I want to check the validity for such a method

Define the digamma function as

$$\psi_0(x)=\frac{d}{dx}\left( \log \Gamma(x)\right)$$


It has the following series representation

$$\tag{2} \psi_0(1+x)=-\gamma +\sum_{n\geq 1}\frac{x}{n(x+n)}$$

we can easily recognize the poles at negative integers using (2).

Using the reflection formula $$\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$$

We can prove $$\tag{3} \psi_0(1-x)-\psi_0(x)=\pi \cot(\pi x)$$ and

$$\tag{4} \psi_0(1+x)-\psi_0(x)=\frac{1}{x}$$

plugging (4) in (2) we have

$$\psi_0(x)+\frac{1}{x}=-\gamma +\sum_{n\geq 1}\frac{x}{n(x+n)}$$

Now if look at the sum on the left

$$\sum_{n\geq 1}\frac{x}{n(x+n)} = \sum_{n\geq 1} \frac{x}{n^2} \frac{1}{1+\frac{x}{n}}$$

Assuming $|\frac{x}{n}| < 1$ and expand using geometric series

$$\sum_{n\geq 1} \frac{x}{n^2}\sum_{k\geq 0}\left(\frac{-x}{n} \right)^k=-\sum_{n\geq 1} \sum_{k\geq 1} \frac{(-x)^k}{n^{k+1}}=-\sum_{k\geq 1}\zeta(k+1)(-x)^k$$

Hence we have

$$\tag{5}\psi_0(x)=-\frac{1}{x}-\gamma-\sum_{k\geq 1}\zeta(k+1)(-x)^k $$

Which is the Laurent expansion of $\psi_0$ around $x=0$ giving a residue of $-1$.


My concern was the geometric expansion and the swapping of two series.

If $|x|<n$ then $|x|<1$ then the expansion is valid for any compact subset of $(-1,1)$ which justifies the uniform convergence of $\frac{(-x)^k}{n^{k+1}}$ by the $\text{M-test}$, hence swapping the two series. Then the Laurent expansion is valid on any annulus inside $(-1,1)$ avoiding $x=0$.

Is that correct ? or am I missing something ?


1 Answer 1


For $\lvert x\rvert < 1$, the nested series is absolutely convergent, hence changing the order of summation is unproblematic.

Thus the series you got is the Laurent expansion of $\psi_0$ in the whole annulus $0 < \lvert z\rvert < 1$, it converges locally uniformly and absolutely there.


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