Evaluation of convergence of $$\int^\infty_1\frac{\log(1+x)}{x^k}dx$$
Where $k\in(0,\infty)$
What i try
Using Integration by parts
$$I=\ln(1+x)\cdot \frac{x^{1-k}}{1-k}\bigg|^{\infty}_{1}-\int^{\infty}_{1}\frac{1}{1+x}\cdot \frac{x^{1-k}}{1-k}dx$$
For $k>1$.I have seems that integral is converge.
I did not understand How to solve it. Help me please . Thanks
Observe that for all $x > 0 $ we have that $$ 1 - \frac{1}{1+x} \leq log(1+x) \leq x. $$
Use this along with monotonicity of the integral to get the bounds you desire.
Use that $\int_1^{\infty}\frac{1}{x^p}\,dx$ converges if and only if $p>1$ and diverges for $p\leq 1$ (this is easy to prove in case you're not familar).
When $k\leq 1$ $$\frac{\log(1+x)}{x^k}\geq \frac{\log(2)}{x^k}\;\text{for}\; x\geq 1 $$ and the divergence of $\log(2)\int_1^{\infty}\frac{1}{x^k}\,dx$ implies the divergence of the given integral.
When $k>1$, there exists an $A>1$ such that $\log(1+x)<x^{(k-1)/2}$ for $x>A$. Hence, $$0<\frac{\log(1+x)}{x^k}<\frac{x^{(k-1)/2}}{x^k}=\frac{1}{x^{(k+1)/2}}\;\text{for}\;x>A $$ and the convergence of $\int_A^{\infty}\dfrac{1}{x^{(k+1)/2}}\,dx$ implies the convergence of the given integral (note $(k+1)/2>1$).
What you've done with integration by parts is also a valid way to prove convergence in the case $k>1$. You have to show that both the limit and the second integral is finite (which basically reduces to the fact I cited above).
