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I've been given two definitions of the Gamma function, the integral defintion:

$\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt$ (for $Re(z)>0$)

and the product definition (for $1/\Gamma$):

$\frac{1}{\Gamma(z)} = ze^{\gamma z}\Pi_{n=1}^\infty ((1+\frac{z}{n})e^{-\frac{z}{n}})$

where $\gamma$ is Euler's constant

My lecturer has asserted that therefore (presumably from the product definition):

$\Gamma'(1) = -\gamma$

but I can't see why this is true. Is this something that follows easily from these definitions? If it is I would appreciate some help or a solution. Thanks :)

Matt Gallagher
  • 471

2 Answers2

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Start from the product form and take the log of both sides, and then differentiate: (Sorry, I have to write out the equations in LaTeX format, I don't know MathML well enough)

Take the log: \begin{equation} - \log (\Gamma(z)) = \log(z) + \gamma z + \sum_{n=1}^{\infty} \left[ \log \left( 1 + \frac{z}{n} \right) - \frac{z}{n} \right] \end{equation}

Take the derivative of both sides with respect to z: \begin{equation} - \frac{\Gamma^{\prime}(z)}{\Gamma(z)} = \frac{1}{z} + \gamma + \sum_{n=1}^{\infty} \left[ \frac{1/n}{1+z/n} - \frac{1}{n} \right] \end{equation}

Now insert z = 1 because we only care about the derivative at z = 1, and simplify the fraction with 1+1/n in the denominator (using $\Gamma(1) = \int_0^\infty e^{-t}dt = 1$): \begin{align} - \Gamma' (1) = 1 + \gamma + \sum_{n=1}^{\infty} \left[ \frac{1}{n+1} - \frac{1}{n} \right] = 1 + \gamma - \sum_{n=1}^{\infty} \frac{ 1}{n(n+1)} = \gamma. \end{align} which is $\Gamma'(1) = - \gamma$.

Matt Gallagher
  • 471
Mark Fischler
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Here if we use the integral definition we get:

$\Gamma'(z)=\int_0^\infty \frac{\partial}{\partial z}(t^{z-1})e^{-t}dt=\int_0^\infty \ln(t)t^{z-1}e^{-t}dt$

thus $\Gamma'(1)=\int_0^\infty \ln(t)e^{-t}dt=-\gamma$

Ellya
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