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How can we prove the following identity? $$\large \frac{\Gamma\left(\frac27\right) \Gamma\left(\frac{11}{42}\right)}{\Gamma\left(\frac1{21}\right)} = \frac{8 \sin\left(\frac\pi7\right) \sqrt{\pi \, \sin\left(\frac\pi{21}\right) \sin\left(\frac{4\pi}{21}\right) \sin\left(\frac{5\pi}{21}\right)}}{\sqrt[42]2 \; \sqrt[3]7 \; \sqrt[28]{19683}}$$ I guess we could use the Gauss multiplication formula, but how?

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  • 1
    $\begingroup$ You might start by squaring both sides (which is OK since each side is obviously nonnegative) and then using $\sin(\pi x)=\pi/\Gamma(x)\Gamma(1-x)$ to convert things into a product of Gammas. Also, note that $19683=3^9$. (Caveat: I haven't thought this all the way through, so it might not work. But it's what I would try first if I were going to post an answer instead of just a comment.) $\endgroup$ – Barry Cipra Jun 1 '18 at 16:09
  • $\begingroup$ Sorry. I guess my "answer" should have been a comment instead, $\endgroup$ – mds Jun 1 '18 at 16:26
  • 3
    $\begingroup$ +1 Very nice identity. $\endgroup$ – Tito Piezas III Jun 1 '18 at 16:58
  • $\begingroup$ By the way, $\Gamma\left(\frac37\right)\,\Gamma\left(\frac2{21}\right)/\,\Gamma\left(\frac1{42}\right)$ is also $\sqrt\pi$ times algebraic. You might enjoy finding its explicit form in trigonometric terms. $\endgroup$ – Vladimir Reshetnikov Jun 2 '18 at 0:32
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The product given by OP can indeed be expressed elementarily. I will give a procedural proof.

I will write $x\sim y$ if $x/y$ is a product of algebraic numbers and a rational power of $\pi$. The following are famous properties of gamma function: $$\tag{1}\Gamma(x) \Gamma(1-x) \sim 1 $$ $$\tag{2}\Gamma(x) \Gamma(x + \frac{1}{2}) \sim \Gamma(2x)$$ $$\tag{3}\Gamma(x) \Gamma(x + \frac{1}{3}) \Gamma(x + \frac{2}{3}) \sim \Gamma(3x)$$ $$\tag{4}\Gamma(x) \Gamma(x + \frac{1}{7}) \cdots \Gamma(x + \frac{6}{7}) \sim \Gamma(7x)$$ The first one is reflection formula, the others are instances of multiplication theorems.


I will omit the $\Gamma$ sign, firstly, use $(2)$ on $11/42$: $$C:=\frac{{\left( {\frac{2}{7}} \right)\left( {\frac{{11}}{{42}}} \right)}}{{\left( {\frac{1}{{21}}} \right)}} \sim \frac{{\left( {\frac{2}{7}} \right)\left( {\frac{{11}}{{21}}} \right)}}{{\left( {\frac{1}{{21}}} \right)\left( {\frac{{16}}{{21}}} \right)}} \tag{*}$$ Use $(3)$ on $2/7$ with $x=2/21$: $$C \sim \frac{{\left( {\frac{2}{{21}}} \right)\left( {\frac{3}{7}} \right)\left( {\frac{{16}}{{21}}} \right)\left( {\frac{{11}}{{21}}} \right)}}{{\left( {\frac{1}{{21}}} \right)\left( {\frac{{16}}{{21}}} \right)}} =\frac{{\left( {\frac{2}{{21}}} \right)\left( {\frac{3}{7}} \right)\left( {\frac{{11}}{{21}}} \right)}}{{\left( {\frac{1}{{21}}} \right)}}$$ Use $(4)$ on $1/21$ with $x=1/21$: $$C\sim \frac{{\left( {\frac{2}{{21}}} \right)\left( {\frac{3}{7}} \right)\left( {\frac{{11}}{{21}}} \right)}}{{\left( {\frac{1}{{21}}} \right)}} \sim \frac{{\left( {\frac{2}{{21}}} \right)\left( {\frac{3}{7}} \right)\left( {\frac{{11}}{{21}}} \right)}}{{\left( {\frac{1}{3}} \right)}}\left( {\frac{4}{{21}}} \right)\left( {\frac{7}{{21}}} \right)\left( {\frac{{10}}{{21}}} \right)\left( {\frac{{13}}{{21}}} \right)\left( {\frac{{16}}{{21}}} \right)\left( {\frac{{19}}{{21}}} \right)$$ Note that $1/3$ in the denominator cancels with $7/21$ in the numerator, some terms cancelled each other via reflection formula, leaving $$C \sim \left( {\frac{3}{7}} \right)\left( {\frac{4}{{21}}} \right)\left( {\frac{{13}}{{21}}} \right)\left( {\frac{{16}}{{21}}} \right) $$ Use $(3)$ on $4/21$ and $13/21$: $$C \sim \left( {\frac{3}{7}} \right)\frac{{\left( {\frac{4}{7}} \right)}}{{\left( {\frac{{11}}{21}} \right)\left( {\frac{6}{7}} \right)}}\frac{{\left( {\frac{{13}}{7}} \right)}}{{\left( {\frac{{20}}{{21}}} \right)\left( {\frac{9}{7}} \right)}}\left( {\frac{{16}}{{21}}} \right) \sim \frac{1}{{\left( {\frac{{11}}{{21}}} \right)}}\frac{1}{{\left( {\frac{{20}}{{21}}} \right)\left( {\frac{2}{7}} \right)}}\left( {\frac{{16}}{{21}}} \right)$$ Finally, use reflection formula to make $20/21$ onto the numerator gives $$C \sim \frac{{\left( {\frac{1}{{21}}} \right)\left( {\frac{{16}}{{21}}} \right)}}{{\left( {\frac{2}{7}} \right)\left( {\frac{{11}}{{21}}} \right)}}$$ Now compare this with $(\ast)$ gives $C \sim \frac{1}{C} $, hence $C$ can be expressed as product of algebraic numbers and a rational power of $\pi$. Note that $C \sim 1/C$ explains why square root appears over sine terms of the result.

