The Laurent series of the digamma function at the negative integers

Solution 1:

A few terms for starters.... $$ \Psi \left( x \right) = -{x}^{-1}-\gamma+1/6\,{\pi }^{2}x-\zeta \left( 3 \right) {x}^{2}+{\frac {1}{90}}\,{\pi }^{4}{x}^{3}-\zeta \left( 5 \right) {x}^{4}+{\frac {1}{945}}\,{\pi }^{6}{x}^{5}+O \left( {x}^{6} \right) $$

$$ \Psi \left( x \right) =- \left( x+1 \right) ^{-1}+1-\gamma+ \left( 1+ 1/6\,{\pi }^{2} \right) \left( x+1 \right) + \left( 1-\zeta \left( 3 \right) \right) \left( x+1 \right) ^{2}+ \left( {\frac {1}{90}}\,{ \pi }^{4}+1 \right) \left( x+1 \right) ^{3}+ \left( 1-\zeta \left( 5 \right) \right) \left( x+1 \right) ^{4}+ \left( {\frac {1}{945}}\,{ \pi }^{6}+1 \right) \left( x+1 \right) ^{5}+O \left( \left( x+1 \right) ^{6} \right) $$

$$ \Psi \left( x \right) =- \left( x+2 \right) ^{-1}+{\frac {3}{2}}- \gamma+ \left( {\frac {5}{4}}+1/6\,{\pi }^{2} \right) \left( x+2 \right) + \left( {\frac {9}{8}}-\zeta \left( 3 \right) \right) \left( x+2 \right) ^{2}+ \left( {\frac {1}{90}}\,{\pi }^{4}+{\frac { 17}{16}} \right) \left( x+2 \right) ^{3}+ \left( {\frac {33}{32}}- \zeta \left( 5 \right) \right) \left( x+2 \right) ^{4}+ \left( { \frac {1}{945}}\,{\pi }^{6}+{\frac {65}{64}} \right) \left( x+2 \right) ^{5}+O \left( \left( x+2 \right) ^{6} \right) $$

$$ \Psi \left( x \right) =- \left( x+3 \right) ^{-1}+{\frac {11}{6}}- \gamma+ \left( {\frac {49}{36}}+1/6\,{\pi }^{2} \right) \left( x+3 \right) + \left( {\frac {251}{216}}-\zeta \left( 3 \right) \right) \left( x+3 \right) ^{2}+ \left( {\frac {1}{90}}\,{\pi }^{4}+{\frac { 1393}{1296}} \right) \left( x+3 \right) ^{3}+ \left( {\frac {8051}{ 7776}}-\zeta \left( 5 \right) \right) \left( x+3 \right) ^{4}+ \left( {\frac {1}{945}}\,{\pi }^{6}+{\frac {47449}{46656}} \right) \left( x+3 \right) ^{5}+O \left( \left( x+3 \right) ^{6} \right) $$

Solution 2:

We will use the Laurent expansion around $x=0$

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

Look the proof here

Then

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

Now we use that

$$\tag{2} \psi_0(x+N)=\psi_0(x)+\sum_{k=0}^{N-1}\frac{1}{x+k}$$

Now we look at the finite sum

\begin{align}\sum_{k=0}^{N-1}\frac{1}{x+k}&=\sum_{k=0}^{N-1}\frac{1}{k-N+x+N}\\&=\sum_{k=0}^{N-1}\frac{1}{k-N}\frac{1}{1+\frac{x+N}{k-N}}\\ &= \sum_{k=0}^{N-1}\frac{1}{k-N}\sum_{m \geq 0}(-1)^m\left(\frac{x+N}{k-N} \right)^m \\&= \sum_{m \geq 0}(-1)^m(x+N)^m \sum_{k=0}^{N-1}\frac{1}{(k-N)^{m+1}}\\&= -\sum_{m \geq 0} H^{(m+1)}_{N} (x+N)^{m}\end{align}

Hence we have

$$\psi_0(x+N)=\psi_0(x)-\sum_{m \geq 0} H^{(m+1)}_{N} (x+N)^{m}$$

Substitute in (1) to obtain

$$\psi_0(x)-\sum_{m \geq 1} H^{(m+1)}_{N} (x+N)^{m}=-\frac{1}{x+N}-\gamma-\sum_{k\geq 1}(-1)^k\zeta(k+1)(x+N)^k$$

$$\psi_0(x)=-\frac{1}{x+N}-\gamma-\sum_{k\geq 1}(-1)^k\zeta(k+1)(x+N)^k+\sum_{m \geq 0} H^{(m+1)}_{N} (x+N)^{m}$$

$$\psi_0(-x)=\frac{1}{x-N}-\gamma-\sum_{k\geq 1}\zeta(k+1)(x-N)^k+\sum_{m \geq 0}(-1)^m H^{(m+1)}_{N} (x-N)^{m}$$

$$\psi_0(-x)+\gamma=\frac{1}{x-N}+H_N+\sum_{k\geq 1}((-1)^kH^{(k+1)}_{N}-\zeta(k+1))(x-N)^k$$

Proof of (2)

$$\psi_0(x+N)=\psi_0(x)+\sum_{k=0}^{N-1}\frac{1}{x+k}$$

By induction for $N=1$ we get

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

Which is true and can be proved using

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

Now for the inductive step , assume

$$\psi_0(x+N)=\psi_0(x)+\sum_{k=0}^{N-1}\frac{1}{x+k}$$

Then

$$\psi_0(x+N+1)=\psi_0(x+N)+\frac{1}{x+N+1}=\psi_0(x)+\sum_{k=0}^{N}\frac{1}{x+k}$$