# Limit of Gamma function becomes Log

I am performing a calculation in Physics and needed to compute the following limit:

$$\lim_{d\to2} \Gamma \left(1-\frac{d}{2} \right) \cdot \left(a^{2-d} - b^{2-d} \right).$$

I typed it into WolframAlpha and it gave me back $$2\log(b/a)$$. Is there somehow a way one can obtain this? I have tried l'Hôpital's rule, but I am not that familiar with derivatives of the $$\Gamma$$ function and was hoping someone could illuminate the answer.

Thanks!

You need to know something about the Gamma function to proceed. Take $$1 - \frac{d}{2} = x$$. Then you are interested in the case when $$x \to 0$$.
$$\lim_{x \to 0}\Gamma(x)(a^{2x} - b^{2x})$$
Now use the series expansion of $$\Gamma(x)$$ about the point at $$x = 0$$.
It is very natural to change coordinates $$x = 1-\frac12 d$$ so that we want to find the limit $$L=\lim_{x\to 0}\Gamma(x)(a^{2x}-b^{2x}) = \lim_{x\to 0}\frac{a^{2x}-b^{2x}}{\Gamma(x)^{-1}}.$$ As you suggest, we can use L'Hospital's rule where the main difficulty is to determine the limit $$K=\lim_{x\to 0}\frac{\Gamma'(x)}{\Gamma(x)^2} \quad\text{and then }L=\log(a^2/b^2)K=2\log(a/b)K$$($$K$$ comes from differentiating the denominator). However, it is a standard result that the gamma function has a simple pole at $$z=0$$ with coefficient 1 on the $$1/z$$ term in its Laurent series expansion. From this it follows $$K=-1$$ and hence your desired result.
Put another way, all you need to know is that the residue of $$\Gamma$$ at $$z=0$$ is $$1$$. This can be obtained by a simple argument using a recurrence relation, see e.g. this Wikipedia page.