Improper integral of natural log over a quadratic

I need to evaluate $$\int\limits_0^{+\infty}\frac{\ln{x}}{x^2+x+1}\,\mathrm{d}x\,.$$

I don't know how to integrate this, and for the most part, I don't even think it is expressible as elementary functions. In that case, how would I even manipulate the integral using some $u$-substitution to transform this into some integrable function? Or can this all be done without actual integration, and just some clever substitution to somehow find a multiple of this integral's value?

Substitute $u=\log(x),$ giving $$\int_0^\infty \frac{\log(x)}{x^2+x+1}dx = \int_{-\infty}^\infty \frac{u}{e^u+1+e^{-u}}du = 0$$ since the integrand is odd. (And the integral exists since the integrand decays exponentially in both directions.)

• I am absolutely shaken. Feb 25, 2017 at 21:26
• @spaceisdarkgreen, how did you know that would work? (It doesn't seem obvious to me to take $u=log(x)$) Oct 18, 2017 at 17:58
• @MrReality Don't remember my exact thought process here... When you see a $\log(x)$ it's a good idea to see what the integral looks like when you rearrange so that you also have a $\frac{dx}{x}$... it's a thing to try. Here it looks promising since the rest has inversion symmetry (which becomes reflection symmetry under $u=\log(x)$ - so my answer's not actually conceptually different from the others). It's also easy to try different things when you're reasonably sure there's a nice answer somewhere. Also it's conceivable that I already knew it was zero so already had symmetry on the brain. Oct 18, 2017 at 23:40
• @spaceisdarkgreen, hey I searched but couldn't find out what 'inversion symmetry' means for an expression (I think I know what 'reflection symmetry' means- it implies $f(x) = f(-x)$, right?). Can you please explain that? Oct 19, 2017 at 15:14
• @mrreality You're right about what reflection symmetry means although here I guess I'm using it ito refer to odd functions, not even. Similarly, inversion symmetry is a relationship between $f(1/x)$ and $f(x)$ rather than $f(-x)$ and $f(x)$. Oct 19, 2017 at 15:19

Considering $$\int_{0}^{\infty}\frac{\ln{x}}{x^2+x+1}\,dx=\int_{0}^{1}\frac{\ln{x}}{x^2+x+1}\,dx+\int_{1}^{\infty}\frac{\ln{x}}{x^2+x+1}\,dx$$ For the second integral, change variable $x=\frac 1y$, simplify and admire !

• What made you try this substitution, is it just experience? Feb 25, 2017 at 7:08
• @mrnovice. May be ! After 60 years of experience, who knows ? Feb 25, 2017 at 7:14
• @mrnovice Also, if you try integrating it numerically with a computer, it is easy to conjecture that the answer should be zero. This is something people should always attempt before asking here, in my opinion. Feb 25, 2017 at 11:36
• @FedericoPoloni Probably something I should start doing then? Feb 25, 2017 at 23:29

First of all, the integral is convergent because :

• $\displaystyle{\frac{\ln(x)}{x^2+x+1}\underset{0}{\sim}\ln(x)}$

• $\displaystyle{\frac{\ln(x)}{x^2+x+1}\underset{+\infty}{=}o\left(\frac1{x^{3/2}}\right)}$

Now consider change of variable $\displaystyle{t=\frac 1x}$

You will get :

$$\int_1^\infty\frac{\ln(x)}{x^2+x+1}\,dx=\int_1^0\frac{-\ln(t)}{\frac1{t^2}+\frac 1t+1}\frac{-dt}{t^2}=-\int_0^1\frac{\ln(t)}{t^2+t+1}\,dt$$

This proves that :

$$\boxed{\int_0^{+\infty}\frac{\ln(x)}{x^2+x+1}\,dx=0}$$

It should be added that, more generally (and for the same reasons) :

$$\forall a\in(-2,+\infty),\,\int_0^{+\infty}\frac{\ln(x)}{x^2+ax+1}\,dx=0$$

• Only if both parts are finite… but I dont see why this should hold
– Gono
Feb 25, 2017 at 7:05
• @Gono: Both parts are finite because of the asymptotic behavior I mentioned at the very beginning of my answer. Feb 25, 2017 at 7:08
• Ah… was it already written in your initial answer? Maybe I've just overseen it. Thx a lot!
– Gono
Feb 25, 2017 at 7:16
• Hey can you tell me why out of infinite possibilities(literally), you chose to split the original limits thus? Was it something obvious or did you already know the integral? Oct 18, 2017 at 17:55
• I knew that $\int_0^{+\infty}\frac{\ln(x)}{x^2+1}\,dx=0$, which can be proved exactly the same way. So it helped me a lot to figure out what to do with this integral. Oct 18, 2017 at 17:57