How do you prove that $$ \Gamma'(1)=-\gamma, $$ where $\gamma$ is the Euler-Mascheroni constant?

  • 5
    $\begingroup$ What definition the the gamma function are you using? $\endgroup$ – Tim Raczkowski Apr 22 '15 at 15:22
  • 3
    $\begingroup$ And what definition of $\gamma$? $\endgroup$ – Chappers Apr 22 '15 at 15:32
  • $\begingroup$ Would this answer your question ? $\endgroup$ – Lucian Apr 22 '15 at 16:01

Consider the integral form of the Gamma function, \begin{align} \Gamma(x) = \int_{0}^{\infty} e^{-t} \, t^{x-1} \, dt \end{align} taking the derivative with respect to $x$ yields \begin{align} \Gamma'(x) = \int_{0}^{\infty} e^{-t} \, t^{x-1} \, \ln(t) \, dt. \end{align} Setting $x=1$ leads to \begin{align} \Gamma'(1) = \int_{0}^{\infty} e^{-t} \, \ln(t) \, dt. \end{align} This is one of the many definitions of the Euler-Mascheroni constant. Hence, \begin{align} \Gamma'(1) = - \gamma = \int_{0}^{\infty} e^{-t} \, \ln(t) \, dt. \end{align}


The Weierstrass product for the $\Gamma$ function gives: $$\Gamma(z+1)=e^{-\gamma z}\cdot\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}\tag{1}$$ hence by considering $\frac{d}{dz}\log(\cdot)$ of both terms we get: $$ \psi(z+1)=\frac{\Gamma'(z+1)}{\Gamma(z+1)}=-\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+z}\right) \tag{2}$$ and by evaluating the previous identity in $z=0$ it follows that: $$ \psi(1) = \Gamma'(1) = -\gamma.\tag{3}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.