# 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.

• I believe that this does not converge for any $x\in\mathbb{R}$. – Peter Foreman May 14 at 18:04

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 $$x$$.