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The question is to find the value values of $p$ and $q$ such that the improper integral converges or diverges. My friend indeed found some values of $p$ and $q$ such that it converges, but after many trials, I am still getting divergent for all values. Via L'Hospital, we have the following:

$$\lim_{x \to 0} x^cln(1+x)^q = \infty \space \text{when} \space c < 0$$ $$\lim_{x \to 0} x^cln(1+x)^q = 0 \space \text{when} \space c > 0$$ $$\lim_{x \to \infty} x^cln(1+x)^q = \infty \space \text{when} \space c > 0$$ $$\lim_{x \to \infty} x^cln(1+x)^q = 0 \space \text{when} \space c < 0$$

Thus, $$\int_1^{\infty} x^pln(1+x)^q dx = \int_1^{\infty} x^{p-c}x^cln(1+x)^q dx > \int_1^{\infty} x^{p-c} dx$$ which diverges when $p > -1$. Here, we set $c = \frac{1}{2}(p+1) > 0$. On the other hand, $$\int_0^1 x^pln(1+x)^q dx = \int_0^1 x^{p-c}x^cln(1+x)^q dx > \int_0^1 x^{p-c} dx$$ which diverges when $p < -1$. Here, we set $c = \frac{1}{2}(p+1) < 0$. Thus, the integral diverges whenever $p \ne –1$. Then, let $p = -1$.

$$\int_0^{\infty} \frac{1}{x}ln(1+x)^q dx = \int_0^1 \frac{1}{x}ln(1+x)^q dx + \int_1^{\infty} \frac{1}{x}ln(1+x)^q dx $$ $$ \ge \int_0^1 \frac{1}{x+1}ln(1+x)^q dx + \int_1^{\infty} \frac{1}{x+1}ln(1+x)^q dx $$

The first summand diverges when $q < -1$ and the second diverges when $q > -1$. Finally, let $ p = -1$ and $q = -1$. The integral $\int_0^{\infty} \frac{1}{xln(1+x)} dx ≥ \int_0^{\infty} \frac{1}{(x+1)ln(1+x)} dx$ diverges on both ends. Did I make a mistake?

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The result seems correct, but check the limits for $x\to 0$.

$\int_1^{+\infty} x^p \log^q (1+x)dx$ diverges if $p>-1$, $\int_0^{1} x^p \log^q (1+x)dx$ diverges if $p<-1$, independently of $q$.

It remains $p=-1$.

$\int_1^{+\infty} x^{-1}\log^q (1+x)dx$ diverges if $q\geq -1$. Note that for small $x$ the behavior of $\log(1+x)\sim x$, hence
$$\int_0^{1} x^{-1}\log^q (1+x)dx\sim \int_0^{1} x^{-1}x^q dx,$$ which diverges for any $q\leq 0$.

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