It says that the roots of $$f(x) = x^{-1} \sin(x^{-1}\log(x))$$ are defined as $$1 > a_{1} > a_{2} > \cdots > 0$$ where $a_{i} = \exp(-b_{i})$ and $b_{i}$ is the unique solution to the equation $b \exp(b) - i\pi = 0$, $1 < b < \infty$.
I am wondering how are formulas of $a_{i}$ and $b_{i}$ calculated?
Because the $x_i>0$, we can find values $b_i$ such that $x_i = \exp(-b_i)$. Plug this into $f(x)$ to get \begin{align} f(\exp(-b_i)) &= \exp(-b_i)^{-1}\sin{(\exp{(-b_i)}^{-1}\log{(\exp{(-b_i)})})}\\ &= \exp(b_i)\sin(-b_i\exp{b_i}) \end{align} Setting this equal to $0$ and using the fact that $\exp(b_i)>0$ we get $$\sin(-b_i\exp(b_i))=0$$ hence, $$-b_i\exp{b_i} = -i\pi$$ for integer $i$. This is equivalent to $$b_i\exp{b_i}-i\pi=0.$$
