Deriving asymptotic for the roots of Digamma Function

Question

So Wikipedia gave these asymptotics for the Digamma function: $$x_n=-n+\frac12+O\left(\frac1{(\ln n)^2}\right)$$$$x_n\approx-n+\frac1\pi\arctan\left(\frac{\pi}{\ln n}\right)$$$$x_n\approx-n+\frac1\pi\arctan\left(\frac{\pi}{\ln n+\frac1{8n}}\right)$$I'm interested in deriving the second one. All Wikipedia says is to use the reflection formula $0=\psi(1-x_n)=\psi(x_n)+\pi\cot(\pi x_n)$ and then substitute the asymptotic expansion of the digamma function. Firstly, because $x_n$ is the $n$th largest root of the Digamma function, $\psi(1-x_n)$ doesn't (and shouldn't) need to be equal to zero. So I think the correct expression to analyze is $0=\psi(1-x_n)-\pi\cot(\pi x_n)$. Substituting the first term in the asymptotic of digamma we get $0\approx\ln(1-x_n)-\pi\cot(\pi x_n)$. I did solve for $x_n$ from the cotangent term and got that $$x_n\approx\frac1{\pi}\arctan\left(\frac\pi{\ln(1-x_n)}\right)$$But I don't know what to do next.

Answer 1

After looking at a similar post, it seems to be a much better idea if we write $\psi(1+x_n)=\psi(-x_n)-\pi\cot(\pi x_n)$ and use the recurrence relation of $\psi$ to get $\frac1{x_n}=\psi(-x_n)-\pi\cot(\pi x_n)\sim\log(-x_n)-\pi\cot(\pi x_n)$. Because $\frac1{x_n}\sim0$ we just get $\log(-x_n)-\pi\cot(\pi x_n)\sim0$. Let $x_n=-n-f(n)$ where $f(n)$ is a bounded function, then we get that $\log(n+f(n))\sim\log(n)$. Plugging this definition into the last asymptotic, we get that $f(n)\sim-\frac1\pi\arctan\left(\frac\pi{\ln n}\right)$ and so $x_n\sim-n-\frac1\pi\arctan\left(\frac\pi{\ln n}\right)$ as desired.