# 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 $$$$\int_0^{\infty}x^p[\ln(1+x)]^q\,dx$$$$ There is already a question regarding this problem (Here), but I think there is a small glitch which needs to be modified. I agree with his argument for the case $$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 $$q$$):

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