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

Question

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.

Answer 1

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.

Answer 2

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