# How to calculate $\displaystyle \lim_{\alpha \to -1}\left(2\Gamma(-\alpha-1)+\Gamma\left(\frac{\alpha + 1}{2}\right) \right)$

How can I compute the following limit? $$\displaystyle \lim_{\alpha \to -1}\left(2\Gamma(-\alpha-1)+\Gamma\left(\frac{\alpha + 1}{2}\right) \right)$$

The answer appears to be about -1.73

I would be really happy if someone could help me.

## 2 Answers

Set $x=\alpha+1$ and recall that for small values of $x$ we have $\Gamma(x) = 1/x - \gamma + O(x)$, to get $$\lim_{x\to0} 2\Gamma(-x)+ \Gamma(x/2)=\lim_{x\to0} - 2/x -2\gamma + 2/x -\gamma = -3\gamma,$$ where $\gamma$ is the Euler-Mascheroni constant.

• thank so much. One question what is $O(x)$ – Argento Aug 25 '15 at 10:26
• @Xision: You're very welcome! That's called Big O notation. – Vincenzo Oliva Aug 25 '15 at 10:58
• @Xision: Please make sure to accept the answer with the approach you liked the most. :) – Vincenzo Oliva Aug 25 '15 at 17:31
• why a missing $O(x)$ .i don't understant. i have $$\displaystyle \lim_{x \to 0}2\Gamma(-x)+\Gamma(\frac{x}{2}) = \lim_{x \to 0} \frac{-2}{x}-2\gamma+2O(-x)+\frac{2}{x}-\gamma+O(\frac{x}{2})$$ – Argento Aug 26 '15 at 16:37

You may use the expansion of $\Gamma(z)$ around one of its poles:

$$\Gamma(z-n)=\frac{(-1)^n}{n!}\left(\frac{1}{z}+\psi(n+1)+\mathcal{O}(z)\right)$$

where $\psi(z):=\frac{\Gamma'(z)}{\Gamma(z)}$

We obtain:

$$\underbrace{\left(\frac{2}{x+1}-\gamma\right)}_{\lim_{x\rightarrow -1}\Gamma(\frac{x+1}{2})}+\underbrace{\left(-\frac{2}{x+1}-2\gamma\right)}_{\lim_{x\rightarrow -1}2\Gamma(-x-1)}=-3\gamma\approx-1.73165$$

Where $\gamma$ is the Euler-Mascheroni constant and we used special values of the digamma function given here

• The terms $\mathcal{O}(x-1)$ will not contribute in this limit so i just didn,t wrote them down... :) – tired Aug 26 '15 at 18:42
• what the terms $O(x-1)$ is come from ? Could you please explain to me... :) – Argento Aug 28 '15 at 18:47
• What exactly don't you understand? are you familiar with $\mathcal{O}$ notation? – tired Sep 1 '15 at 16:55