TITLE: For what values of p does the improper integral converge?
QUESTION [5 upvotes]: I have: 
$$\int^\infty_0\frac{\log(1+x)}{x^p}dx\qquad, \qquad p>0$$
I know the integral diverges if $\;0<p\le1\;$ by the comparison test, but need help with other values of $p$. Ideas? thanks!

REPLY [3 votes]: Near $0^+$, the integrand is equivalent to $$\frac {1}{x^{p-1}} $$
it converges if $p-1 <1$.
Near $+\infty $, if we put $x+1=t $, the new integrand is equivalent to
$$\frac {ln (t)}{t^p} $$
it is Bertrand type which converges if $p>1$
finally, the integral converges $\iff 1 <p <2$.