# Digamma function

For Barnes's gamma function of 2 variables, see double gamma function.
The color representation of the Digamma function, $\psi(z)$, in a rectangular region of the complex plane.

In mathematics, the digamma function is defined as the logarithmic derivative of the gamma function:[1][2]

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

It is the first of the polygamma functions.

## Relation to harmonic numbers

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

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

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

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

## Integral representations

If the real part of x is positive then the digamma function has the following integral representation

$\psi(x) = \int_0^{\infty}\left(\frac{e^{-t}}{t} - \frac{e^{-xt}}{1 - e^{-t}}\right)\,dt$.

This may be written as

$\psi(s+1)= -\gamma + \int_0^1 \frac {1-x^s}{1-x} dx$

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

## Series formula

Digamma can be computed in the complex plane outside negative integers (Abramowitz and Stegun 6.3.16),[1] using

$\psi(z+1)= -\gamma +\sum_{n=1}^\infty \frac{z}{n(n+z)} \qquad z \neq -1, -2, -3, \ldots$

or

$\psi(z)=-\gamma+\sum_{n=0}^{\infty}\frac{z-1}{(n+1)(n+z)}=-\gamma+\sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+z}\right)\qquad z\neq0,-1,-2,-3,\ldots$

This can be utilized to evaluate infinite sums of rational functions, i.e.,

$\sum_{n=0}^{\infty}u_{n}=\sum_{n=0}^{\infty}\frac{p(n)}{q(n)},$

where p(n) and q(n) are polynomials of n.

Performing partial fraction on un in the complex field, in the case when all roots of q(n) are simple roots,

$u_{n} =\frac{p(n)}{q(n)}=\sum_{k=1}^{m}\frac{a_{k}}{n+b_{k}}.$

For the series to converge,

$\lim_{n\to\infty}nu_{n}=0,$

or otherwise the series will be greater than the harmonic series and thus diverges.

Hence

$\sum_{k=1}^{m}a_{k}=0,$

and

\begin{align} \sum_{n=0}^{\infty}u_{n} &= \sum_{n=0}^{\infty}\sum_{k=1}^{m}\frac{a_{k}}{n+b_{k}} \\ &=\sum_{n=0}^{\infty}\sum_{k=1}^{m}a_{k}\left(\frac{1}{n+b_{k}}-\frac{1}{n+1}\right) \\ &=\sum_{k=1}^{m}\left(a_{k}\sum_{n=0}^{\infty}\left(\frac{1}{n+b_{k}}-\frac{1}{n+1}\right)\right)\\ &=-\sum_{k=1}^{m}a_{k}\left(\psi(b_{k})+\gamma\right) \\ &=-\sum_{k=1}^{m}a_{k}\psi(b_{k}). \end{align}

With the series expansion of higher rank polygamma function a generalized formula can be given as

$\sum_{n=0}^{\infty}u_{n}=\sum_{n=0}^{\infty}\sum_{k=1}^{m}\frac{a_{k}}{(n+b_{k})^{r_{k}}}=\sum_{k=1}^{m}\frac{(-1)^{r_{k}}}{(r_{k}-1)!}a_{k}\psi^{(r_{k}-1)}(b_{k}),$

provided the series on the left converges.

## Taylor series

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

$\psi(z+1)= -\gamma -\sum_{k=1}^\infty \zeta (k+1)\;(-z)^k$,

which converges for |z| < 1. Here, ζ(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:

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

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

## Reflection formula

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

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

## Recurrence formula and characterization

The digamma function satisfies the recurrence relation

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

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

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

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

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

where γ is the Euler-Mascheroni constant.

More generally, one has

$\psi(x+1) = -\gamma + \sum_{k=1}^\infty \left( \frac{1}{k}-\frac{1}{x+k} \right).$

Actually, ψ is the only solution of the functional equation

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

that is monotone on R+ and satisfies F(1) = −γ. This fact follows immediately from the uniqueness of the Γ function given its recurrence equation and convexity-restriction. This implies the useful difference equation:

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

## Some finite sums involving the digamma function

There are numerous finite summation formulas for the digamma function. Basic summation formulas, such as

$\sum_{r=1}^m \psi \left(\frac{r}{m}\right) =-m(\gamma+\ln m),$
$\sum_{r=1}^{m}\psi \left(\frac{r}{m}\right) \cdot\exp\dfrac{2\pi rk i}{m} = m\ln\!\left(\!1-\exp\dfrac{2\pi k i}{m}\!\right) \,, \qquad\quad k\in\mathbb{Z}\,, \qquad m\in\mathbb{N}\,, \qquad k\neq m.$
$\sum_{r=1}^{m-1} \psi \left(\frac{r}{m}\right) \cdot\cos\dfrac{2\pi rk}{m} \,=\, m\ln\!\left(\!2\sin\frac{\,k\pi\,}{m}\!\right) + \, \gamma \,, \qquad\qquad\qquad k=1, 2,\ldots, m-1$
$\sum_{r=1}^{m-1}\psi \left(\frac{r}{m}\right) \cdot\sin\dfrac{2\pi rk}{m} =\frac{\pi}{2} (2k-m) \,, \qquad\qquad\qquad k=1, 2,\ldots, m-1$

