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

Question

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.

Answer 1

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 $I(1)=0$.

Answer 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, $I=\frac12\ln n\cdot\int_0^\infty\frac{dx}{x^2+nx+n^2}$.