I am reading the book "Special functions: an introduction to classical functions of mathematical physics" by Nico M. Temme and I'm having trouble understanding how to find a constant (in page 62).
I understand why
$$
\log\Gamma(z+1)=(z+\frac{1}{2})\log z – z + \int_{0}^{\infty}g(t)e^{-zt}dt + C,
$$
where $g$ is bounded for $t\gt0$ and $C$ is a constant of integration, which has to be determined. I know that Legendre's duplication formula can be written in the form
$$
\log \frac{2^{2z}\Gamma(z+1)\Gamma(z+\frac{1}{2})}{\Gamma(2z+1)} = \frac{1}{2}\log \pi.
$$
Now, the book asserts the following: Letting $z\to\infty$ and using Legendre's duplication formula we obtain $C=\frac{1}{2}\log 2\pi.$ But I don't understand how to use the duplication formula, when I let $z\to\infty$ the integral vanishes (since $g$ is bounded). Thus
$$
C=\lim_{z\to\infty} \log\Gamma(z+1)-(z+\frac{1}{2})\log z + z.
$$
Now, using Stirling's approximation, the above limit equals $\frac{1}{2}\log 2\pi$, but I don't want to use Stirling's approximation, I want to know how to use the duplication formula.
Best Answer
First, define the two functions $$ f(z) := (z + \frac12)\log(z)-z,\quad F(z) := f(z) + f(z - \frac12) - f(2z). $$ Second, find that the power series expansion at infinity of $\ F(z) \ $ is $$ F(z) = -2\log(2)\ z - \frac12 \log(2) - \frac18 z^{-1} + O(z^{-2}). $$ The rest is a simple exercise.