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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?

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  • $\begingroup$ 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 $0$? $\endgroup$ – Stinking Bishop Feb 5 at 12:44
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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?

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  • $\begingroup$ But the integral goes from $0$ to $\infty$ not $1$ to $\infty$ . $\endgroup$ – Future Math person Feb 5 at 14:46
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    $\begingroup$ 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. $\endgroup$ – José Carlos Santos Feb 5 at 14:55
  • $\begingroup$ What about the 2nd integral? How does that converge? $\endgroup$ – Future Math person Feb 5 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. $\endgroup$ – José Carlos Santos Feb 5 at 18:18
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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

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