Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:

     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

share|cite|improve this question
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
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 $x<-1$, $\Re\left( \arcsin(x) \right) = -\frac{\pi}{2}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.