Mathworld's discussion of the Gamma function has the pleasant formula:
$$ \frac{\Gamma(\frac{1}{24})\Gamma(\frac{11}{24})}{\Gamma(\frac{5}{24})\Gamma(\frac{7}{24})} = \sqrt{3}\cdot \sqrt{2 + \sqrt{3}} $$
This may have been computed algorithmically, according to the page. So I ask how one might derive this?
My immediate thought was to look at $(\mathbb{Z}/24\mathbb{Z})^\times
= \big( \{ 1,5,7,11 \big| 13, 17, 19 , 23 \}, \times \big)$ where $1,5,7,11$ are relatively prime to 24. And the other half?
We could try to use the mirror formula
$$ \Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin(\pi z)} $$
or the Euler beta integral but nothing has come up yet:
$$ \int_0^1 x^a (1-x)^b \, dx = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)} $$
I am lucky the period integral of some Riemann surface will be the ratio of Gamma functions:
$$ \int_0^1 (x – a)^{1/12}
(x – 0)^{11/12}
(x – 1)^{-5/12}
(x – d)^{-7/12} \, dx
$$
these integrals appear in the theory of hypergeometric function
In light of comments, I found a paper of Benedict Gross and the paper of Selberg and Chowla
$$ F( \tfrac{1}{4},\tfrac{1}{4};1;\tfrac{1}{64}) = \sqrt{\frac{2}{7\pi}} \times \left[\frac{
\Gamma(\frac{1}{7})\Gamma(\frac{2}{7})\Gamma(\frac{4}{7})
}{
\Gamma(\frac{3}{7})\Gamma(\frac{5}{7})\Gamma(\frac{6}{7})
}\right]^{1/2} $$
so in our case we are looking at quadratic residues mod 12. However, however it does not tell us that LHS evaluates to RHS.
Best Answer
This formula can actually be proved using only properties of the Gamma function already known to Gauss, with no need to invoke special values of Dirichlet series. The relevant identities are $$ \Gamma(z) \, \Gamma(1-z) = \frac\pi{\sin(\pi z)}, $$ already cited by john mangual as the "mirror formula", and the triplication formula for the Gamma function, i.e. the case $k=3$ of Gauss's multiplication formula: $$ \Gamma(z) \, \Gamma\bigl(z+\frac13\bigr) \, \Gamma\bigl(z+\frac23\bigr) = 2\pi \cdot 3^{\frac12-3z} \Gamma(3z) $$ [the $k=2$ case is the more familiar duplication formula $\Gamma(z) \, \Gamma(z+\frac12) = 2^{1-2z} \sqrt{\pi}\, \Gamma(2z)$].
Take $z=1/24$ and $z=1/8$ in the triplication formula, multiply, and remove the common factors $\Gamma(1/8) \, \Gamma(3/8)$ to deduce $$ \Gamma\bigl(\frac{1}{24}\bigr) \Gamma\bigl(\frac{11}{24}\bigr) \Gamma\bigl(\frac{17}{24}\bigr) \Gamma\bigl(\frac{19}{24}\bigr) = 4 \pi^2 \sqrt{3}. $$ Take $z=5/24$ and $z=7/24$ in the mirror formula and multiply to deduce $$ \Gamma\bigl(\frac{5}{24}\bigr) \Gamma\bigl(\frac{7}{24}\bigr) \Gamma\bigl(\frac{17}{24}\bigr) \Gamma\bigl(\frac{19}{24}\bigr) = \frac{\pi^2}{ \sin (5\pi/24) \sin (7\pi/24) }. $$ Hence $$ \frac{\Gamma(1/24) \, \Gamma(11/24)} {\Gamma(5/24) \, \Gamma(7/24)} = 4 \sqrt{3} \sin (5\pi/24) \sin (7\pi/24), $$ which is soon reduced to the radical form $\sqrt3 \cdot \sqrt{2+\sqrt3}$.