Radical simplification of Gamma function expression

This one popped up and I have no idea what motivated my idea that there was, in fact, a way to radically simplify the god-awful Gamma functions. But, well…you’ll see. The problem asks whether the expression

$$\frac{\Gamma\left(\frac{1}{10}\right)}{\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right)}$$

has any chance of being simplified somehow. Recall that a gamma function is like a factorial, but defined for non-integers. In fact,

$$\Gamma(z) = \int_0^{\infty} dt \, t^{z-1} \, e^{-t}$$

Gammas of rationals, while being possible that some non-gamma expression exists, are typically dreadful. Amazingly, however, it turns out that this one can be greatly simplified. I’ll state the result first:

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

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

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

or, with $z=2/15$:

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

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 $z=2/5$:

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

Putting this all together, we get

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

And now, we may use the reflection formula:

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

With $z=1/10$, and recognizing 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 Mathematica/WA.

2 Comments

  • Very nice and clever, Ron.

  • rlgordonma@yahoo.com wrote:

    Thanks man. Funny, that was the first comment I got, although I guess 29 upvotes is enough of a statement. Yeah, that was one of those problems where I just went down this path blindly and it just got better and better. I wonder where the ratio came from.

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