Expansion of two Gamma function

I would like to expand

$$ \frac{\Gamma[\theta]}{\Gamma[1-\theta]} $$

up to order $\mathcal{O}(\theta)$.

Now, it's clear, that we just have to expand both parts of the fraction independently and compare the terms.

$$ \Gamma[\theta]=\frac{1}{\theta}-\gamma +\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right) \theta+\mathcal{O}\left(\theta^2\right)~. $$

$$ \Gamma[1-\theta]=1+\gamma \theta+\mathcal{O}\left(\theta^2\right) $$

However, in my source I have the following result, which doesn't makes sense to me. According to it, the fraction would be

$$ \frac{\Gamma[\theta]}{\Gamma[1-\theta]}=\frac{1}{\theta}\left(1-2 \gamma \theta+2 \gamma^{2} \theta^{2}\right)+\mathcal{O}\left[\theta^{2}\right]~. $$

Solution 1:

You need to take one more term $$\Gamma[\theta]=\frac{1}{\theta }-\gamma +\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right) \theta +\frac{1}{6} \theta ^2 \left(-\gamma ^3-\frac{\gamma \pi ^2}{2}+\psi ^{(2)}(1)\right)+O\left(\theta ^3\right)$$ $$\Gamma[1-\theta]=1+\gamma \theta +\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right) \theta ^2+O\left(\theta ^3\right)$$

Now, using long division $$\frac{\Gamma[\theta]}{\Gamma[1-\theta]}=\frac 1 \theta\Big[1-2 \gamma \theta +2 \gamma ^2 \theta ^2+O\left(\theta ^3\right)\Big]$$ which is your formula.