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I was wondering if there is a closed form for $$\prod_{i=0}^{N}\big(i!\big)^{{N}\choose{i}}$$ I know that for $$\prod_{i=0}^{N}\big(i!\big)=G(N+2)$$ where we have expressed it as Barnes G-function. Yet I am not sure how to go about this with the binomial involved?

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    $\begingroup$ What requirements should a "closed form" fulfil? $\endgroup$
    – user
    Commented Dec 19, 2019 at 19:23
  • $\begingroup$ @user there is no requirement per se. I wonder if one can write the closed form in the manner I represented in the second equation. The only thing is $N$ is positive and finite. $\endgroup$
    – Wiliam
    Commented Dec 19, 2019 at 19:51
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    $\begingroup$ I just would say that a finite sum is a closed form per se. And of course you can define a function $ W(N):=\sum(i!)^\binom Ni$. Is it a satisfactory closed form? $\endgroup$
    – user
    Commented Dec 19, 2019 at 20:53
  • $\begingroup$ It would have been better if I could find the closed form in terms of some known function such as Gamma, Barnes G and etc. $\endgroup$
    – Wiliam
    Commented Dec 19, 2019 at 21:15
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    $\begingroup$ This was the point of my question about the requirements on the closed form. $\endgroup$
    – user
    Commented Dec 20, 2019 at 6:35

1 Answer 1

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An asymptotic formula can be derived that is better than 0.1% accurate for $n>7.$ I'll work with the logarithm and sketch a proof that

$$ (*) \quad S_n :=\sum_{k=0}^n \binom{n}{k} \log \Gamma(k+1) \sim 2^n\Big( \log \Gamma(n/2+1) + 1/4 - 1/(8n) + ... \Big) $$

Use the following formula (it can be found on the wiki for $\Gamma$ ) $$ \log \Gamma(k+1) = -\gamma \, k + \int_0^\infty \frac{e^{-k\,t} -1 + k \,t}{e^t-1} \frac{dt}{t} $$

Insert the previous equation into the left-hand side of (*), switch the finite sum with the integral, do the sum, and factor out a $2^n,$

$$ S_n = 2^n\Big(-\gamma \frac{n}{2} + \int_0^\infty \frac{e^{-n\,t/2}\cosh^n(t/2) -1 + \frac{n}{2} \,t}{e^t-1} \frac{dt}{t} \Big)$$ Note I've used a hyperbolic trig ID. The reason is because, for $t$ small enough, $$ \cosh^n(t/2) = 1 + \frac{n\,t^2}{8} + \frac{(3n-2)n\,t^4}{384}+...$$ Therefore, $$ S_n \sim 2^n\Big( -\gamma \frac{n}{2} + \int_0^\infty \frac{e^{-n\,t /2} -1 + \frac{n}{2} \,t}{e^t-1} \frac{dt}{t} +\int_0^\infty \frac{dt}{t} \frac{e^{-n\,t/2 }}{e^t-1} \big(\frac{n\,t^2}{8} + \frac{(3n-2)n\,t^4}{384} \big) \Big)$$ $$ = 2^n\Big(\log\Gamma(1+n/2)+\frac{n}{8}\int_0^\infty \frac{e^{-n\,t/2 }}{e^t-1}t\,dt + \frac{(3n-2)n}{384}\int_0^\infty \frac{e^{-n\,t/2 }}{e^t-1}t^3\,dt + ... \Big)$$

We can split the integral up because each converges (the $t$ is denominator is cancelled). More importantly the terms shown, and the subsequent ones, form an asymptotic sequence in $n$. In particular, the integrals can be explicitly written in terms of odd derivatives of the polygamma function, and since the polygamma function asymptotics is well-known, you can derive (by hand, if you want) the terms given in (*).

The remark about the accuracy of the expansion is from some numerical calculations. Take the $LHS(*)/RHS(*)$ and for n=7 the ratio is 0.99949 and for n=20, the ratio is 0.999993

Incidentally, the asymptotics proves that the product form will not be a Barnes G-function.

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