On determining the constant term in Stirling's approximation

Question

It is well-known that the discrete version of Stirling's formula is as follows

$$ \log N!=\left(N+\frac12\right)\log N-N+\log C+\int_N^\infty{P_1(t)\over t}\mathrm dt\tag1 $$

where $P_1(t)=B_1(\{t\})$, and $C$ is some constant term emerged while applying the Euler-Maclaurin formula. Using the limit definition of gamma function that

$$ \Gamma(s+1)=\lim_{N\to\infty}{N^sN!\over(s+1)(s+2)\cdots(s+N)} $$

I am able to obtain an identity similar to (1) but with respect to continuous variable $s$:

$$ \log\Gamma(s+1)=\left(s+\frac12\right)\log s-s+\log C+\int_0^\infty{P_1(t)\over s+t}\mathrm dt\tag2 $$

I already know that $C=\sqrt{2\pi}$ can be obtained by replacing factorial terms in Wallis product, but I wonder if there is an alternative way to figure out the constant.

Answer 1

Indeed, it is possible to determine $C$ by employing Legendre's duplication formula, which states

$$ \Pi(s)\Pi\left(s-\frac12\right)=2^{-2s}\sqrt\pi\cdot\Pi(2s) $$

where we set $\Pi(s)=\Gamma(s+1)$ for convenience

Taking logarithms, we have

$$ \color{orange}{\log\Pi(s)+\log\Pi\left(s-\frac12\right)}-\color{purple}{[\log\Pi(2s)-2s\log2]}=\frac12\log\pi $$

Now, due to (2), we get

$$ \begin{aligned} \color{orange}{\log\Pi(s)+\log\Pi\left(s-\frac12\right)} &=\left(s+\frac12\right)\log s-s \\ &+s\log\left(s-\frac12\right)-s+\frac12+2\log C+\mathcal O\left(\frac1s\right) \\ &=s\log\left[s\left(s-\frac12\right)\right]-2s+\frac12\log s \\ &+\frac12+2\log C+\mathcal O\left(\frac1s\right) \end{aligned} $$

and for the purple part we also have

$$ \begin{aligned} \color{purple}{\log\Pi(2s)-2s\log2} &=\left(2s+\frac12\right)\log2s-2s-2s\log2+\log C+\mathcal O\left(\frac1s\right) \\ &=s\log(s^2)+\frac12\log2+\frac12\log s-2s+\log C+\mathcal O\left(\frac1s\right) \end{aligned} $$

Combining these results gives

$$ \frac12\log\pi=s\log\left(1-{1\over2s}\right)-\log2+\frac12+\log C+\mathcal O\left(\frac1s\right) $$

Finally, if we were to take $s\to\infty$, we get $C=\sqrt{2\pi}$.