4

I was working on some integrals, and I proved the following really beautiful identity:

$$\int_0^{\infty}\frac{\ln(x)}{x^2+nx+n^2}dx=\frac{2\pi}{3\sqrt{3}}\frac{\ln(n)}{n}$$

I found it by solving $\int_0^{\infty}\frac{x^{k-1}}{ax^2+bx+c}dx$, and then differentiating my result with respect to $k$ and evaluating at $k=1$. When I saw that it yielded a nice formula for the above integral, I was surprised, which is why I posted this question.

But my proof is very long and messy, and since the result is so elegant, I was wondering if there was a simple and elegant proof of this identity.

user21820
  • 57,693
  • 9
  • 98
  • 256
Milo Moses
  • 2,517
  • 2
    What is the idea of your proof? – Math Dec 15 '19 at 18:45
  • 2
    Can you tell in short terms what your proof consisted of? There is quite a short proof for it and I can't see a long one. – Zacky Dec 15 '19 at 18:45
  • It's a duplicate of https://math.stackexchange.com/q/2473132/515527 too. – Zacky Dec 15 '19 at 18:53
  • I had found $\int_0^{\infty}\frac{\ln(x)}{ax^2+bx+c}dx$, for a general $a,b,c$, and then I plugged in $a=1$, $b=n$ and $c=n^2$ to get this equation, which I know is not at all optimal. – Milo Moses Dec 15 '19 at 19:05
  • Well, how did you find that general integral in a long and messy way? – Zacky Dec 15 '19 at 19:07
  • I found it by solving $\int_0^{\infty}\frac{x^{k-1}}{ax^2+bx+c}dx$, and then differentiating my result with respect to $k$ and evaluating at $k=1$. – Milo Moses Dec 15 '19 at 19:08
  • Ouch no, that is definitely not a nice way to go about it. Dx – Simply Beautiful Art Dec 15 '19 at 19:09
  • Yeah, my original goal was to solve $\int_0^{\infty}\frac{x^k}{ax^2+bx+c}dx$, and then when I saw that it yielded such a since formula for $\int_0^{\infty}\frac{\ln(x)}{x^2+nx+n^2}dx$, I was surprised which is why I posted this question. – Milo Moses Dec 15 '19 at 19:11
  • Thank you! Indeed it's definetly messy to approach it that way. – Zacky Dec 15 '19 at 19:17
  • Since someone wants to delete this topic, I'll open it. – Michael Rozenberg Jan 26 '20 at 14:08

2 Answers2

5

Let $I(n)$ be the integral in question. Then, using the substitution $x=ny$, we get $$I(n)=\int_0^\infty\frac{\ln(nx)}{n(x^2+x+1)}\,dx=I(1)+ \frac{\ln n}n\int_0^\infty\frac{dy}{y^2+y+1}.$$ That latter integral can be done by a arctangent substitution. But using the substitution $y=1/x$ gives $I(1)=-I(1)$ so that $I(1)=0$.

Angina Seng
  • 158,341
2

Call your integral $I$. With $x\mapsto\frac{n^2}{x}$, $I=\int_0^\infty\frac{\ln(n/x)dx}{x^2+nx+n^2}$. Averaging, $I=\frac12\ln n\cdot\int_0^\infty\frac{dx}{x^2+nx+n^2}$.

J.G.
  • 115,835