Edit of original post:
I've been noodling around with non-trivial zeroes of the Riemann zeta function, and in particular with the expression
$$\sum_{\rho}R(x^\rho)$$
where $R(x)$ is Riemann's famous approximation to the prime counting function, and $\rho=\sigma+it$ represents all instances of non-trivial zeroes of the zeta function.
Since the non-trivial zeroes come in conjugate pairs, this expression can be expanded to
$$\sum_{\rho}\bigl(R(x^{\sigma+it})+R(x^{\sigma-it})\bigr)$$
Henrik has explained (comments below) why the complex components cancel.
This means that the sum $R(x^{\sigma+it}) + R(x^{\sigma - it})$ must equal $2$ times some real value. How do I arrive at this real value?
Original post, for the record:
I've been noodling around with non-trivial zeroes of the Riemann zeta function, and in particular with the expression
$$\sum_{\rho}R(x^\rho)$$
where $R(x)$ is Riemann's famous approximation to the prime counting function, and $\rho=\sigma+it$ represents all instances of non-trivial zeroes of the zeta function.
Since the non-trivial zeroes come in conjugate pairs, this expression can be expanded to
$$\sum_{\rho}\bigl(R(x^{\sigma+it})+R(x^{\sigma-it})\bigr)$$
What I can't figure out is why the imaginary components of $\bigl(R(x^{\sigma+it})+R(x^{\sigma-it})\bigr)$ cancel.
I know that the complex exponent of a real base is given by
$$x^{\sigma+it}=x^\sigma \biggl(\cos \bigl(t \ln(x)\bigr)+i\sin \bigl(t \ln(x)\bigr) \biggr)$$
and I tried plugging this into the equation
$$R(x)=1+\sum_{k=1}^\infty \frac{\ln(x)^k}{kk!\zeta(k+1)}$$
but then I got tangled up in summing powers of logarithms, and didn't get anywhere.
Could someone help me understand why the complex components cancel?
Turning a comment into an answer.
Is $x$ suppose to be real and nonnegative? In that case this would follow from the fact that $R$ is meromorphic (so that $R(\bar z)= \overline{R(z)}$) and from $x^{\sigma−\operatorname{i}t} = \overline{x^{\sigma+\operatorname{i}t}}$.