Digamma 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]

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

It is the first of the polygamma functions.

The digamma function is often denoted as ψ₀(x), ψ⁰(x) or $\digamma$ (after the archaic Greek letter Ϝ digamma).

Relation to harmonic numbers

The gamma function obeys the equation

$\Gamma(z+1)=z\Gamma(z).$

Taking the derivative with respect to z gives:

$\Gamma'(z+1)=z\Gamma'(z)+\Gamma(z)$

Dividing by Γ(z+1) or the equivalent zΓ(z) gives:

$\frac{\Gamma'(z+1)}{\Gamma(z+1)}=\frac{\Gamma'(z)}{\Gamma(z)}+\frac 1z$

or:

$\psi(z+1)=\psi(z)+\frac 1z$

Since the harmonic numbers are defined as

$H_n=\sum_{k=1}^n\frac 1k$

the digamma function is related to it by:

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

where H_n 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

$\psi(x)=\int\limits_0^\infty \left(\frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}}\right)dt$.

This may be written as

$\psi(s+1)=-\gamma+\int\limits_0^1 \left(\frac{1-x^s}{1-x}\right)dx$

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

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

or

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

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 u_n 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,

$\psi(1-x)-\psi(x)=\pi\ \cot(\pi x)$

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

$\sum_{r=1}^m \psi\left(\frac{r}{m}\right)=-m(\gamma+\ln(m)),$

$\sum_{r=1}^m \psi\left(\frac{r}{m}\right) \cdot\exp\dfrac{2\pi rki}{m}=m\ln\left(1-\exp\dfrac{2\pi ki}{m}\right)\ , \qquad\quad k\in\Z\ ,\qquad m\in\N\ ,\qquad k\ne m.$

$\sum_{r=1}^{m-1} \psi\left(\frac{r}{m}\right) \cdot\cos\dfrac{2\pi rk}{m}\ =\ m\ln\left(2\sin\left(\frac{k\pi}{m}\right)\right)+\gamma\ , \qquad\qquad\qquad k=1, 2,\ldots, m-1$

$\sum_{r=1}^{m-1}\psi \left(\frac{r}{m}\right) \cdot\sin\dfrac{2\pi rk}{m} =\frac{\pi}{2} (2k-m) \,, \qquad\qquad\qquad k=1, 2,\ldots, m-1$

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 an asymptotic series for $1/\exp(\psi(x))$. The series matches the overall behaviour well, that is, it behaves asymptotically as it should for large arguments, and has a zero of unbounded multiplicity at the origin too.

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

This can be considered a Taylor expansion of $\exp(-\psi(1/y))$ at y = 0, but it does not converge.^[8]

Another 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 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^[9]

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

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

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

See also

• Polygamma function
• Trigamma function
• Chebyshev expansions of the Digamma function in Wimp, Jet (1961). "Polynomial approximations to integral transforms". Math. Comp. 15: 174–178. doi:10.1090/S0025-5718-61-99221-3.

References

1. Abramowitz, M.; Stegun, I. A., eds. (1972). "6.3 psi (Digamma) Function.". Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables (10th ed.). New York: Dover. pp. 258–259.
2. Weisstein, Eric W., "Digamma function", MathWorld.
3. R. Campbell. Les intégrales eulériennes et leurs applications, Dunod, Paris, 1966.
4. H.M. Srivastava and J. Choi. Series Associated with the Zeta and Related Functions, Kluwer Academic Publishers, the Netherlands, 2001.
5. Iaroslav V. Blagouchine A theorem for the closed-form evaluation of the first generalized Stieltjes constant at rational arguments and some related summations Journal of Number Theory (Elsevier), vol. 148, pp. 537–592, 2015. arXiv PDF
6. Bernardo, José M. (1976). "Algorithm AS 103 psi(digamma function) computation" (PDF). Applied Statistics 25: 315–317.
7. Beal, Matthew J. (2003). Variational Algorithms for Approximate Bayesian Inference (PDF) (PhD thesis). The Gatsby Computational Neuroscience Unit, University College London. pp. 265–266.
8. If it converged to a function f(y) then ln(f(y)/y) would have the same Maclaurin series as $\ln(1/y)-\phi(1/y).$ But this does not converge because the series given earlier for φ(x) does not converge.
9. I. Mező (2014). "A note on the zeros of the Digamma function and the derivative of the log-Barnes function". arXiv:1409.2971v1.

External links

•  A020759 psi(1/2),  A047787 psi(1/3),  A200064 psi(2/3),  A020777 psi(1/4),  A200134 psi(3/4),  A200135 to  A200138 psi(1/5) to psi(4/5).