Is there any way to extend the domain of this function through analytic continuation? $\prod_{k=2}^x \log k=F(x)$
It looks a lot like the gamma function (a sort of logarithmic factorial), and I wonder if it can be similarly expressed as an integral or something. Any ideas?
 A: Taking log and using Abel summation formula we have$$\log\left(\prod_{k=2}^{x}\log\left(k\right)\right)=\sum_{k=2}^{x}\log\left(\log\left(k\right)\right)=\left(x-1\right)\log\left(\log\left(x\right)\right)-\int_{2}^{x}\frac{\left[t\right]-1}{t\log\left(t\right)}dt=\left(x-1\right)\log\left(\log\left(x\right)\right)-\int_{2}^{x}\frac{\left[t\right]}{t\log\left(t\right)}dt+\log\left(\log\left(x\right)\right)-\log\left(\log\left(2\right)\right)=$$
 $$=x\log\left(\log\left(x\right)\right)-\int_{2}^{x}\frac{\left[t\right]}{t\log\left(t\right)}dt-\log\left(\log\left(2\right)\right)$$
 where $\left[t\right]$
  is the integer part of $t$
 . Now if we use $\left[t\right]=t-\left\{ t\right\} $
  where $0\leq\left\{ t\right\} <1$
  is the fractional part of $t$
  we have $$\sum_{k=2}^{x}\log\left(\log\left(k\right)\right)=x\log\left(\log\left(x\right)\right)-\log\left(\log\left(2\right)\right)-\int_{2}^{x}\frac{1}{\log\left(t\right)}dt+\int_{2}^{x}\frac{\left\{ t\right\} }{t\log\left(t\right)}dt$$
 now we can bound$$\int_{2}^{x}\frac{\left\{ t\right\} }{t\log\left(t\right)}dt\leq\int_{2}^{x}\frac{1}{t\log\left(t\right)}dt=O\left(\log\left(\log\left(x\right)\right)\right)$$
 so$$\sum_{k=2}^{x}\log\left(\log\left(k\right)\right)=x\log\left(\log\left(x\right)\right)-\textrm{Li}\left(x\right)+O\left(\log\left(\log\left(x\right)\right)\right)$$
 where $$\textrm{Li}\left(x\right)=\int_{2}^{x}\frac{1}{\log\left(t\right)}dt$$
  is the logarithmic integral. Hence$$\prod_{k=2}^{x}\log\left(k\right)=e^{x\log\left(\log\left(x\right)\right)-\textrm{Li}\left(x\right)+O\left(\log\left(\log\left(x\right)\right)\right)}.$$
