Wonderful Limit

**HallsofIvy** · August 16th 2011, 09:21 AM

So $\Gamma'(t)= (t-1)\int_0^\infty y^{t- 2}e^{-y}dy$

**chisigma** · August 16th 2011, 11:38 AM

We start with the well known 'infinite product'...

$\Gamma(x)= \frac{e^{-\gamma x}}{x}\ \prod_{n=1}^{\infty} \frac{e^{\frac{x}{n}}}{1+\frac{x}{n}}$ (1)

... where $\gamma$ is the so called Euler's constant, then from (1)...

$\ln \Gamma(x)= -\ln x - \gamma x + \sum_{n=1}^{\infty} \{\frac{x}{n}-\ln (1+\frac{x}{n}) \}$ (2)

... and then from (2)...

$\frac{d}{dx} \ln \Gamma (x) = -\frac{1}{x} - \gamma + \sum_{n=1}^{\infty} \frac{x}{n\ (n+x)}$ (3)

Now is...

$\frac{d}{dx} \ln \Gamma (x) = \frac{\Gamma^{'}(x)}{\Gamma(x)}$ (4)

... so that is...

$\Gamma^{'}(x)= \Gamma(x)\ \{-\frac{1}{x} - \gamma + \sum_{n=1}^{\infty} \frac{x}{n\ (n+x)}\}$ (5)

... and, taking into account (1) and (5), we can conclude that...

$\lim_{x \rightarrow 0+} x^{2}\ \Gamma^{'}(x) = -1$ (6)

Kind regards

$\chi$ $\sigma$

**uniquesailor** · August 16th 2011, 05:26 PM

Originally Posted by HallsofIvy

So $\Gamma'(t)= (t-1)\int_0^\infty y^{t- 2}e^{-y}dy$

This wrong

**uniquesailor** · August 16th 2011, 05:28 PM

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