What this constant is can be figured out by performing above steps, I hope someone with more computational stamina than me can find it out explicitly.

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It looks like something (in particular, the RHS denominator root terms) that you would obtain from a special constant involving hypergeometric series. Sometimes products of gamma functions show up as special constants arising in theta function series (the Ramanujan theta functions, for example). Have you seen this article? It might give you some hints as to go about finding products of related values.

Another possibility for relating the gamma function products to the trig function values seen on the RHS of your identity would be to check for special trigonometric values that give, respectively, $2/7$, $11/42$, and $1/21$, like is $\sin(\vartheta) = \frac{11}{42}$ for any special values? This would allow you to get some special constants from the Weierstrass product formula for the gamma function (see here).

Where did you find this identity to begin with?

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  • 1
    $\begingroup$ There is Rohrlich–Lang conjecture saying that if a product of rational powers of Γ-functions with rational arguments happens to be an algebraic number, then it is provable by an elementary composition of shift, reflection and multiplication identities for the Γ-function. The conjecture is open, but no counterexamples are known. Is some cases the multiplication formula can be applied several times, resulting in a (sometimes unexpected) cancellation of most Γ-terms. $\endgroup$ – Vladimir Reshetnikov Jun 1 '18 at 19:56
  • $\begingroup$ Hm... this looks like it's answering more questions than it answers... $\endgroup$ – Simply Beautiful Art Jun 2 '18 at 21:57
  • $\begingroup$ I really do not think I'm that far off... There is the identity $${_2F_1}(1/12,5/12;1/2;121/125)=2^{-11/6}{15}^{1/4}(1+\sqrt{3})\Gamma^3(1/3)/\Gamma^2(1/4),$$ listed on Math World which relates these functions to products of the gamma function. And the product in question is given by $$\frac{B(2/7,11/42)\sqrt{\pi}}{B(1/21,1/2)}.$$ There are known integral representations for both the beta function and ${_2F_1}$ which effectively relate the two, so it seems reasonable to expect the product is somehow related to hypergeometric functions $\endgroup$ – mds Jun 5 '18 at 7:26
  • $\begingroup$ Also, $\Gamma\left(\frac{11}{42}\right)\Gamma\left(\frac{2}{7}\right)/\Gamma\left(\frac{1}{21}\right)={_2F_1}\left(\frac{3}{14},-\frac{5}{21}; \frac{11}{42}; 1\right)\sqrt{\pi}$, and this is also equal to $\left(\frac{1+z}{z}\right)^{5/21}\sqrt{\pi} \times \frac{\int_0^1 \frac{dt}{t^{5/7}(1-t)^{31/42}(t+z)^{23/42}}}{\int_0^1 \frac{dt}{t^{20/21}(1-t)^{1/2}(t+z)^{23/42}}},$ for any $\Re(z) > 0$ $\endgroup$ – mds Jun 5 '18 at 8:19
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I am typing some (informal) book on sagemath, with focus on structural application, and this post was an ideal occasion / setting to use sage in a natural, non-trivial way, here i am copy+pasting the full (computer aided, quick) proof, so that in similar situations one has a straight line to the solution.


