Closed form for $ \prod_{k=1}^n (a+k^2) $

Question

I have come across the following product: $$ \prod_{k=1}^n (a+k^2) $$ where $a$ is a positive constant. Could anyone suggest a closed form for this product? I need to approximate this for large $n$, but the problem is that $a$ is also of order of $n$ (it does not change with $n$, but it is also a large number).

Answer 1

Hint. Here is an approach.

Recall that, for $z \in \mathbb{C}$, the Pochhammer symbol may be defined as $$(z)_n = \prod_{k=0}^{n-1} (z+k)=z(z + 1)(z + 2) \cdots (z + n - 1), \quad n > 1, \quad (z)_0 = 1.$$ Then, for $a\geq0$, write $$ \begin{align} \prod_{k=0}^{n-1} (a+k^2) &=\prod_{k=0}^{n-1} (k+i\sqrt{a})\prod_{k=0}^{n-1} (k-i\sqrt{a}) =(i\sqrt{a})_n(-i\sqrt{a})_n \quad (i^2=-1) \end{align} $$ and the given product may be expressed as

$$ \prod_{k=1}^{n} (a+k^2) =(1+i\sqrt{a})_{n}(1-i\sqrt{a})_{n} \tag1 $$

Added. If you write the Pochhammer symbol in terms of the Euler $\Gamma$ function, then $(1)$ rewrites

$$ \prod_{k=1}^{n} (a+k^2)\! =\!\frac{\Gamma(n+1+i\sqrt{a})}{\Gamma(i\sqrt{a})}\!\frac{\Gamma(n+1-i\sqrt{a})}{\Gamma(-i\sqrt{a})}\!=\!\left|\Gamma(n+1+i\sqrt{a})\right|^2\frac{\sinh(\pi\sqrt{a})}{\pi\sqrt{a}} \tag2 $$

as noticed by Antinous' answer, where we have used eulerian infinite products involving $\Gamma$.

Then, as $n$ is great, by applying Stirling formula (see here or here), you obtain the following asymptotic term:

$$ \prod_{k=1}^{n} (a+k^2) = \frac{2\pi \: n^{2n+1}e^{-2n}}{|\Gamma(1+i\sqrt{a})|^2}\!\left(\! 1 + \mathcal{O}\!\left(\frac1n\right)\!\right)\! \sim \frac{2\sinh(\pi\sqrt{a})}{\sqrt{a}} n^{2n+1}e^{-2n} \tag3 $$

Answer 2

Your product has the nice closed form $$\prod_{k=1}^n(a+k^2)=\left|\Gamma(n+1+i\sqrt{a})\right|^2\frac{\sinh(\pi\sqrt{a})}{\pi\sqrt{a}}.$$