show that $ \frac{\Gamma(\frac{1}{24})\Gamma(\frac{11}{24})}{\Gamma(\frac{5}{24})\Gamma(\frac{7}{24})} = \sqrt{3}\cdot \sqrt{2 + \sqrt{3}} $

Question

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?

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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

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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.

Answer 1

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 $\sqrt3 \cdot \sqrt{2+\sqrt3}$.

Answer 2

After reading the original paper of Chowla-Selberg, I think it is a problem which can be solved with Kummer's Fourier Expansion of Gamma Function

$$\log\Gamma(x)=(\frac{1}{2}-x)(\gamma+\log2)+(1-x)\log\pi-(\log\sin\pi x)/2+\sum_{m=1}^{\infty}\frac{\sin 2\pi m x}{\pi m}\log m,$$

where $0<x<1$.

We do not need to worry about the first and the second term in Kummer's expansion because they cancel each other in addition. And it is not hard to show that $$\log\frac{\sin(\pi/24)\sin(11\pi/24)}{\sin(5\pi/24)\sin(7\pi/24)}=\log(2-\sqrt{3}).$$

We also need to figure out the sum $$S(m)={\sin(2\pi m/24)+\sin(2\pi m\times 11 /24)-\sin(2\pi m\times 5 /24)-\sin(2\pi m\times 7 /24)}.$$ It is nothing more than half of the Gauss sum for some Dirichlet characters $\chi$ modulo 24, which can be constructed like this:

$$\chi(1)=\chi(11)=\chi(17)=\chi(19)=1$$ and $$\chi(5)=\chi(7)=\chi(13)=\chi(23)=-1.$$

It is not hard to get $S(m)=0$ when $m$ is even, $S(m)=-\sqrt{2}\chi(m)$ when $m$ is prime to 24, and $S(3m)=2\sqrt{2}\chi_1(m)$ when $m$ is odd, where $\chi_1$ is a Dirichlet character modulo 8 and $$\chi_1(1)=\chi_1(3)=1,\chi_1(5)=\chi_1(7)=-1.$$ So we need to figure out the sum $$-\frac{\sqrt{2}}{\pi}\sum_{n=1}^{\infty}\frac{\chi(n)}{n}\log n+\frac{2\sqrt{2}}{\pi}\sum_{n=1}^{\infty}\frac{\chi_1(n)}{3n}\log (3n).$$ We notice that $\chi(n)=\chi_1(n)$ when $n$ is prime to 24. So the sum is equal to $$\frac{\sqrt{2}\log 3}{\pi}\sum_{n=1}^{\infty}\frac{\chi_1(n)}{n}.$$ We know that $$\sum_{n=1}^{\infty}\frac{\chi_1(n)}{n}=\frac{\pi}{2\sqrt{2}}$$ from class number formula, and we are done.

Answer 3

This follows from the discussion at and preceding page 31 in Campbell's book.

Answer 4

As a consequence of Euler's refection and Gauss's multiplication formulas (mentioned in the Elkies answer), all values $\Gamma(a)$ with $24a\in\mathbb{Z}$ can be expressed algebraically in terms of these values: $$\Gamma\!\left(\frac12\right)=\sqrt{\pi},\quad \Gamma\!\left(\frac13\right), \quad \Gamma\!\left(\frac14\right), \quad \Gamma\!\left(\frac18\right), \quad \Gamma\!\left(\frac1{24}\right).$$ For example, $$\Gamma\!\left(\frac5{24}\right)=\frac{\sqrt{\pi}\,\sqrt{\sqrt2-1}\,\sqrt{\sqrt3-1}}{2^{1/6}\,\sqrt{3}}\,\Gamma\left(\frac13\right)^{\!-1}\,\Gamma\!\left(\frac1{24}\right),\\ \Gamma\!\left(\frac7{24}\right)=\frac{\sqrt{\pi}\,\sqrt{\sqrt3-1}\,\sqrt{\sqrt3-\sqrt2}}{2^{1/4}\,3^{3/8}}\,\Gamma\left(\frac14\right)^{\!-1}\,\Gamma\!\left(\frac1{24}\right).$$ These expressions can be found in this paper, along with basic expressions of $\Gamma(a)$ with $60a\in\mathbb{Z}$ that additionally involve
$$\Gamma\!\left(\frac15\right),\quad \Gamma\!\left(\frac25\right), \quad \Gamma\!\left(\frac1{15}\right), \quad \Gamma\!\left(\frac1{20}\right), \quad \Gamma\!\left(\frac1{60}\right), \quad \Gamma\!\left(\frac7{60}\right).$$ Those formulas give $$\frac{\Gamma(\frac{1}{24})\,\Gamma(\frac{11}{24})}{\Gamma(\frac{5}{24})\,\Gamma(\frac{7}{24})} = \frac{\sqrt{6}}{\sqrt3-1}=\sqrt{\frac32}\,(\sqrt{3}+1).$$ This is a simpler answer! Consequently, $$\sqrt{2+\sqrt3}=\frac{\sqrt3+1}{\sqrt2}.$$