What is the formula for the first Riemann zeta zero?

Can you improve on the formula above? -- Yes: $$ \text{Log}\left[\frac{2}{3} e^{-5 \pi /2}+\frac{e^{7 \pi }}{\frac{7}{2} e^{-7 \pi /2}+\frac{5}{2} e^{-5 \pi /2}+\frac{3}{2} e^{-3 \pi /2}+e^{5 \pi /2}+2 \pi }\right] = 14.1347251417343... $$

Let $n=1$, then the formula where $\vartheta (t)$ is the Riemann Siegel theta function and $\text{sgn}$ is the sign function:

$$ 14.13472514173469...=\int _0^{16}\frac{1}{2} \left(1-\text{sgn}\left(\frac{\vartheta (t)+\Im\left(\log \left(\zeta \left(i t+\frac{1}{2}\right)\right)\right)}{\pi }-n+\frac{3}{2}\right)\right)dt$$

gives the first Riemann zeta zero.

Mathematica:

Clear[n, t, gamma];
gamma = 15; 
Quiet[
 Do[gamma = 
   N[NIntegrate[(1/2)*(1 - 
        Sign[(RiemannSiegelTheta[t] + Im[Log[Zeta[I*t + 1/2]]])/Pi - 
          n + 3/2]), {t, 0, gamma + 15}, PrecisionGoal -> 45, 
     MaxRecursion -> 220, WorkingPrecision -> 50], 40]; 
  Print[gamma], {n, 1, 1}]]

14.13472514173469379045725198356247027078

which gives the same result as:

N[Im[ZetaZero[1]], 40]

14.13472514173469379045725198356247027078

Update 16.8.2017:

The integral below might be solvable through Fourier series approximations of the Floor and Sign functions, and repeated integration by parts, since there is no function in the denominators of the power series for the Fourier series.

$n=1$ $$14.1347251417346937904572\text{...}=\int _0^{16}\frac{1}{2} \left(1-\text{sgn}\left(\left(\left\lfloor \frac{\vartheta (t)}{\pi }+1\right\rfloor +\frac{1}{2} \left(-1+\text{sgn}\left(\Im\left(\zeta \left(i t+\frac{1}{2}\right)\right)\right)\right)\right)-n+\frac{3}{2}\right)\right)dt$$

The convergence of the symbolic solution is probably not great though, since the powers in the power series cause the complexity of the symbolic expression to grow rapidly.

Mathematica program:

Clear[n, t, gamma]
gamma = 15;
Quiet[Do[gamma = 
   N[NIntegrate[(1/2)*(1 - 
        Sign[((Floor[
              RiemannSiegelTheta[t]/Pi + 
               1]) + (Sign[Im[Zeta[1/2 + I*t]]] - 1)/2) - n + 
          3/2]), {t, 0, gamma + 16}, PrecisionGoal -> 45, 
     MaxRecursion -> 220, WorkingPrecision -> 50], 40];
  Print[gamma], {n, 1, 10}]]