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

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