# Can this expression be simplified and is the real part of the left hand side equal to minus pi?

By starting with some Dirichlet series similar to logarithms, or somewhat similar to logarithm Dirichlet series, I arrived at this expression:

$$\left(\text{ArcSin}\left[\text{Sqrt}[2]*\text{Cos}[\text{Im}[\text{ZetaZero}[2]]*\text{Log}[2]]\left/\left(\frac{1}{18} \left(\sqrt{3} \pi -\sqrt{6} \pi \text{Cos}[\text{Im}[\text{ZetaZero}[1]] \text{Log}[2]]+9 \text{Log}[3]+9 \sqrt{2} \text{Cos}[\text{Im}[\text{ZetaZero}[1]] \text{Log}[2]] \text{Log}[3]\right)\right)\right.\right]*2\right)= -\pi$$

Can it be simplified and is the real part the left hand side equal to $-\pi$ ?

I apologize for the bad Latex formatting.

As a Mathematica command the the real part of the left hand side is:

N[Re[(ArcSin[
Sqrt[2]*Cos[
Im[ZetaZero[2]]*Log[2]]/(1/
18 (Sqrt[3] \[Pi] -
Sqrt[6] \[Pi] Cos[Im[ZetaZero[1]] Log[2]] + 9 Log[3] +
9 Sqrt[2] Cos[Im[ZetaZero[1]] Log[2]] Log[3]))]*2)], 90]


= -3.1415926535897932384626433832795028841971693993751058209749445923078\ 1640628620899862803483

Edit 31.8.2012:

This is what the Latex should look like:

= -3.1415926535897932384626433832795028841971693993751058209749445923078\ 1640628620899862803483

-
The expression in the Latex does not seem to be correct. The Mathematica program however gives this answer, -Pi = 3.14... –  Mats Granvik Aug 31 '12 at 15:55
Looking at it now, I start to believe that this is trivial. Sometimes the real part of an expression is some kind of identity. –  Mats Granvik Aug 31 '12 at 15:59

Notice that for $x>1$, $z =\arcsin(x)$ is complex. Let $z=u+i v$, then $$x = \sin(u+i v) = \cosh(v) \sin(u) + i \sinh(v) \cos(u)$$ Since $x$ is real $\cos(u) = 0$, and since $x>1$, $\sin(u) > 0$.
Restricting to the principal branch-cut of $\arcsin$, we thus have: $$\Re\left( \arcsin(x) \right) = u =\frac{\pi}{2}$$ Similarly for $x<-1$, $\Re\left( \arcsin(x) \right) = -\frac{\pi}{2}$.