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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:

minus pi expression

= -3.1415926535897932384626433832795028841971693993751058209749445923078\ 1640628620899862803483

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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
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1 Answer 1

up vote 2 down vote accepted

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 $x<-1$, $\Re\left( \arcsin(x) \right) = -\frac{\pi}{2}$.

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