4
$\begingroup$

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 $\Gamma(x)$, 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.

$\endgroup$
  • $\begingroup$ int log gamma(1/2+x/2^n)-log(sqrt(pi))dx, from x=0 to 1 $\endgroup$ – user142929 Aug 2 at 16:13
  • $\begingroup$ We have $\int_0^1 \log(x!)dx = \log(\sqrt{2\pi})-1$ See : math.stackexchange.com/questions/575644/… $\endgroup$ – LAGRIDA Aug 2 at 17:29
  • $\begingroup$ @JeremyRouse my English is bad, that I am asking is if it is possible to prove that $((1/2)\log(2\pi)-1)$ is equals to $$\log 2+\frac{1}{24}\sum_{n=1}^\infty\left(-2^{n}(6(6\log A)+\log \pi)-24\psi^{(-2)}(\frac{1}{2}+2^{-n})+\log(32))-12\log\pi\right)$$ but without using directly that my deduction is equals to $\int_0^1 \log(x!)dx$. I mean thus if it is possible to analyze the series to get its sum. Many thanks. $\endgroup$ – user142929 Aug 2 at 18:43
  • $\begingroup$ Many thanks @LAGRIDA , I am asking if one can to analyze the series to get its sum directly. $\endgroup$ – user142929 Aug 2 at 18:51
  • 1
    $\begingroup$ @user142929 the link [1] also has a simple proof via integration by series in a comment $\endgroup$ – Pietro Majer Aug 2 at 19:44
4
$\begingroup$

Details of the simple integration by series for $\int_0^1\log(x!)dx$ mentioned above (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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.