are due to Gauss.[3][4] More complicated formulas, such as

$\sum_{r=0}^{m-1} \psi \left(\frac{2r+1}{2m}\right)\cdot\cos\dfrac{(2r+1)k\pi }{m} = m\ln\left(\tan\frac{\,\pi k\,}{2m}\right) \,, \qquad\qquad\qquad k=1, 2,\ldots, m-1$
$\sum_{r=0}^{m-1} \psi \left(\frac{2r+1}{2m}\right)\cdot\sin\dfrac{(2r+1)k\pi }{m} = -\frac{\pi m}{2} \,, \qquad\qquad\qquad k=1, 2,\ldots, m-1$
$\sum_{r=1}^{m-1} \psi\left(\frac{r}{m}\right)\cdot\cot\frac{\pi r}{m}= -\frac{\pi(m-1)(m-2)}{6}$
$\sum_{r=1}^{m-1}\psi \left(\frac{r}{m}\right)\cdot \frac{r}{m}=-\frac{\gamma}{2}(m-1)-\frac{m}{2}\ln m -\frac{\pi}{2}\sum_{r=1}^{m-1} \dfrac{r}{m}\cdot\cot\dfrac{\pi r}{m}$
$\sum_{r=1}^{m-1}\psi \left(\frac{r}{m}\right) \cdot\cos\dfrac{(2l+1)\pi r}{m}= -\frac{\pi}{m}\sum_{r=1}^{m-1} \frac{r \cdot\sin\dfrac{2\pi r}{m}}{\,\cos\dfrac{2\pi r}{m} -\cos\dfrac{(2l+1)\pi }{m} \,} , \qquad\quad l\in\mathbb{Z}$
$\sum_{r=1}^{m-1}\psi \left(\frac{r}{m}\right) \cdot\sin\dfrac{(2l+1)\pi r}{m}= -(\gamma+\ln2m)\cot\frac{(2l+1)\pi}{2m} + \sin\dfrac{(2l+1)\pi }{m}\sum_{r=1}^{m-1} \frac{\ln\sin\dfrac{\pi r}{m}} {\,\cos\dfrac{2\pi r}{m} -\cos\dfrac{(2l+1)\pi }{m} \,} , \qquad\quad l\in\mathbb{Z}$
$\sum_{r=1}^{m-1} \psi^2\!\left(\frac{r}{m}\right)= (m-1)\gamma^2 + m(2\gamma+\ln4m)\ln{m} -m(m-1)\ln^2 2 +\frac{\pi^2 (m^2-3m+2)}{12} +m\sum_{l=1}^{ m-1 } \ln^2 \sin\frac{\pi l}{m}$

are due to works of certain modern authors (see e.g. Appendix B in[5]).

## Gauss's digamma theorem

For positive integers r and m (r < m), the digamma function may be expressed in terms of Euler's constant and a finite number of elementary functions

$\psi\left(\frac{r}{m}\right) = -\gamma -\ln(2m) -\frac{\pi}{2}\cot\left(\frac{r\pi}{m}\right) +2\sum_{n=1}^{\lfloor \frac{m-1}{2} \rfloor} \cos\left(\frac{2\pi nr}{m} \right) \ln\sin\left(\frac{\pi n}{m}\right)$

which holds, because of its recurrence equation, for all rational arguments.

## Computation and approximation

According to the Euler–Maclaurin formula applied to[6]

$\sum_{n=1}^x \frac{1}{n}$

the digamma function for x, also a real number, can be approximated by

$\psi(x) = \ln(x) - \frac{1}{2x} - \frac{1}{12x^2} + \frac{1}{120x^4} - \frac{1}{252x^6} + \frac{1}{240x^8} - \frac{5}{660x^{10}} + \frac{691}{32760x^{12}} - \frac{1}{12x^{14}} + O\left(\frac{1}{x^{16}}\right)$

which is the beginning of the asymptotical expansion of ψ(x). The full asymptotic series of this expansions is

$\psi(x) = \ln(x) - \frac{1}{2x} + \sum_{n=1}^\infty \frac{\zeta(1-2n)}{x^{2n}} = \ln(x) - \frac{1}{2x} - \sum_{n=1}^\infty \frac{B_{2n}}{2n\, x^{2n}}$

where $B_k$ is the k-th Bernoulli number and ζ is the Riemann zeta function. Although the infinite sum converges for no x, this expansion becomes more accurate for larger values of x and any finite partial sum cut off from the full series. To compute ψ(x) for small x, the recurrence relation

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

can be used to shift the value of x to a higher value. Beal[7] suggests using the above recurrence to shift x to a value greater than 6 and then applying the above expansion with terms above $x^{14}$ cut off, which yields "more than enough precision" (at least 12 digits except near the zeroes).

\begin{align} \psi(x) &\in [\ln(x-1), \ln x] \\ \exp(\psi(x)) &\approx \begin{cases} \frac{x^2}{2} & x\in[0,1] \\ x - \frac{1}{2} & x>1 \end{cases} \end{align}

From the above asymptotic series for ψ, one can derive asymptotic series for $\exp \circ\, \psi$ that contain only rational functions and constants. The first series matches the overall behaviour of $\exp \circ\, \psi$ well, that is, it behaves asymptotically identically for large arguments and has a zero of unbounded multiplicity at the origin, too. It can be considered a Taylor expansion of $\exp(-\psi(1/y))$ at y = 0.

