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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Nov 14, 2010 at 21:19 history edited Blue CC BY-SA 2.5
added reference to another MathOverflow post for context
Nov 14, 2010 at 12:10 comment added J. M. isn't a mathematician After days of playing around, I have to agree that it's very tantalizing; I suspect some weird identity for the Lerch transcendent may be the key (since Dirichlet series and the polylogarithms are special cases of the Lerch transcendent).
Nov 12, 2010 at 20:14 comment added Suvrit Yes, I also played around; first you have $16^k$; if you had $(-16)^k$ then things would simplify dramatically I think.
Nov 10, 2010 at 20:43 comment added Blue @Suvrit: Interestingly, Adamchik uses that trick in Entry 30 here cs.cmu.edu/~adamchik/articles/catalan/catalan.htm to get a series for Catalan's constant that's almost like the series for $T(1/2)$, except my $16^k$ is his $8^k$. (So close!) Applying the trick to the $T(1/2)$ series gives an integral that Mathematica immediately evaluates to what is effectively the dilogarithm formula.
Nov 9, 2010 at 22:27 comment added Suvrit does the standard trick of writing $1/(2k+1) = \int_0^1 x^{2k}dx$ help?
Nov 9, 2010 at 20:26 history edited Blue CC BY-SA 2.5
added dilogarithm rep of T(1/2) and hypergeom rep of G
Nov 9, 2010 at 15:49 comment added J. M. isn't a mathematician Another integral representation: $2\int_0^{\pi/6}u\cot\;u\mathrm{d}u$
Nov 9, 2010 at 15:20 comment added J. M. isn't a mathematician From functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/… , we get an integral representation for $T(1/2)$: $\int_0^1 \frac1{u}\arcsin\left(\frac{\sqrt{u}}{2}\right)\mathrm{d}u$.
Nov 9, 2010 at 15:13 comment added J. M. isn't a mathematician $T(1/2)$ is expressible in terms of dilogarithms: functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/…
Nov 9, 2010 at 10:33 history edited Blue CC BY-SA 2.5
added focus on r=1/2 and catalan constant
Nov 9, 2010 at 10:10 history edited Blue CC BY-SA 2.5
fixed embarrassing typo in title
Nov 9, 2010 at 10:02 comment added Blue @J. M.: Checking them now ... :)
Nov 9, 2010 at 10:00 history edited Blue CC BY-SA 2.5
added "3F2" representations
Nov 9, 2010 at 9:38 comment added J. M. isn't a mathematician Mathematica says they're both ${}_3 F_2$ functions. I'll take a look, but in the meantime, have you checked dlmf.nist.gov/16 and functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2 ?
Nov 9, 2010 at 9:30 history asked Blue CC BY-SA 2.5