Digamma function

File:Complex Polygamma 0.jpg
Digamma function ${\displaystyle \psi (s)}$ in the complex plane. The color of a point ${\displaystyle s}$ encodes the value of ${\displaystyle \psi (s)}$. Strong colors denote values close to zero and hue encodes the value's argument.

In mathematics, the digamma function is defined as the logarithmic derivative of the gamma function:

${\displaystyle \psi (x)={\frac {d}{dx}}\ln {\Gamma (x)}={\frac {\Gamma '(x)}{\Gamma (x)}}.}$

It is the first of the polygamma functions.

Relation to harmonic numbers

The digamma function, often denoted also as ψ0(x), ψ0(x) or ${\displaystyle \digamma }$ (after the shape of the obsolete Greek letter Ϝ digamma), is related to the harmonic numbers in that

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

where Hn is the n 'th harmonic number, and γ is the Euler-Mascheroni constant. For half-integer values, it may be expressed as

${\displaystyle \psi \left(n+{\frac {1}{2}}\right)=-\gamma -2\ln 2+\sum _{k=1}^{n}{\frac {2}{2k-1}}}$

Integral representations

It has the integral representation

${\displaystyle \psi (x)=\int _{0}^{\infty }\left({\frac {e^{-t}}{t}}-{\frac {e^{-xt}}{1-e^{-t}}}\right)\,dt}$

This may be written as

${\displaystyle \psi (s+1)=-\gamma +\int _{0}^{1}{\frac {1-x^{s}}{1-x}}dx}$

which follows from Euler's integral formula for the harmonic numbers.

Taylor series

The digamma has a rational zeta series, given by the Taylor series at z=1. This is

${\displaystyle \psi (z+1)=-\gamma -\sum _{k=1}^{\infty }\zeta (k+3)\;(-z)^{k}}$,

which converges for |z|<1. Here, ${\displaystyle \zeta (n)}$ is the Riemann zeta function. This series is easily derived from the corresponding Taylor's series for the Hurwitz zeta function.

Newton series

The Newton series for the digamma follows from Euler's integral formula:

${\displaystyle \psi (s+1)=-\gamma -\sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k}}{s \choose k}}$

where ${\displaystyle \textstyle {s \choose k}}$ is the binomial coefficient.

Reflection formula

The digamma function satisfies a reflection formula similar to that of the Gamma function,

${\displaystyle \psi (1-x)-\psi (x)=\pi \,\!\cot {\left(\pi x\right)}}$

Recurrence formula

The digamma function satisfies the recurrence relation

${\displaystyle \psi (x+1)=\psi (x)+{\frac {1}{x}}}$

Thus, it can be said to "telescope" 1/x, for one has

${\displaystyle \Delta [\psi ](x)={\frac {1}{x}}}$

where Δ is the forward difference operator. This satisfies the recurrence relation of a partial sum of the harmonic series, thus implying the formula

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

where ${\displaystyle \gamma }$ is the Euler-Mascheroni constant.

More generally, one has

${\displaystyle \psi (x+1)=-\gamma +\sum _{k=1}^{\infty }\left({\frac {1}{k}}-{\frac {1}{x+k}}\right)}$

Gaussian sum

The digamma has a Gaussian sum of the form

${\displaystyle {\frac {-1}{\pi k}}\sum _{n=1}^{k}\sin \left({\frac {2\pi nm}{k}}\right)\psi \left({\frac {n}{k}}\right)=\zeta \left(0,{\frac {m}{k}}\right)=-B_{1}\left({\frac {m}{k}}\right)={\frac {1}{2}}-{\frac {m}{k}}}$

for integers ${\displaystyle 0. Here, ζ(s,q) is the Hurwitz zeta function and ${\displaystyle B_{n}(x)}$ is a Bernoulli polynomial. A special case of the multiplication theorem is

${\displaystyle \sum _{n=1}^{k}\psi \left({\frac {n}{k}}\right)=-k(\gamma +\log k),}$

and a neat generalization of this is

${\displaystyle \sum _{p=0}^{q-1}\psi (a+p/q)=q(\psi (qa)-\ln(q)),}$

in which it is assumed that q is a natural number, and that 1-qa is not.

Gauss's digamma theorem

For positive integers m and k (with m < k), the digamma function may be expressed in terms of elementary functions as

${\displaystyle \psi \left({\frac {m}{k}}\right)=-\gamma -\ln(2k)-{\frac {\pi }{2}}\cot \left({\frac {m\pi }{k}}\right)+2\sum _{n=1}^{\lfloor (k-1)/2\rfloor }\cos \left({\frac {2\pi nm}{k}}\right)\ln \left(\sin \left({\frac {n\pi }{k}}\right)\right)}$

Special values

The digamma function has the following special values:

${\displaystyle \psi (1)=-\gamma \,\!}$
${\displaystyle \psi \left({\frac {1}{2}}\right)=-2\ln {2}-\gamma }$
${\displaystyle \psi \left({\frac {1}{3}}\right)=-{\frac {\pi }{2{\sqrt {3}}}}-{\frac {3}{2}}\ln {3}-\gamma }$
${\displaystyle \psi \left({\frac {1}{4}}\right)=-{\frac {\pi }{2}}-3\ln {2}-\gamma }$
${\displaystyle \psi \left({\frac {1}{6}}\right)=-{\frac {\pi }{2}}{\sqrt {3}}-2\ln {2}-{\frac {3}{2}}\ln(3)-\gamma }$
${\displaystyle \psi \left({\frac {1}{8}}\right)=-{\frac {\pi }{2}}-4\ln {2}-{\frac {1}{\sqrt {2}}}\left\{\pi +\ln(2+{\sqrt {2}})-\ln(2-{\sqrt {2}})\right\}-\gamma }$