Gamma function of $4/3$

Question

I know how to calculate the value of $\big(\frac12\big)!$ using the gamma function, but I don't know how to find the value of $\big(\frac13\big)!$ or $\big(\frac16\big)!$ using the gamma function. I got an answer with this formula $\Gamma(n+1/p)=\frac{\Gamma(1/p)(pn−p+1)!^p}{p^n}$, to calculate it but I still don't understand it. How can I use it to get the value in this case of gamma function of $\frac43$. Can any one put all the steps to follow?

Answer 1

In fact, there are very few values for which $$\Gamma \left(1+\frac{1}{p}\right)\qquad \text{with} \qquad 0 \leq p$$ has a known value.

However, we can develop quite good approximations in the same spirit as in this question of mine.

Let $x=\frac{1}{p}$ and write $$\Gamma \left(1+x\right)\sim 1+x(1-x) \sum_{k=0}^6 d_k\, x^k$$ and the $d_k$'s would be computed in order to match the function, first and second derivative values at $x=0$, $x=\frac 12$ and $x=1$. This leads to $$d_0=-\gamma \qquad d_1=-\gamma +\frac{\gamma ^2}{2}+\frac{\pi ^2}{12}$$ $$d_2=-246+38 \gamma -\frac{9 \gamma ^2}{2}-\frac{3 \pi ^2}{4}+2 \pi ^{5/2}+4 \sqrt{\pi } \left(24+\log ^2(4)+\gamma (\gamma +\log (16))\right)$$ $$d_3=1211-125 \gamma +16 \gamma ^2+\frac{8 \pi ^2}{3}-12 \pi ^{5/2}-8 \sqrt{\pi } (68+(-2+\gamma +\log (4)) (2+3 \gamma +\log (64)))$$ $$d_4=-2386+2 (79-14 \gamma ) \gamma -\frac{14 \pi ^2}{3}+26 \pi ^{5/2}+4 \sqrt{\pi } \left(268+\psi \left(\frac{3}{2}\right) \left(13 \psi \left(\frac{3}{2}\right)-20\right)\right)$$ $$d_5=4 \left(529+\gamma (6 \gamma -19)+\pi ^2-6 \pi ^{5/2}-4 \sqrt{\pi } \left(60+\psi \left(\frac{3}{2}\right) \left(3 \psi \left(\frac{3}{2}\right)-2\right)\right)\right)$$ $$d_6=-696-8 (\gamma -1) \gamma -\frac{4 \pi ^2}{3}+8 \pi ^{5/2}+16 \sqrt{\pi } \left(20+\psi\left(\frac{3}{2}\right)^2\right)$$ Making them rational to high accuracy, the $d_k$'s are $$\left\{-\frac{33841}{58628},\frac{19541}{47448},-\frac{12261}{24970},\frac{30367}{6 8303},-\frac{45695}{134763},\frac{21117}{128032},-\frac{6099}{165595}\right\}$$ and the maximum error is less that $3 \times 10^{-6}$.

Listing for a few values of $p$ $$\left( \begin{array}{ccc} p & \text{approximation} & \text{exact} \\ 1 & 1.00000000 & 1.00000000 \\ 2 & 0.88622693 & 0.88622693 \\ 3 & 0.89298092 & 0.89297951 \\ 4 & 0.90640551 & 0.90640248 \\ 5 & 0.91817214 & 0.91816874 \\ 6 & 0.92772247 & 0.92771933 \\ 7 & 0.93544026 & 0.93543756 \\ 8 & 0.94174496 & 0.94174270 \\ 9 & 0.94696723 & 0.94696535 \\ 10 & 0.95135233 & 0.95135077 \\ 11 & 0.95508084 & 0.95507953 \\ 12 & 0.95828678 & 0.95828568 \\ 13 & 0.96107099 & 0.96107006 \\ 14 & 0.96351044 & 0.96350965 \\ 15 & 0.96566474 & 0.96566406 \\ 16 & 0.96758065 & 0.96758007 \\ 17 & 0.96929538 & 0.96929487 \\ 18 & 0.97083880 & 0.97083836 \\ 19 & 0.97223524 & 0.97223485 \\ 20 & 0.97350461 & 0.97350427 \end{array} \right)$$