$\frac{1}{\exp(\psi(x))} = \frac{1}{x}+\frac{1}{2\cdot x^2}+\frac{5}{4\cdot3!\cdot x^3}+\frac{3}{2\cdot4!\cdot x^4}+\frac{47}{48\cdot5!\cdot x^5} - \frac{5}{16\cdot6!\cdot x^6} + \cdots$

The other expansion is more precise for large arguments and saves computing terms of even order.

$\exp(\psi(x+\tfrac{1}{2})) = x + \frac{1}{4!\cdot x} - \frac{37}{8\cdot6!\cdot x^3} + \frac{10313}{72\cdot8!\cdot x^5} - \frac{5509121}{384\cdot10!\cdot x^7} + O\left(\frac{1}{x^9}\right)\qquad\mbox{for } x>1$

## Special values

The digamma function has values in closed form for rational numbers, as a result of Gauss's digamma theorem. Some are listed below:

\begin{align} \psi(1) &= -\gamma \\ \psi\left(\tfrac{1}{2}\right) &= -2\ln{2} - \gamma \\ \psi\left(\tfrac{1}{3}\right) &= -\tfrac{\pi}{2\sqrt{3}} -\tfrac{3}{2}\ln{3} - \gamma \\ \psi\left(\tfrac{1}{4}\right) &= -\tfrac{\pi}{2} - 3\ln{2} - \gamma \\ \psi\left(\tfrac{1}{6}\right) &= -\tfrac{\pi}{2}\sqrt{3} -2\ln{2} -\tfrac{3}{2}\ln(3) - \gamma \\ \psi\left(\tfrac{1}{8}\right) &= -\tfrac{\pi}{2} - 4\ln{2} - \frac{1}{\sqrt{2}} \left\{\pi + \ln \left (2 + \sqrt{2} \right ) - \ln \left (2 - \sqrt{2} \right ) \right \} - \gamma. \end{align}

Moreover, by the series representation, it can easily be deduced that at the imaginary unit,

\begin{align} \Re\left(\psi(i)\right) &= -\gamma-\sum_{n=0}^\infty\frac{n-1}{n^3+n^2+n+1}, \\ \Im\left(\psi(i)\right) &= \sum_{n=0}^\infty\frac{1}{n^2+1} = \frac12+\frac{\pi}{2}\coth(\pi). \end{align}

## Roots of the digamma function

The roots of the digamma function are the saddle points of the complex-valued gamma function. Thus they lie all on the real axis. The only one on the positive real axis is the unique minimum of the real-valued gamma function on R+ at $x_0 = 1.461632144968\ldots$. All others occur single between the poles on the negative axis:

\begin{align} x_1 &= -0.504083008..., \\ x_2 &= -1.573498473..., \\ x_3 &= -2.610720868..., \\ x_4 &= -3.635293366..., \\ &\qquad \cdots \end{align}

Already in 1881, Hermite observed[citation needed] that

$x_n = -n + \frac{1}{\ln n} + o\left(\frac{1}{\ln^2 n}\right)$

holds asymptotically. A better approximation of the location of the roots is given by

$x_n \approx -n + \frac{1}{\pi}\arctan\left(\frac{\pi}{\ln n}\right)\qquad n \ge 2$

and using a further term it becomes still better

$x_n \approx -n + \frac{1}{\pi}\arctan\left(\frac{\pi}{\ln n + \frac{1}{8n}}\right)\qquad n \ge 1$

which both spring off the reflection formula via

$0 = \psi(1-x_n) = \psi(x_n) + \frac{\pi}{\tan(\pi x_n)}$

and substituting $\psi(x_n)$ by its not convergent asymptotic expansion. The correct 2nd term of this expansion is of course $\tfrac1 {2n}$, where the given one works good to approximate roots with small index n.

Regarding the zeros, the following infinite sum identities were recently proved by Mező[8]

\begin{align} \sum_{n=0}^\infty\frac{1}{x_n^2}&=\gamma^2+\frac{\pi^2}{2}, \\ \sum_{n=0}^\infty\frac{1}{x_n^4}&=\gamma^4+\frac{\pi^4}{9}+\frac23\gamma^2\pi^2+4\gamma\zeta(3). \end{align}

Here $\gamma$ is the Euler–Mascheroni constant.

## Regularization

The Digamma function appears in the regularization of divergent integrals

$\int_{0}^{\infty} \frac{dx}{x+a},$

this integral can be approximated by a divergent general Harmonic series, but the following value can be attached to the series

$\sum_{n=0}^{\infty} \frac{1}{n+a}= - \psi (a).$