Find the values of p and q that make an improper integral converge. Q. Determine the values of p and q for which the improper integral converges
\begin{equation}
 \int_0^{\infty}x^p[\ln(1+x)]^q\,dx
\end{equation}
 There is already a question regarding this problem (Here), but I think there is a small glitch which needs to be modified. I agree with his argument for the case $p = -1$ or $p > -1$. When proving that this integral diverges for $p < -1$, however, he used$$\lim_{x \to 0} x^cln(1+x)^q = \infty \space \text{when} \space  c < 0$$ But for the case $ c = -1$ and $q = 1$ , corresponding limit approaches $1$, not $\infty$. So I think we need a modification for the case $p < -1$. Maybe I'm missing some context, too. Summarizing, what I'm curious about is as follows: 
 1. Is his argument correct for the case $p = -1$ and $p > -1\ ?$
 2. How can we prove for the case $p < -1 $ correctly?
 A: It's probably best to organize our thoughts by considering a $pq$ plane of values. Note that both $x$ and $\ln(1+x)$ have zeros at $x=0$.
If both $p$ and $q$ are positive, this integral diverges horribly due to the infinite bound, so our entire first quadrant is out. Similarly, if $p$ and $q$ were both negative, then the integral would diverge on the other bound at $0$, so our third quadrant is gone, too. The last easy rule-outs are the axes, where either $p=0$ or $q=0$.
Split the integral into three parts:
$$= \int_0^\epsilon x^p \ln^q(1+x) dx + \int_\epsilon^T x^p \ln^q(1+x) dx + \int_T^\infty x^p \ln^q(1+x) dx$$
for some $\epsilon$ and $T$ small enough and large enough to apply our approximations, respectively. Near $0$, $\ln(1+x)\approx x$, so the integrand looks like $\frac{1}{x^{-(p+q)}}$. The integral test near $0$ tells us that $$-(p+q) < 1 \implies p+q > -1$$
The integral in the middle is always finite so we can ignore it. On the other hand, for large $x$, $\ln(1+x) \approx \ln(x)$. So consider the the right integral with that asymptotic substitution and the change of variable $u = \ln(x)$:
$$\int_T^\infty x^p \ln^q(x) dx = \int_{\ln T}^\infty u^q e^{(p+1)u}du$$
The integral only converges either if $p+1 < 0$, or if $p+1 = 0$ and $q < -1$ because the exponential dominates the behavior of the integral. However, the second case is impossible because those points $(-1,q)$ exist in the third quadrant of the $pq$ plane.
Combining our inequalities, we definitively have that $p < -1$ and $p+q > -1$. Here is a graph from Desmos showing the region of allowed values (with the horizontal axis being $p$ and the vertical axis being $q$): 

