I need Help in evaluating an integral. I have a function $$\rho_0(x) = \min_{ y\in\mathbb R} \rho( x,y) = \min_{0\leq y < 1} \rho(x, y), $$ which is $1$- periodic.
$\rho_0(x) $ can be explicitly given on 0$\leq$ x <1 by
$$\rho_0(x) = \begin{cases} 0 & \phantom{1/}0\leq x <1/3, \\ 1 & 1/3 \leq x<1/2, \\ 2 & 1/2 \leq x <2/3, \\ 3& 2/3 \leq x < 5/6, \\ 4& 5/6 \leq x < 1. \end{cases}$$
Integral to be proved is $\int _{0}^1 \rho_0(t) d (\psi(t) + 1/t) $ is approximately equal to 1.29564. ( here $\psi(t) $ = $\frac{ \Gamma'(t) } { \Gamma(t) } $ .
I would be really thankful for any help.
By the properties of the Riemann-Stieltjes integral we have that $$\int_{0}^{1}\rho_{0}\left(t\right)d\left(\psi\left(t\right)+1/t\right)=\int_{0}^{1}\rho_{0}\left(t\right)\left(\psi\left(t\right)+1/t\right)^{\prime}dt$$ hence $$\int_{0}^{1}\rho_{0}\left(t\right)d\left(\psi\left(t\right)+1/t\right)=\int_{1/3}^{1/2}\left(\psi\left(t\right)+1/t\right)^{\prime}dt+2\int_{1/2}^{2/3}\left(\psi\left(t\right)+1/t\right)^{\prime}dt$$ $$+3\int_{2/3}^{5/6}\left(\psi\left(t\right)+1/t\right)^{\prime}dt+4\int_{5/6}^{1}\left(\psi\left(t\right)+1/t\right)^{\prime}dt$$ $${\color{blue}{=\left[\psi\left(t\right)+1/t\right]_{1/3}^{1/2}+2\left[\psi\left(t\right)+1/t\right]_{1/2}^{2/3}+3\left[\psi\left(t\right)+1/t\right]_{2/3}^{5/6}+4\left[\psi\left(t\right)+1/t\right]_{5/6}^{1}}}\approx {\color{red}{1.29564}}.$$