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This question is somewhat informative. In Wikipedia the definition of double factorial continued in the complex arguments is provided $$ k!!=\sqrt{\frac{2}{\pi}}2^{\frac{k}{2}}\Gamma[k/2+1] $$ Clearly, this definition does not work for positive even numbers. Is there any double factorial formula in terms of Gamma functions which works for both odd and even?

Tanmoy Pati
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  • What type of properties do you want? There's no uniqueness property to these types of expressions, so there are a lot of ways to get an expression that equals $k!!$ for all non-negative integers. I'm not sure why you're wanting it though. – Brian Moehring Oct 02 '23 at 19:02
  • I need it from physics point of view. This is when in dimensional regularization, we take $k=d-\epsilon$ , where $\epsilon \in \mathbb{C}$, I am encountering double factorial of $k$. To evaluate it, in terms of series with respect to $\epsilon$, it is necessary to convert it into Gamma functions. In that sense, I need a consistent formula, like $n!=\Gamma[n+1]$ – Tanmoy Pati Oct 02 '23 at 19:07
  • I've written an answer for the question you've asked, but I don't think it's the question you need to ask. Unfortunately, other than saying "this works but it's probably not what you wanted", I can't comment on whether the function you want exists because I don't know the additional requirements from the physics context. – Brian Moehring Oct 02 '23 at 19:18
  • I'm not sure how you would get a double factorial to be honest. Are you sure there isn't one with just Gamma functions (e.g. the volume of balls)? – user10354138 Oct 02 '23 at 19:24

2 Answers2

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Note that for $k \geq 0$ even, we have $$k!! = 2^{k/2} (k/2)! = 2^{k/2} \Gamma(k/2+1)$$ so that the difference only appears in the coefficient.

To generate the coefficient, it suffices to note that $\sin^2(k\pi/2)$ generates the sequence $0,1,0,1,\ldots$, so that your answer is $$\left(1 + \left(\sqrt{2/\pi}-1\right)\sin^2(k\pi/2)\right)2^{k/2}\Gamma(k/2+1)$$

Does this generate $k!!$ for every non-negative integer $k$? Yes. We can check this.

Does it have the properties that you want? Probably not.

Brian Moehring
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$$(x)!!=\lim_{n\rightarrow\infty}\left(2n-\sin^{2}\left(\frac{\pi x}{2}\right)\right)^{\frac{x+\sin^{2}\left(\frac{\pi x}{2}\right)}{2}}\prod_{k=1}^{n}\frac{2k-\sin^{2}\left(\frac{\pi x}{2}\right)}{2k+x}$$Let $b(x)$ be Brian's continuation and $k(x)$ be mine. Then for all $x\in\mathbb{R}\backslash\mathbb{Z}^-$, $|k(x)|\ge|b(x)|$. I.e. our continuations aren't the same except in the positive integers.

Kamal Saleh
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