It is handy, and it makes typing easier to use a better notation for $\Gamma(k/42)$ for all potentially relevant values of $k$, so let us set formally $$ \color{red}{ [k] = \Gamma\left(\frac k{42}\right) }\ ,\qquad k=1,2,\dots, 20,21, 22, \dots, 41,\dots\ \ . $$ Then a duplication formula is $$ [k]\;[k+21]=\text{(Known constant)}\cdot[2k]\ , $$ Note: In the programming section we consider the purely multiplicative functional equations (FEs), which are the Gauss multiplication FEs, and the reflexion FEs, written additively, then use notationally $x_k$ instead of $[k]$ (in the additive world), and the above relation reads $x_k+x_{k+21}-x_{2k}$ is "known", this leads to a system of linear equations, and for our purposes it is enough to pass to the homogeneous version and search for a writing of $x_{11}+x_{12}-x_2$ in terms of the collected equations.


Sage brings us straightforward to the finish, and suggests then (code in the sequel) to look at the relations:

$$ \begin{aligned}{} [4] \cdot[4+6] \cdot[4+12] \cdot[4+18] \cdot[4+24] \cdot[4+30] \cdot[4+36] \cdot[7\cdot 4]^{-1} &=(2\pi)^{6/2}\cdot 7^{1/2-7\cdot 4/42} &&\text{ to power $+1$}\ , \\ % [6] \cdot\color{magenta}{[6+6]} \cdot[6+12] \cdot[6+18] \cdot[6+24] \cdot[6+30] \cdot[6+36] \cdot[7\cdot 6]^{-1} &=(2\pi)^{6/2}\cdot 7^{1/2-7\cdot 6/42} &&\text{ to power $+1$}\ , \\[3mm] % \color{blue}{[2]} \cdot[2+14] \cdot[2+28] \cdot[3\cdot 2]^{-1} &=(2\pi)^{2/2}\cdot 3^{1/2-3\cdot 2/42} &&\text{ to power $-1$}\ , \\ % [4] \cdot[4+14] \cdot[4+28] \cdot\color{magenta}{[3\cdot 4]^{-1}} &=(2\pi)^{2/2}\cdot 3^{1/2-3\cdot 4/42} &&\text{ to power $-1$}\ , \\ % [8] \cdot[8+14] \cdot[8+28] \cdot[3\cdot 8]^{-1} &=(2\pi)^{2/2}\cdot 3^{1/2-3\cdot 8/42} &&\text{ to power $+1$}\ , \\[3mm] % \color{red}{[11]} \cdot[11+21] \cdot[2\cdot 11]^{-1} &=(2\pi)^{1/2}\cdot 2^{1/2-2\cdot 11/42} &&\text{ to power $+2$}\ , \\[3mm] % \color{blue}{[2]} \cdot[42-2] &=\pi\cdot \left(\sin\frac{2\pi}{42}\right)^{-1} &&\text{ to power $-1$}\ , \\ % [6] \cdot[42-6] &=\pi\cdot \left(\sin\frac{6\pi}{42}\right)^{-1} &&\text{ to power $-2$}\ , \\ % [8] \cdot[42-8] &=\pi\cdot \left(\sin\frac{8\pi}{42}\right)^{-1} &&\text{ to power $-1$}\ , \\ % [10] \cdot[42-10] &=\pi\cdot \left(\sin\frac{10\pi}{42}\right)^{-1} &&\text{ to power $-1$}\ . % \end{aligned} $$ We multiply the equations above, taken to the mentioned powers, only the colored factors survive: $$ \begin{aligned} \color{red}{[11]^2} \cdot \color{magenta}{[12]^2} \cdot \color{blue}{[2]^{-2}} &= (2\pi)^6 \cdot \pi^{-5} \cdot 2^{1-22/21} \cdot 3^{-1/2-1/7} \cdot 7^{-2/3} \cdot \\ &\qquad\qquad\cdot \sin^2\frac{6\pi}{42} \cdot \sin\frac{2\pi}{42} \cdot \sin\frac{8\pi}{42} \cdot \sin\frac{10\pi}{42} \end{aligned} $$ which is our relation. One can separate the algebraic number, and rewrite putting in evidence $ \color{red}{\frac{11}{42}} + \color{magenta}{\frac{12}{42}} = \color{blue}{\frac{2}{42}} + \color{green}{\frac{21}{42}} $: $$ \begin{aligned} \frac { \color{red}{\Gamma\left(\frac{11}{42}\right)} \color{magenta}{\Gamma\left(\frac{12}{42}\right)} } { \color{blue}{\Gamma\left(\frac{2}{42}\right)} \color{green}{\Gamma\left(\frac{21}{42}\right)} } &= \frac {\displaystyle 2^3\cdot\sin\frac{\pi}7 \sqrt{ \sin\frac{\pi}{21} \cdot \sin\frac{4\pi}{21} \cdot \sin\frac{5\pi}{21}} } {\displaystyle2^{1/42}\cdot 3^{9/28}\cdot 7^{1/3}} \\ &= \frac {\displaystyle 2^{5/4} (5-\sqrt{21})^{1/4} \sin\frac{\pi}7 } {\displaystyle 2^{1/42}\cdot 3^{9/28}\cdot 7^{1/3}} \ . \end{aligned} $$ That's all.

