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?


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 $q$):


  • $\begingroup$ Sorry for being late!! Thanks for the answer~ I'd really appreciate it! $\endgroup$ – Kim Oct 19 at 4:42

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