Formulas for prime-counting functions $$\tag{1.a}\psi _{0}(x)=x-\sum _{\rho }{\frac {x^{\rho }}{\rho }}-\ln 2\pi -{\frac {1}{2}}\ln(1-x^{-2})$$
$$\tag{2.a}\Pi _{0}(x)=\operatorname {li} (x)-\sum _{\rho }\operatorname {li} (x^{\rho })-\ln 2+\int _{x}^{\infty }{\frac {dt}{t(t^{2}-1)\ln t}}$$
$$\tag{3.a}\pi _{0}(x)=\operatorname {R} (x)-\sum _{\rho }\operatorname {R} (x^{\rho })-{\frac {1}{\ln x}}+{\frac {1}{\pi }}\arctan {\frac {\pi }{\ln x}}$$
Thus corresponding sums over trivial zeros are: $$\tag{1.b}-\sum _{n=1}^{\infty}\frac{x^{-2 n}}{-2 n}=-{\frac {1}{2}}\ln(1-x^{-2})$$ $$\tag{2.b}-\sum _{n=1}^{\infty}\operatorname {li} (x^{-2 n})=\int _{x}^{\infty }{\frac {dt}{t(t^{2}-1)\ln t}}$$ $$\tag{3.b}-\sum _{n=1}^{\infty}\operatorname {R} (x^{-2 n})=-{\frac {1}{\ln x}}+{\frac {1}{\pi }}\arctan {\frac {\pi }{\ln x}}$$
I can verify $(1.b)$ by series expansion of $\log (x+1)=\sum _{n=1}^{\infty } \frac{(-1)^{n+1} x^n}{n}$ around x=0 and then substituting $x\to-x^{-2}$.
How do we verify or derive $(2.b)$ and $(3.b)?$
Another related question:
I can compute residue at s=0 in $(1.a)$ which is equal $-\ln 2\pi$ by using Taylor expansion.
But how can we compute residues at s=0 which appear in $(2.a)$ as $-\ln 2$ and in $(3.a)$ as $0$?
So here is my attempt on $(2.b)$ (which I do not consider very obvious): $$-\sum _{n=1}^{\infty}{\operatorname {li} (x^{-2 n})}=-\sum_{n=1}^{\infty}{\int _{0}^{x^{-2 n}}{\frac {1}{\ln t}\,dt}}$$
Using substitution $t=u^{-2 n}$ so that $dt=-2 n\,u^{-2 n-1} du$:
$$=-\sum_{n=1}^{\infty}{\int _{\infty}^{x}{\frac {-2 n\,u^{-2 n-1}}{\ln(u^{-2 n})}\,du}}=-\sum_{n=1}^{\infty}{\int _{\infty}^{x}{\frac {-2 n\,u^{-2 n-1}}{-2 n \ln u}\,du}}=-\sum_{n=1}^{\infty}{\int _{x}^{\infty}{\frac {-u^{-2 n-1}}{\ln u}\,du}}$$
$$=\sum_{n=1}^{\infty}{\int _{x}^{\infty}{\frac {u^{-2 n}}{u \ln u}\,du}}=\int _{x}^{\infty}{\frac {\sum_{n=1}^{\infty}{u^{-2 n}}}{u \ln u}\,du}=\int _{x}^{\infty}{\frac {1/(u^2-1)}{u \ln u}\,du}=\int _{x}^{\infty}{\frac {1}{u (u^2-1) \ln u}\,du}$$
So $(3.b)$ still remains open, so feel free to comment or answer on this one. ...and also the "residue question" still open.