Prove or disprove: \[ \Gamma'(3/2)=0. \]
Solution
The assertion is false. In fact \[ \Gamma'(3/2) =\frac{\sqrt\pi}{2}\bigl(2-\gamma-2\log 2\bigr)>0, \] where \(\gamma\) is Euler’s constant.
1. If\[ \psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}, \]then \(\psi(1)=-\gamma\).
The Weierstrass product formula is \[ z\Gamma(z) =e^{-\gamma z} \prod_{n=1}^{\infty} \left(1+\frac zn\right)^{-1}e^{z/n}. \] Logarithmic differentiation yields \[ \frac1z+\psi(z) =-\gamma+ \sum_{n=1}^{\infty} \left(\frac1n-\frac1{n+z}\right). \] At \(z=1\) the series telescopes to \(1\). Hence \[ 1+\psi(1)=-\gamma+1, \] so \(\psi(1)=-\gamma\).
2. One has\[ \psi(3/2)=2-\gamma-2\log2. \]
The Legendre duplication formula is \[ 2\sqrt\pi\,\Gamma(2z) =2^{2z}\Gamma(z)\Gamma(z+1/2). \] Taking logarithmic derivatives gives \[ 2\psi(2z) =2\log2+\psi(z)+\psi(z+1/2). \] Setting \(z=1/2\) and using 1 gives \[ 2\psi(1) =2\log2+\psi(1/2)+\psi(1), \] and therefore \[ \psi(1/2)=-\gamma-2\log2. \]
On the other hand, logarithmic differentiation of \(\Gamma(z+1)=z\Gamma(z)\) gives \[ \psi(z+1)=\psi(z)+\frac1z. \] Taking \(z=1/2\) now yields \[ \psi(3/2) =\psi(1/2)+2 =2-\gamma-2\log2. \]
3. The number \(2-\gamma-2\log2\) is strictly positive.
Euler’s constant is \[ \gamma =\lim_{N\to\infty} \left(\sum_{k=1}^N\frac1k-\log(N+1)\right). \] Since \[ \log(N+1)=\sum_{k=1}^N\log\left(1+\frac1k\right), \] this is equivalently \[ \gamma =\sum_{k=1}^{\infty} \left(\frac1k-\log\left(1+\frac1k\right)\right). \] The first three summands are \[ \sum_{k=1}^3\frac1k-\log4 =\frac{11}{6}-2\log2. \]
For every \(k\ge4\), \[ \frac1k-\log\left(1+\frac1k\right) =\int_0^{1/k}\frac{t}{1+t}\,dt <\int_0^{1/k}t\,dt =\frac1{2k^2}. \] Consequently \[ \begin{aligned} \gamma-\left(\frac{11}{6}-2\log2\right) &<\frac12\sum_{k=4}^{\infty}\frac1{k^2}\\ &<\frac12\int_3^\infty\frac{dx}{x^2} =\frac16. \end{aligned} \] Thus \[ \gamma+2\log2<2, \] which is the desired strict positivity.
4. Therefore \(\Gamma'(3/2)\neq0\).
The functional equation and \(\Gamma(1/2)=\sqrt\pi\) give \[ \Gamma(3/2)=\frac12\Gamma(1/2)=\frac{\sqrt\pi}{2}>0. \] Combining this with 2 and 3, \[ \Gamma'(3/2) =\Gamma(3/2)\psi(3/2) =\frac{\sqrt\pi}{2} \bigl(2-\gamma-2\log2\bigr)>0. \] Hence the proposed equality is false.