Is it possible to simplify $\frac{\Gamma\left(\frac{1}{10}\right)}{\Gamma\left(\frac{2}{15}\right)\ \Gamma\left(\frac{7}{15}\right)}$?

Is it possible to simplify this expression? $$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\ \Gamma\left(\frac{7}{15}\right)}$$ Is there a systematic way to check ratios of Gamma-functions like this for simplification possibility?

Amazingly, this can be greatly simplified. I'll state the result first:

$$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right)} = \frac{\sqrt{5}+1}{3^{1/10} 2^{6/5} \sqrt{\pi}}$$

The result follows first from a version of Gauss's multiplication formula:

$$\displaystyle\Gamma(3 z) = \frac{1}{2 \pi} 3^{3 z-1/2} \Gamma(z) \Gamma\left(z+\frac13\right) \Gamma\left(z+\frac{2}{3}\right)$$

or, with $z=2/15$:

$$\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right) = 2 \pi \,3^{1/10} \frac{\displaystyle\Gamma\left(\frac{2}{5}\right)}{\displaystyle\Gamma\left(\frac{4}{5}\right)}$$

Now use the duplication formula

$$\Gamma(2 z) = \frac{1}{\sqrt{\pi}}\, 2^{2 z-1} \Gamma(z) \Gamma\left(z+\frac12\right)$$

or, with $z=2/5$:

$$\frac{\displaystyle\Gamma\left(\frac{2}{5}\right)}{\displaystyle\Gamma\left(\frac{4}{5}\right)} = \frac{\sqrt{\pi} \, 2^{1/5}}{\displaystyle\Gamma\left(\frac{9}{10}\right)}$$

Putting this all together, we get

$$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right)} = \frac{\displaystyle\Gamma\left(\frac{1}{10}\right) \Gamma\left(\frac{9}{10}\right)}{\sqrt{\pi^3} \, 2^{6/5} \, 3^{1/10}}$$

And now, we may use the reflection formula:

$$\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin{\pi z}}$$

With $z=1/10$, and noting that

$$\sin{\left(\frac{\pi}{10}\right)} = \frac{\sqrt{5}-1}{4} = \frac{1}{\sqrt{5}+1}$$

the stated result follows. This has been verified numerically in Wolfram|Alpha.

• Wonderful! Wouldn't it look even better as $$\frac{\sqrt{5}+1}{\sqrt[10]{12288}\sqrt{\pi}}$$? Maybe not... – Kieren MacMillan Sep 17 '14 at 21:00
• @KimPeek: Why...because you say so? I say it is true. I have analysis and numerics on my side. Prove me wrong. – Ron Gordon Jan 19 '16 at 20:53
• – YoTengoUnLCD Jan 19 '16 at 20:57
• Nice solution. I particular like that software like Mathematica is not able to simplify it directly. The expression can also be written $\frac{\phi }{\sqrt[10]{12}\sqrt{\pi}}$ where $\phi$ is the golden ratio. – Winther Jan 19 '16 at 21:02
• @Winther: lots of ways to express, but the important thing is that we can write this horrific-looking beast in terms of things a pocket calculator can produce. – Ron Gordon Jan 19 '16 at 21:03