from http://www.cs.cmu.edu/~adamchik/articles/catalan/catalan.htm In[1]:=Integrate[ArcSin[Sinh[t]], {t, 0, ArcSinh[1]}]
Out[1]=-Catalan + (1/4)*Pi*Log[3 + 2*Sqrt[2]]
(Not the trivial changevar of the ArcSinh[Sin[]].)
rcs>Cosmetic change:
I'd restate the pi term as log(1+sqrt2) pi /2. --Rich
Maybe Mma 7.0 with have CosmeticForm.-) Actually, it usually takes the lazy approach of brutally canonicalizing. NoninvasiveForm would be ArcSinh[1] Pi/2 . WFL> [...]but it is disconcerting to discover how often [fatuous arrogance] arises
in a subject lacking any obvious reliance on experimental interpretation.
Can anyone supply a non-experimental proof or refutation of inf n + 1 ==== ==== \ \ j > > j (- 1) log(j) binomial(n, j - 1) = 0 ? / / ==== ==== n = 0 j = 1 Failing that, let's experiment with it and quarrel over the results! Mike Stay> What's the definition of q-deformed pi, pi_q? Just take that infinite product following (d37) as the definition--it's a q-extension of Wallis's product. (-> pi as q -> 1). rwg>> [Re the display labeled (d183)]
near the end of www.tweedledum.com/rwg/idents.htm, suppose a,b>0. How can the 4th equand be negative, but not the 5th?
Mike Stay>If a,b > 0 then W(-e^{-a} b) is negative, so the entire expression is
positive. The only questionable equand in the sequence is, as you noticed, the fourth. But the trick there is that the k = -1 term in the sum outweighs the nonnegative terms.
Spot on, as usual. OK, cruel demystifier (or anyone else puzzle-prone), why does n (n + 1) --------- n 2 (sqrt(2) + 1) pi (- 1) sec(-----------------) 4 4 limit ------------------------------------- = -- ? n -> inf n pi (sqrt(2) + 1) --rwg