# Wonderful Limit

• Aug 16th 2011, 07:22 AM
uniquesailor
Wonderful Limit
• Aug 16th 2011, 09:21 AM
HallsofIvy
Re: Wonderful Limit
So $\displaystyle \Gamma'(t)= (t-1)\int_0^\infty y^{t- 2}e^{-y}dy$
• Aug 16th 2011, 11:38 AM
chisigma
Re: Wonderful Limit

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

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

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

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

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

Now is...

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

... so that is...

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Aug 16th 2011, 05:26 PM
uniquesailor
Re: Wonderful Limit
Quote:

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

This wrong
• Aug 16th 2011, 05:28 PM
uniquesailor
Re: Wonderful Limit