I was interested in an integral that I known from [1], it is


$$\int\_0^1 \log(x!)dx.$$


I tried to get such closed-form using myself ideas and symbolic calculations, also with the help of Wolfram Alpha online calculator. But I don't know how get the sum of certain series involving a special function.


My first step was to invoke the formula $(3.13)$ from [2], taking the logarithm and integrating one has that $$\int\_0^1 \log(x!)dx$$
is equals to
$$\log 2+\frac{1}{24}\sum\_{n=1}^\infty\left(-2^{n}(6(6\log A)+\log \pi)-24\psi^{(-2)}(\frac{1}{2}+2^{-n})+\log(32))-12\log\pi\right),$$
where $A$ is the Glaisher-Kinkelin constant and $\psi^{(n)}(s)$ denotes the $n^{th}$ derivative of the digamma function.


See the first comment that you can to evaluate in Wolfram Alpha online calculator. I've some computational evidence, for example the following code (it is a line)


`sum 1/24 (-2^n (6 (6 log(Glaisher) + log(π)) - 24 polygamma(-2, 1/2 + 2^(-n)) + log(32)) - 12 log(π)), from n=1 to 100`


and 


`log(2)-0.7742086473552725676369-(1/2log(2pi)-1)`


but I can not to prove the corresponding closed-form. Of course I know how to get the sum of geometric series but the problem here is different.



> 
> **Question.** Please, prove that $$\log 2+\frac{1}{24}\sum\_{n=1}^\infty\left(-2^{n}(6(6\log A)+\log \pi)-24\psi^{(-2)}(\frac{1}{2}+2^{-n})+\log(32))-12\log\pi\right)$$
>  is equals to the closed-form for $\int\_0^1 \log(x!)dx$ given in [1]. 
>  What I ask is if you can analyze the series to calculate its sum. I've deduced/considered previous expression and I would like to know if it is possible to prove that previous expression equals to $\frac{1}{2}\log(2\pi )-1$ analizing the series to get its sums (without invoking that it is equals to $\int\_0^1 \log(x!)dx$). **Many thanks.**
> 
> 
> 


I have no intuition/knowledges to know if it is easy to get the sum of the series.


References:
-----------


[1] [*Muliplicative integral of* $\Gamma(x)$](https://mathoverflow.net/questions/32954/multiplicative-integral-of-gammax), this MathOverflow (July of 2010).


[2] Manuel Benito, Luis M. Navas and Juan Luis Varona, *Möbius inversion from the point of view of arithmetical semigroup flows*, Biblioteca de la Revista Matemática Iberoamericana, Proceedings of the "Segundas Jornadas de Teoría de Números", (2008), pages 61-81.



On the integral $\int_0^1\log(x!)dx$ revisited
Details of the simple [integration by series](https://mathoverflow.net/questions/32954/multiplicative-integral-of-gammax#comment75191_32954) for $\int\_0^1\log(x!)dx$ mentioned [above](https://mathoverflow.net/questions/337500/on-the-integral-int-01-logxdx-revisited#comment843736_337500) (hopefully yours may be treated analogously, if you wish to try it).


Start from the series of the logarithm of the infinite product for $x!$
$$\log(x!)=- \gamma x + \sum\_{k=1}^m\bigg(\frac{x}{k}-\log\big(1+\frac{x}{k}\big)\bigg)+o(1),$$
uniformly on $[0,1]$ as $m\to+\infty$.
Integrating on $[0,1]$ we get 


$$\int\_0^1\log(x!)dx= -\frac{\gamma}{2}+\frac{1}{2}\sum\_{k=1}^m \frac{1}{k}+m-\sum\_{k=1}^m\Big((k+1)\log(k+1)-k\log k\Big) +\sum\_{k=1}^m \log k +o(1)$$
Simplifying the telescopic sum, and by the definition of the Euler-Mascheroni constant, this is


$$= \frac{1}{2}\log m +m -(m+1)\log(m+1)+ \log(m!) +o(1)$$
$$= \frac{1}{2}\log m +m -\Big(m\log m + \log m + 1\Big)+ \log(m!) +o(1)$$
$$=- \frac{1}{2}\log m +m -m\log m -1 + \log(m!) +o(1)$$
Finally by the (logarithm of the) Stirling formula this is exactly
$$\frac{1}{2}\log(2\pi)-1+ o(1),$$
ending the computation.


