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 $$a_1 = 1$$, $$a_2 = \gamma$$,
 * $$a_k = k a_1 a_k - a_2 a_{k-1} + \sum_{j=2}^k (-1)^j \, \zeta(j) \, a_{k-j} \;\;\; \mathrm{for}\; k > 2$$

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