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.



2 Answers 2


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.


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