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As promised, some computer art work support the above.


  • (1)

Numerical check of the equalities above:

LHS = gamma(11/42) * gamma(12/42) / gamma(2/42) / gamma(21/42)
RHS1 = 8 * sin(pi/7) * sqrt(prod([sin(k*pi/21) for k in [1,4,5]])) / 2^(1/42) / 3^(9/28) / 7^(1/3)
RHS2 = 2^(5/4) * sin(pi/7) * (5-sqrt(21))^(1/4) / 2^(1/42) / 3^(9/28) / 7^(1/3)

print "LHS =", LHS.n()
print "RHS1 =", RHS1.n()
print "RHS2 =", RHS2.n()

which gives

LHS = 0.299625085598959
RHS1 = 0.299625085598959
RHS2 = 0.299625085598959
  • (2)

Minimal polynomial of the expression (sine product) under the square root:

K.<u> = CyclotomicField(84)        # 84 = 7*4*3
u42, u21, u4 = u^2, u^4, u^21
s1 = (u42^1 - 1/u42^1) / 2 / u4    # sine of 2.1pi/42 =  pi/21
s4 = (u42^4 - 1/u42^4) / 2 / u4    # sine of 2.4pi/42 = 4pi/21
s5 = (u42^5 - 1/u42^5) / 2 / u4    # sine of 2.5pi/42 = 5pi/21
print (8*s1*s4*s5).minpoly()

which gives, leading then immediately to the factor $(5-\sqrt{21})$:

x^4 - 5*x^2 + 1
  • (3)

As promised, the code that found the needed "linear" relation to be used:

R = PolynomialRing(QQ, 'x', 42 )     # quick init, we do not use x0
v = R.gens()
for k in range(42):
    exec('x%s = v[%s]' % (k, k) )    # so x1 is indeed x1, x2 is indeed x2...

def normalized_variable(k):
    return v[ k % 42 ]

def eq(times, k):
    """times should divide 42 in the caller, Gauss FE linearized, no constant."""
    inc = ZZ(42/times)
    return ( sum([ normalized_variable(k + j*inc)
                   for j in range(times) ])
             - normalized_variable(times*k) )

equations = (
    [ eq(times, k)
      for times in [7, 3, 2] # ZZ(42).divisors()[ ::-1 ]
      for k in [ 1..ZZ(42/times) ]
      if  times > 1
      # and ( times != 7 or k == 1 )
    ]
    +
    [ v[k] + v[42-k] for k in [1..21]
      if k in [2, 6, 8, 10] ]
    # last is a psychological condition, knowing the result,
    # exactly this condition makes us go straightforward to the result,  
    # else we have a hard work in the cyclotomic field of all sine values,
    # which is harder than parking the gamma values in each other...
)

J = R.ideal(equations)
f = (x11 + x12 - x2)

print "Is f = %s in J? %s" % ( f, f in J )
lif = f.lift(J)

for r in range(len(equations)):
    if lif[r]:
        print '%+4s TIMES %s' % (lif[r], equations[r])

The above gives:

Is f = -x2 + x11 + x12 in J? True

 1/2 TIMES x4 + x10 + x16 + x22 + x34 + x40
 1/2 TIMES x6 + x12 + x18 + x24 + x30 + x36
-1/2 TIMES x2 - x6 + x16 + x30
-1/2 TIMES x4 - x12 + x18 + x32
 1/2 TIMES x8 + x22 - x24 + x36
   1 TIMES x11 - x22 + x32
-1/2 TIMES x2 + x40
  -1 TIMES x6 + x36
-1/2 TIMES x8 + x34
-1/2 TIMES x10 + x32
sage: 

which "is" the solution.

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