2 typo fix

I decided to take the opposite approach to what Matt Y. asked for (sorry!), which was to see exactly how his proof depends on trig function identities. As it turns out, you can use trig identities to produce additional Gamma function identities like the original formula, though usually without being quite as clean and symmetric.

To see how it goes, let's look at Matt's proof. The classical reflection and doubling formulas for the the Gamma function produce the formulas used in the proof: $$\Gamma(s)\sin(\pi s) = \frac{\pi}{1-s} frac{\pi}{\Gamma(1-s)}$$ and $$\Gamma(s)\cos \Bigl( \frac{\pi s}{2}\Bigr) = \pi^{1/2} 2^{s-1} \frac{\Gamma \bigl( \frac{s}{2} \bigr)}{\Gamma \bigl( \frac{1-s}{2} \bigr)}.$$ Consider the trig identity, $$\frac{ \sin(x) + \sin(y) }{\sin(x+y)} + 1 = \frac{2\cos \bigl(\frac{x}{2} \bigr) \cos \bigl( \frac{y}{2} \bigr)}{\cos \bigl( \frac{x+y}{2} \bigr)},$$ which is easy enough to verify (and is floating around in the proof already). If we take $x=\pi a$ and $y=\pi b$ and multiply this identity by $\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$, then the previous $\Gamma$-function identities yield Matt's formula.

So I wondered how easy it would be to use known trig identities to produce additional formulas of this type, say with different denominators, and here is an example. Again using the reflection and multiplication formulas, you can show that $$\Gamma(s)\biggl( \frac{1}{2} + \cos\biggl(\frac{ 2\pi s}{3} \biggr) \biggr) = \pi \frac{3^{s-1/2} \Gamma\bigl(\frac{s}{3} \bigr)}{\Gamma \bigl( \frac{1-s}{3} \bigr) \Gamma \bigl( \frac{2-s}{3} \bigr)}.$$ The trig identity I want to use (coming from the triple angle formulas for cosine) is $$2 \biggl( \frac{1}{2} + \cos\biggl( \frac{2x}{3} \biggr) \biggr)^3 - 3 \biggl( \frac{1}{2} + \cos \biggl( \frac{2x}{3} \biggr) \biggr)^2 + \sin^2(x) = 0.$$ Take $x=\pi s$ and multiply by $\Gamma(s)^3$. Then after the dust settles this yields the identity $$2\pi 3^{3s-3/2}\frac{ \Gamma\bigl( \frac{s}{3} \bigr)^3}{ \Gamma\bigl( \frac{1-s}{3} \bigr)^3 \Gamma \bigl( \frac{2-s}{3} \bigr)^3} - 3^{2s} \frac{ \Gamma(s) \Gamma\bigl( \frac{s}{3} \bigr)^2}{\Gamma \bigl( \frac{1-s}{3} \bigr)^2 \Gamma \bigl( \frac{2-s}{3} \bigr)^2} + \frac{ \Gamma(s)}{\Gamma(1-s)^2} = 0.$$ So yeah, it's not that pretty and we've only obtained a one-variable identity, but presumably one can find more examples of multivariable formulas in a similar fashion --- though you need to get the stars to align properly. What is nice about Matt Y.'s formula is that the powers of 2 that could be there (much like the powers of 3 here) nicely cancel away, leaving only $\Gamma$-values and a single power of $\pi$.

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I decided to take the opposite approach to what Matt Y. asked for (sorry!), which was to see exactly how his proof depends on trig function identities. As it turns out, you can use trig identities to produce additional Gamma function identities like the original formula, though usually without being quite as clean and symmetric.

To see how it goes, let's look at Matt's proof. The classical reflection and doubling formulas for the the Gamma function produce the formulas used in the proof: $$\Gamma(s)\sin(\pi s) = \frac{\pi}{1-s}$$ and $$\Gamma(s)\cos \Bigl( \frac{\pi s}{2}\Bigr) = \pi^{1/2} 2^{s-1} \frac{\Gamma \bigl( \frac{s}{2} \bigr)}{\Gamma \bigl( \frac{1-s}{2} \bigr)}.$$ Consider the trig identity, $$\frac{ \sin(x) + \sin(y) }{\sin(x+y)} + 1 = \frac{2\cos \bigl(\frac{x}{2} \bigr) \cos \bigl( \frac{y}{2} \bigr)}{\cos \bigl( \frac{x+y}{2} \bigr)},$$ which is easy enough to verify (and is floating around in the proof already). If we take $x=\pi a$ and $y=\pi b$ and multiply this identity by $\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$, then the previous $\Gamma$-function identities yield Matt's formula.

So I wondered how easy it would be to use known trig identities to produce additional formulas of this type, say with different denominators, and here is an example. Again using the reflection and multiplication formulas, you can show that $$\Gamma(s)\biggl( \frac{1}{2} + \cos\biggl(\frac{ 2\pi s}{3} \biggr) \biggr) = \pi \frac{3^{s-1/2} \Gamma\bigl(\frac{s}{3} \bigr)}{\Gamma \bigl( \frac{1-s}{3} \bigr) \Gamma \bigl( \frac{2-s}{3} \bigr)}.$$ The trig identity I want to use (coming from the triple angle formulas for cosine) is $$2 \biggl( \frac{1}{2} + \cos\biggl( \frac{2x}{3} \biggr) \biggr)^3 - 3 \biggl( \frac{1}{2} + \cos \biggl( \frac{2x}{3} \biggr) \biggr)^2 + \sin^2(x) = 0.$$ Take $x=\pi s$ and multiply by $\Gamma(s)^3$. Then after the dust settles this yields the identity $$2\pi 3^{3s-3/2}\frac{ \Gamma\bigl( \frac{s}{3} \bigr)^3}{ \Gamma\bigl( \frac{1-s}{3} \bigr)^3 \Gamma \bigl( \frac{2-s}{3} \bigr)^3} - 3^{2s} \frac{ \Gamma(s) \Gamma\bigl( \frac{s}{3} \bigr)^2}{\Gamma \bigl( \frac{1-s}{3} \bigr)^2 \Gamma \bigl( \frac{2-s}{3} \bigr)^2} + \frac{ \Gamma(s)}{\Gamma(1-s)^2} = 0.$$ So yeah, it's not that pretty and we've only obtained a one-variable identity, but presumably one can find more examples of multivariable formulas in a similar fashion --- though you need to get the stars to align properly. What is nice about Matt Y.'s formula is that the powers of 2 that could be there (much like the powers of 3 here) nicely cancel away, leaving only $\Gamma$-values and a single power of $\pi$.