UPDATED:
(I originally asked for a proof that the expression was real; I meant to ask for the real algebraic expression that the sum equates to - i.e., an expression in terms of $x$, $\sigma$ and $t$ but not $i$.)
Let $z = \sigma + i t$ be a complex number and let $\bar{z} = \sigma - i t$ be its conjugate. Let $x$ be a positive real number. I want to find the real expression corresponding to
$$\frac{x^z}{z} + \frac{x^{\bar{z}}}{\bar{z}}$$
(I am generalising from an expression in von Mangoldt's explicit formula for the Chebyshev $\psi$ function here.)
I realise that since $\frac{x^{\bar{z}}}{\bar{z}}$ is the conjugate of $\frac{x^z}{z}$, their sum must be real - but what form does it take?
This should involve a fairly simple piece of algebra, but I can't get there. I begin as follows:
$$ \begin{aligned} \frac{x^z}{z} + \frac{x^{\bar{z}}}{\bar{z}} &= \frac{x^{\sigma +i t}}{\sigma +i t}+\frac{x^{\sigma -i t}}{\sigma -i t} \\ &= \frac{e^{(\sigma +i t) \log (x)}}{\sigma +i t}+\frac{e^{(\sigma -i t) \log (x)}}{\sigma -i t} \\ &= \frac{(\sigma +i t) e^{(\sigma -i t) \log (x)}+(\sigma -i t) e^{(\sigma +i t) \log (x)}}{\sigma ^2+t^2} \end{aligned} $$
But I don't know how to proceed. I'd like to apply Euler's identity, but as far as I'm aware, the identity $e^{a+ib} = e^a e^{ib}$ does not apply because the original exponent is complex, so I cannot split out the complex component of $e^{(\sigma \pm i t) \log (x)}$. And even if it is valid to write $e^{a+ib} = e^a e^{ib}$, I still end up with a factor of $-2 i \sin (t \log (x))$ in the numerator.
Could someone show me how to arrive at the purely real expression that $\frac{x^z}{z} + \frac{x^{\bar{z}}}{\bar{z}}$ resolves to?
Your calculation is almost done. $$ 2\,{\frac {{{\rm e}^{\sigma\,\ln x }} \big( \sigma\cos \left( t\ln x \right)+t \sin \left( t\ln x \right) \big) }{{\sigma}^{2}+{t}^{2}}} $$