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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?

Mike Pierce
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Saketh Malyala
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3 Answers3

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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.)

spaceisdarkgreen
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  • I am absolutely shaken. – Saketh Malyala Feb 25 '17 at 21:26
  • @spaceisdarkgreen, how did you know that would work? (It doesn't seem obvious to me to take $u=log(x)$) – HeWhoMustBeNamed Oct 18 '17 at 17:58
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    @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. – spaceisdarkgreen Oct 18 '17 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? – HeWhoMustBeNamed Oct 19 '17 at 15:14
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    @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 $f(x)$. – spaceisdarkgreen Oct 19 '17 at 15:19
  • Ok, understood and I just realized that when you said this : "looks promising since the rest has inversion symmetry"$-$you were talking about the integrand as a whole (by 'the rest' I thought you were referring only to the denominator) so that was part of my confusion too! So, thank you, I think I understand it now. ;) – HeWhoMustBeNamed Oct 19 '17 at 16:00
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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 !

Claude Leibovici
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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) :

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

Adren
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  • Only if both parts are finite… but I dont see why this should hold – Gono Feb 25 '17 at 07:05
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    @Gono: Both parts are finite because of the asymptotic behavior I mentioned at the very beginning of my answer. – Adren Feb 25 '17 at 07:08
  • Ah… was it already written in your initial answer? Maybe I've just overseen it. Thx a lot! – Gono Feb 25 '17 at 07:16
  • 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? – HeWhoMustBeNamed Oct 18 '17 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. – Adren Oct 18 '17 at 17:57
  • @Adren, Thanks!! – HeWhoMustBeNamed Oct 19 '17 at 15:17