I'm asked to determine if the following integral converges or diverges.

$$\int_{0}^{\infty} \frac{\ln(x)}{x^2+x+1} dx $$.

I know I have to use the comparison test for improper integrals (since there is no elementary anti derivative), but I'm not sure what to use as my comparison.

INTUITIVELY, it seems that the function will converge since the denominator grows very fast, but what can I use for my comparison test here? I need a function that's larger than the integrand in order to determine convergence or divergence of this integral.

Can someone suggest what kind of function I can pick as my comparison?

  • $\begingroup$ Hint: You need to use the comparison test separately on intervals $(0,c]$ and $[c,+\infty)$ as this integral is improper "on both sides". Towards $+\infty$ you can indeed use $\frac{x^\alpha}{x^2+x+1}$, where $0<\alpha<1$. What would you use towards $0$? $\endgroup$ Feb 5, 2021 at 12:44

2 Answers 2


Hint: You can deal with$$\int_1^\infty\frac{\ln(x)}{x^2+x+1}\,\mathrm dx$$doing $x=\frac1y$ and $\mathrm dx=-\frac{\mathrm dy}{y^2}$, thereby getting$$\int_1^0-\frac{\ln(1/y)}{y^2+y+1}\,\mathrm dy=-\int_0^1\frac{\ln(y)}{y^2+y+1}\,\mathrm dy.$$Can you take it from here?

  • $\begingroup$ But the integral goes from $0$ to $\infty$ not $1$ to $\infty$ . $\endgroup$ Feb 5, 2021 at 14:46
  • 1
    $\begingroup$ I know that! Note that$$\int_0^\infty\frac{\ln(y)}{y^2+y+1}\,\mathrm dy=\int_0^1\frac{\ln(y)}{y^2+y+1}\,\mathrm dy+\int_1^\infty\frac{\ln(y)}{y^2+y+1}\,\mathrm dy$$and that therefore what I did reduces the problem of deciding whether your integral converges or not to the problem of deciding whether the integral $\int_0^1\frac{\ln(y)}{y^2+y+1}\,\mathrm dy$ converges or not. But this one is easy, since, near $0$, $\frac{\ln(y)}{y^2+y+1}$ behaves as $\ln y$ and it is easy to see that the integral $\int_0^1\ln(y)\,\mathrm dy$ converges. $\endgroup$ Feb 5, 2021 at 14:55
  • $\begingroup$ What about the 2nd integral? How does that converge? $\endgroup$ Feb 5, 2021 at 17:57
  • $\begingroup$ You have$$\int_0^1-\ln(y)\,\mathrm dy=\lim_{y\to0^+}1-1\times\ln(1)-\bigl(y-y\ln(y)\bigr)=1.$$So, since$$\lim_{y\to0^+}\frac{\frac{-\ln(y)}{y^2+y+1}}{-\ln(y)}=1$$and since the integral $\int_0^1\ln(y)\,\mathrm dy$ converges, then the integral $\int_0^1\frac{-\ln(y)}{y^2+y+1}\,\mathrm dy$ converges, and so the integral $\int_0^1\frac{\ln(y)}{y^2+y+1}\,\mathrm dy$ converges too. $\endgroup$ Feb 5, 2021 at 18:18

for the convergence around $0$, you can notice that $\left|\frac{\ln(x)}{x^2+x+1} \right|\leq |ln(x)|$ which is convergent around $0$.

Then, for the upper bound, you can note that for $x$ large enough, $ln(\sqrt{x})\leq \sqrt{x} $

So you get :

$$\left|\frac{\ln(x)}{x^2+x+1} \right|\leq \dfrac{2\sqrt{x}}{x^2+x+1}\sim_{\infty} 2x^{-3/2} $$ Which is integrable according to the Reimman test


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.