These functions are very similar. Is there an explanation?

Question

Blue: $x \ln x-x$

Brown: $\ln \frac{\Gamma(x+1/2)}{\sqrt{\pi}}$

Answer 1

Using Sterling’s approximation for the Gamma function, $$\Gamma(z)\sim \sqrt{\frac{2\pi}{z}}\left(\frac{z}{e}\right)^z$$ In this case \begin{align*} \Gamma\left(x+\frac{1}{2}\right)&\sim \sqrt{\frac{2\pi}{x+\frac{1}{2}}}\left(\frac{x+\frac{1}{2}}{e}\right)^{x+\frac{1}{2}}\\ \ln\Gamma\left(x+\frac{1}{2}\right)&\sim \frac{1}{2}\ln 2+\ln \sqrt \pi -\frac{1}{2}\ln \left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)\ln\left(x+\frac{1}{2}\right)-x-\frac{1}{2}\\ &=\frac{1}{2}\ln 2+\ln \sqrt \pi +x\ln \left(x+\frac{1}{2}\right)-x-\frac{1}{2}\\ &\sim \ln \sqrt\pi +x\ln x-x\\ \ln \left(\frac{\Gamma\left(x+\frac{1}{2}\right)}{\sqrt\pi}\right)&\sim x\ln x-x \end{align*}

Answer 2

Well, I post my own answer. The real reason is that (up to some constant terms and factors), the first function is $D^{-1}D^{-1}[\frac1x]$ and the second function is $D^{-1}\Delta^{-1}[\frac1x]$. So, in one case we replace one normal integration operator with discrete integration. There is no surprise, the functions should be similar.