Can Euler's reflection formula be used to calculate $\Gamma (1/3)$?

Question

Can Euler's reflection formula be used to calculate $\Gamma (1/3)$? (That is, to express in in some closed form.) When trying to apply it, I am always falling in a loop.

Answer 1

Although Euler's reflection formula is satisfied by the gamma function:

$$ \Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin \pi x}, \;\; \forall x \not\in \mathbb{Z} $$

and this suffices to determine $\pm\;\Gamma(\frac{1}{2})$, it does not define all values of the function $\Gamma(x)$. In particular there are many functions that satisfy the reflection formula above, but which have differing values at $x=\frac{1}{3}$.

One easily sees that setting $x=\frac{1}{2}$ in the reflection formula gives us $\Gamma(\frac{1}{2})^2 = \frac{\pi}{1}$, so $\Gamma(\frac{1}{2}) = \sqrt{\pi}$ if we know this value is positive (clear from the integral definition of the gamma function).

However consider any positive real base $a \gt 0$ and define $f_a(x) = a^{x - \frac{1}{2}}$.

Then $f_a(x) f_a(1-x) = a^0 = 1$, and one easily verifies that $f_a(x) \Gamma(x)$ will satisfy Euler's reflection formula for noninteger $x$. By varying $a$ we can obtain any positive value for $f_a(\frac{1}{3}) \Gamma(\frac{1}{3})$ that we like.

Note also that substituting $-\Gamma(x)$ for $\Gamma(x)$ in the reflection formula leaves the product $\Gamma(x) \Gamma(1-x)$ unchanged, so this is another demonstration that the reflection formula alone does not define the gamma function.

The integral definition of $\Gamma(\frac{1}{3})$ can give precise approximations as the link in Raymond Manzoni's Comment explains, and Grigory M's Comment pointing you to Wikipedia should provide more rapid approximations of this transcendental number.