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Can Euler's generalization of factorial be done for double factorial?

Euler's generalization of factorial to non-integer values is

$$t! =\lim_{n \to \infty} \dfrac{n!n^t}{\prod_{k=1}^n(t+k)}.$$

I decided to see if I could come up with a similar generalization for double factorial, defined by $$n!! =\prod_{k \ge 0, 2k < n} (n-2k) =n(n-2)(n-4) \cdots. $$

My result is

$$(2t)!! =\lim_{n \to \infty}\dfrac{(2n)^t(2n)!!}{2^n\prod_{k=1}^{n} (t+k)}. $$

My questions:

Is this correct (there is a moderate amount of messy algebra involved)?

Is there a similar version for double factorial of Euler's formula $t! =\int_0^{\infty} x^t e^{-x} dx $?

I am quite confident that my formula generalizes to the m-factorial defined by $$n!_{(m)} =\prod_{k \ge 0, mk < n} (n-mk) =n(n-m)(n-2m) \cdots $$ just by replacing $2$ by $m$ everywhere.

I will post my proof in two days if there are no answers.

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The limit you suggest for $(2t)!!$ is actually

$$\lim_{n \to \infty}\dfrac{(2n)^t(2n)!!}{2^n\prod_{k=1}^{n} (t+k)}=2^t\Gamma(t+1)\ge(2t)!!,$$

and only equals $(2t)!!$ for integer $t$, but not for half-integer. You can take this as an upper bound for the generalization of $n!!$, while the lower bound would be

$$\sqrt{\frac2\pi}2^t\Gamma(t+1)\le(2t)!!,$$

which becomes an equality for half-integer $t$ (cf. equation $(2)$ from Wolfram MathWorld Double Factorial page).

Full extension of double factorial to complex arguments can be done as follows (from MathWorld again):

$$z!!=2^{(1+2z-\cos(\pi z))/4}\pi^{(\cos(\pi z)-1)/4}\Gamma\left(\frac z2+1\right).$$

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