How to find interval of convergence in integral? Can someone explain how to find interval of convergence for$$F(x)=\int_{0}^{+\infty}\frac{\ln(1+y)}{y^x}dy$$
I tried to integrate then find convergence interval of it, but I couldn't integrate it correctly (I guess). Also I tried to calculate it with Maxima Antiderivative but it didn't work aswell.
I don't have my workbooks with me unfortunately so I don't know if it is possible to find interval with using integral test.

For $\Re({x})>1$, the integrand $\frac{\ln(1+y)}{y^x}$ diverges at $y=0$ since $\frac{\ln(1+y)}{y^x}\asymp \frac{1}{y^{x-1}}$ as $y\to0$. For $\Re(x)\le0$, the integrand does not go to $0$ as $y\to\infty$ and thus the whole integral is divergent. That leaves $\Re(x)\in(0,1]$. We notice that $$F(x)=\int_0^\infty \frac{\ln(1+y)}{y^x}dy>\int_1^\infty \frac{dy}{y^x}$$
Now for ${x}\ne 1$ we have $$\int_1^\infty \frac{dy}{y^x}=\frac{y^{1-x}}{1-x}|_{y=1}^{y=\infty}$$
which diverges. 
At $x=1$, the integral evaluates to $$\int_1^\infty \frac{dy}{y}=\ln(y)|_1^\infty=\infty$$ Thus, $F(x)$ diverges for any given $x$.
