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The $\Gamma$ function is the analytic continuation of the factorial function. Is there a similar analog for multifactorials?

I am particularly interested in the double factorial. All Google has given me is the following formula relating the $\Gamma$ function to the double factorial for half integer values: $$\Gamma\left(n+\frac{1}{2}\right)=\frac{(2n-1)!!}{2^n}\sqrt{\pi}$$ But I want to double factorial nonintegers, so this is not really helpful.

Alexander Gruber
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    The Wikipedia article on "factorial" has some answers: http://en.wikipedia.org/wiki/Factorial#Alternative_extension_of_the_double_factorial – bradhd Feb 01 '13 at 04:20
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    "The Γ function is the analytic continuation of the factorial function." This makes it sound as if $\Gamma$ is the unique analytic function that interpolates factorial, but this is far from being the case. What criteria are you using to single out $\Gamma$ from all possible analytic interpolants? (Or maybe you mean something different.) – Andrés E. Caicedo Feb 01 '13 at 06:20

2 Answers2

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Why is this not helpful? If you write the identity as $$(2n-1)!! = \frac{\Gamma(n + \frac{1}{2}) 2^n}{\sqrt{\pi}}$$ and then let $n = (x+1)/2$ you get $$x!! = \frac{\Gamma(\frac{x+2}{2}) 2^{(x+1)/2}}{\sqrt{\pi}}$$ The right side is now defined for any complex $x$ as long as the argument for the Gamma function is defined.

Ted
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Despite this being a 10-year old post, it does come up when searching "double factorial analytic continuation", so I thought I'd expand on Ted's answer. (I'm new to StackExchange so hopefully I don't do anything wrong!)

The equation given by Ted (and Brad's linked Wikipedia page) does work for any complex input $x$, but it is specifically based on the double-factorial definition that starts with an odd integer: $$(2n - 1)!! = \prod_{k=1}^n (2k - 1) = 3 \cdot 5 \cdot 7 \cdots (2n - 1)$$ However, it seems quite common to extend the definition to even numbers like so: $$(2n)!! = \prod_{k=1}^n (2k) = 2 \cdot 4 \cdot 6 \cdots (2n)$$

It turns out that the analytic continuation for the 'even version' is different to the 'odd version'! I'll denote the odd version $F_-(x)$ and the even version $F_+(x)$.

The formula for $F_+(x)$ is actually quite simple to derive. We simply notice that the even double-factorial is a normal factorial except with each term doubled: $$(2n)!! = \prod_{k=1}^n 2k = 2(1) \cdot 2(2) \cdot 2(3) \cdots 2(n) = 2^n \cdot n!$$ Then to define $F_+$ we just set $2n = x$ and substitute in the gamma function: $$F_+(x) = 2^{x/2} \cdot \Gamma \left(\frac x2 + 1 \right)$$ Recalling the formula for $F_-(x)$: $$F_-(x) = \frac{2^{(x+1)/2}}{\sqrt \pi} \cdot \Gamma \left( \frac{x + 2}{2} \right)$$ We can then find the relationship between $F_+$ and $F_-$: $$F_+(x) = F_-(x) \cdot \sqrt{\pi / 2}$$ So one is a scaled version of the other!

It should be possible to define a more complicated function that interpolates between these two cases -- matching $F_+$ for even integers and $F_-$ for odd ones -- but at that point I expect it would be difficult to call any particular choice of interpolation 'the most intuitive'.

For 'higher-order' multifactorials, I expect they require more functions -- $3$ for $x!!!$, $4$ for $x!!!!$, etc. -- though I have not yet tried to confirm this.

Esran Douma
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  • For anyone curious, here is a Desmos plot of these functions together with one choice of interpolant $F(x) = \sin^2(\pi x/2) F_-(x) + \cos^2(\pi x/2) F_+(x)$ https://www.desmos.com/calculator/od1qhvjhv7 – Calvin Khor Aug 26 '23 at 06:34
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    You're correct in your final comment about multifactorials, in general one needs to interpolate $k$ functions for $x(!^k)$. I worked it out in this answer. – pregunton Aug 26 '23 at 08:15