Polygamma function

In mathematics, the polygamma functions are the logarithmic derivatives of the gamma function. The polygamma function of order $m$ is defined as

$\psi ^{(m-1)}(z)={\frac {d^{m}}{dz^{m}}}\log \Gamma (z).$

The case $\psi =\psi ^{(0)}\,\;$ is called the digamma function.

Taking the logarithm of Weierstrass's product for the gamma function, we can write the logarithm of the gamma function in the form of a series

$\log \Gamma (z)=-\log z-\gamma z+\sum _{k=1}^{\infty }\left[{\frac {z}{k}}-\log \left(1+{\frac {z}{k}}\right)\right].$

Differentiating termwise renders the logarithmic derivative of the gamma function

$\left(\log \Gamma (z)\right)'=-{\frac {1}{z}}-\gamma +\sum _{k=1}^{\infty }\left({\frac {1}{k}}-{\frac {1}{k+z}}\right)$

which is called the digamma function, denoted by $\psi (z)\,\!$. The expression for the digamma function is simpler than the series we started with, all logarithms notably absent. If $z=n$, a positive integer, all but finitely many terms in the series cancel and we are left with

$\psi (n)=H_{n-1}-\gamma \,\!$

where $H_{n}$ is the harmonic number $\textstyle 1+{\frac {1}{2}}+{\frac {1}{3}}+\cdots +{\frac {1}{n}}$. As it generalizes harmonic numbers to non-integer indices, we could say that the digamma function is to harmonic numbers as the gamma function is to the factorials. Since $(\log f)'=f'/f$, we can recover the ordinary derivative of the gamma function as $\Gamma '(z)=\Gamma (z)\psi (z)$; the derivative at an integer is then $\Gamma '(n)=(H_{n-1}-\gamma )(n-1)!$ and in particular, $\Gamma '(1)=-\gamma$, providing a geometric interpretation of Euler's constant as the slope of the gamma function's graph at 1.

Continuing in a similar manner, rational series for the polygamma functions of higher order can be derived.

The polygamma functions are related to the Riemann zeta function: it can be shown that the polygamma function at an integer value is expressible in terms of the zeta function at integers. With the higher-order derivatives available, it becomes possible to calculate the Taylor series of the gamma function around any integer in terms of Euler's constant and the zeta function. Choosing $1/\Gamma (z)$ instead of $\Gamma (z)$ for convenience, since the former is an entire function and hence has an everywhere convergent Taylor series in the simple point $z=0$, we can compute

${\frac {1}{\Gamma (z)}}=z+\gamma z^{2}+\left({\frac {\gamma ^{2}}{2}}-{\frac {\pi ^{2}}{12}}\right)z^{3}+\cdots =\sum _{k=1}^{\infty }a_{k}z^{k}$

where $a_{1}=1$, $a_{2}=\gamma$,

$a_{k}=ka_{1}a_{k}-a_{2}a_{k-1}+\sum _{j=2}^{k}(-1)^{j}\,\zeta (j)\,a_{k-j}\;\;\;\mathrm {for} \;k>2$

and $\zeta (s)\,\!$ is the Riemann zeta function.