# Application of Legendre's duplication formula

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.

• It has nothing obvious that $\log\Gamma(z+1)=(z+\frac{1}{2})\log z - z + \int_{-\infty}^{\infty}g(t)e^{-zt}dt + C$ even less that $g(t) = 0$ for $t< 0$. And it depends on the tools you want to use : complex analysis, Fourier analysis, some theorems like duplication, reflection, Stirling – reuns Mar 24 at 19:10
• @reuns I only want to use the duplication formula – Sebitas Mar 24 at 19:14
• Once you know that $\log\Gamma(z+1)=(z+\frac{1}{2})\log z - z + \int_{0}^{\infty}g(t)e^{-zt}dt + C$ then the duplication formula gives $C = \lim_{z \to \infty}\log 2^{2z} + \log\Gamma(z+1)+\log \Gamma(z+\frac{1}{2})-\log\Gamma(2z+1)$. But again $\log\Gamma(z+1)=(z+\frac{1}{2})\log z - z + \int_{0}^{\infty}g(t)e^{-zt}dt + C$ has nothing obvious and you should probably look again at the steps they used to find it. – reuns Mar 24 at 19:35
• I know it is not obvious, but after some trouble I was able to prove that $\log\Gamma(z+1)=(z+\frac{1}{2})\log z-z + \int_{0}^{\infty} g(t)e^{-zt}dt + C$ but from this I really don't see how the duplication formula gives $C=\lim_{z\to\infty} \log 2^{2z} + \log \Gamma(z+1) + \log \Gamma(z+\frac{1}{2})-\log\Gamma(2z+1).$ Could you please elaborate that line? – Sebitas Mar 24 at 20:24

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.