1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I believe that this does not converge for any $x\in\mathbb{R}$. $\endgroup$ – Peter Foreman May 14 at 18:04
1
$\begingroup$

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 $x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.