$ and $q$ is the integral convergent and absolutely convergent : $\int_1^{\infty} \frac{x^p\,dx}{1+x^q}$ 
For which values of $p$ and $q$ is the integral convergent and absolutely convergent :
$$\int_1^{\infty} \frac{x^p\,dx}{1+x^q}$$

My try :
Let $$f(x) = \frac{x^p}{1+x^q}$$
For $q \gt 0$ :
Let $g(x) = \frac{1}{x^{q-p}}$ on $[1, \infty]$.
Then
$$
\lim_{x\to \infty} \frac{f(x)}{g(x)} 
= \lim_{x\to \infty} \frac{x^q}{1+x^q} 
= 1 \ne 0.
$$
Thus $\int g$ and $\int f$ converge together.
Now, $\int g$ converges iff $q-p \gt 1$ i.e. $q \gt 1+p$, so does $\int f$.
I'm not sure that if this concludes the case $q \gt 0$.
Also what can I do when $q \le 0$.
I neee some help please.
Also since the integrand is positive in the given interval so I guess whenever it's convergent, it is absolutely convergent. So I don't need to work it out separately. Am I right?

*

*You are correct about absolute convergence: the function being integrated is positive in this interval so convergence and absolute convergence are the same thing here.


*Your calculation for $q>0$ is correct.


*For $q<0$ the denominator of the integrand tends to $1$ as $x$ tends to infinity, which means that the integrand $\frac{x^p}{1+x^q}$ behaves roughly like $x^p$, and since $\int_1^{\infty} x^p dx$ converges if and only if $p<-1$ then the same is true for the original integral. Of course to prove this rigorously we use the comparison test: for $q<0$ we have
$$\lim_{x\to\infty} \frac{\frac{x^p}{1+x^q}}{x^p}=\lim_{x\to\infty} \frac{1}{1+x^q}=1$$
so that for $q<0$ the integral converges if and only if $p<-1$.


*For $q=0$ we get in a similar way that the integrand behaves like $x^p$ and we prove it in the same way using the comparison test.
