Find the values of $m$, so the integral converges I need to find the values of $m$, so that the following integral converges:
$$\int_{1}^\infty \frac{\ln(1+x)}{x^m}dx$$
I need to choose between two options $m>1$ and $m>0$.
I know that for any $x>e-1,$ $\ln(1+x)>1$, so by the comparison test the integral divergent if $m<1.$
However, I don't know how to prove that it converges for $m>1$.
 A: For $m> 1$, you have when $x$ tends to $+\infty$
$$\frac{\ln(1+x)}{x^m} = o \left( \frac{1}{x^{\frac{m+1}{2}}}\right)$$
Because $\frac{1+m}{2} > 1$, then $\frac{1}{x^{\frac{m+1}{2}}}$ is integrable on $[1, +\infty)$, therefore your integral converges.
For $m \leq 1$,
$$\frac{\ln(1+x)}{x^m} \geq \frac{1}{x^m}$$
which is not integrable, therefore your integral diverges.
A: We have that for $m>1$
$$\frac{\left(\frac{\ln(1+x)}{x^m}\right)}{\left(\frac{1}{x^{\frac{1+m}2}}\right)}=\frac{\ln(1+x)}{x^\frac{m-1}2}\to 0$$
therefore the integral converges by limit comparison test with $\int_1^\infty \frac{dx}{x^{\frac{1+m}2}}$.
For the case $m\le 1$ we have
$$\frac{\left(\frac{\ln(1+x)}{x^m}\right)}{\left(\frac{1}{x^m}\right)}=\ln(1+x)\to \infty$$
therefore the integral diverges by limit comparison test with $\int_1^\infty \frac{dx}{x^{m}}$.
A: If $ \  0<m<1$, for example, $m=$$1 \over 2$, then we have $\int_1^{\infty} \frac {\ln(x)}{\sqrt{x}}dx$.
But $\forall x > \frac32$ we know that $\frac {\ln(x)}{\sqrt{x}}>\frac1x$ so the integral does not conference.
