The color representation of the Digamma function,
ψ
(
z
)
{\displaystyle \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]
ψ
(
x
)
=
d
d
x
ln
(
Γ
(
x
)
)
=
Γ
′
(
x
)
Γ
(
x
)
.
{\displaystyle \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 ψ 0 (x ) , ψ 0 (x ) or Ϝ (the uppercase form of the archaic Greek consonant digamma ) meaning Double-Gamma .
Relation to harmonic numbers
The gamma function obeys the equation
Γ
(
z
+
1
)
=
z
Γ
(
z
)
.
{\displaystyle \Gamma (z+1)=z\Gamma (z).}
Taking the derivative with respect to z gives:
Γ
′
(
z
+
1
)
=
z
Γ
′
(
z
)
+
Γ
(
z
)
{\displaystyle \Gamma '(z+1)=z\Gamma '(z)+\Gamma (z)}
Dividing by Γ(z + 1) or the equivalent z Γ(z ) gives:
Γ
′
(
z
+
1
)
Γ
(
z
+
1
)
=
Γ
′
(
z
)
Γ
(
z
)
+
1
z
{\displaystyle {\frac {\Gamma '(z+1)}{\Gamma (z+1)}}={\frac {\Gamma '(z)}{\Gamma (z)}}+{\frac {1}{z}}}
or:
ψ
(
z
+
1
)
=
ψ
(
z
)
+
1
z
{\displaystyle \psi (z+1)=\psi (z)+{\frac {1}{z}}}
Since the harmonic numbers are defined as
H
n
=
∑
k
=
1
n
1
k
{\displaystyle H_{n}=\sum _{k=1}^{n}{\frac {1}{k}}}
the digamma function is related to it by:
ψ
(
n
)
=
H
n
−
1
−
γ
{\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
ψ
(
n
+
1
2
)
=
−
γ
−
2
ln
2
+
∑
k
=
1
n
2
2
k
−
1
{\displaystyle \psi \left(n+{\tfrac {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
ψ
(
x
)
=
∫
0
∞
(
e
−
t
t
−
e
−
x
t
1
−
e
−
t
)
d
t
{\displaystyle \psi (x)=\int _{0}^{\infty }\left({\frac {e^{-t}}{t}}-{\frac {e^{-xt}}{1-e^{-t}}}\right)\,dt}
.
This may be written as
ψ
(
s
+
1
)
=
−
γ
+
∫
0
1
(
1
−
x
s
1
−
x
)
d
x
{\displaystyle \psi (s+1)=-\gamma +\int _{0}^{1}\left({\frac {1-x^{s}}{1-x}}\right)\,dx}
which follows from Leonhard Euler 's integral formula for the harmonic numbers.
The digamma function can be computed in the complex plane outside negative integers (Abramowitz and Stegun 6.3.16),[ 1] using
ψ
(
z
+
1
)
=
−
γ
+
∑
n
=
1
∞
z
n
(
n
+
z
)
z
≠
−
1
,
−
2
,
−
3
,
…
{\displaystyle \psi (z+1)=-\gamma +\sum _{n=1}^{\infty }{\frac {z}{n(n+z)}}\qquad z\neq -1,-2,-3,\ldots }
or
ψ
(
z
)
=
−
γ
+
∑
n
=
0
∞
z
−
1
(
n
+
1
)
(
n
+
z
)
=
−
γ
+
∑
n
=
0
∞
(
1
n
+
1
−
1
n
+
z
)
z
≠
0
,
−
1
,
−
2
,
−
3
,
…
{\displaystyle \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\neq 0,-1,-2,-3,\ldots }
This can be utilized to evaluate infinite sums of rational functions, i.e.,
∑
n
=
0
∞
u
n
=
∑
n
=
0
∞
p
(
n
)
q
(
n
)
,
{\displaystyle \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
=
p
(
n
)
q
(
n
)
=
∑
k
=
1
m
a
k
n
+
b
k
.
{\displaystyle u_{n}={\frac {p(n)}{q(n)}}=\sum _{k=1}^{m}{\frac {a_{k}}{n+b_{k}}}.}
For the series to converge,
lim
n
→
∞
n
u
n
=
0
,
{\displaystyle \lim _{n\to \infty }nu_{n}=0,}
otherwise the series will be greater than the harmonic series and thus diverge. Hence
∑
k
=
1
m
a
k
=
0
,
{\displaystyle \sum _{k=1}^{m}a_{k}=0,}
and
∑
n
=
0
∞
u
n
=
∑
n
=
0
∞
∑
k
=
1
m
a
k
n
+
b
k
=
∑
n
=
0
∞
∑
k
=
1
m
a
k
(
1
n
+
b
k
−
1
n
+
1
)
=
∑
k
=
1
m
(
a
k
∑
n
=
0
∞
(
1
n
+
b
k
−
1
n
+
1
)
)
=
−
∑
k
=
1
m
a
k
(
ψ
(
b
k
)
+
γ
)
=
−
∑
k
=
1
m
a
k
ψ
(
b
k
)
.
{\displaystyle {\begin{aligned}\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}{\big (}\psi (b_{k})+\gamma {\big )}\\&=-\sum _{k=1}^{m}a_{k}\psi (b_{k}).\end{aligned}}}
With the series expansion of higher rank polygamma function a generalized formula can be given as
∑
n
=
0
∞
u
n
=
∑
n
=
0
∞
∑
k
=
1
m
a
k
(
n
+
b
k
)
r
k
=
∑
k
=
1
m
(
−
1
)
r
k
(
r
k
−
1
)
!
a
k
ψ
r
k
−
1
(
b
k
)
,
{\displaystyle \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
ψ
(
z
+
1
)
=
−
γ
−
∑
k
=
1
∞
ζ
(
k
+
1
)
(
−
z
)
k
{\displaystyle \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:
ψ
(
s
+
1
)
=
−
γ
−
∑
k
=
1
∞
(
−
1
)
k
k
(
s
k
)
{\displaystyle \psi (s+1)=-\gamma -\sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k}}{\binom {s}{k}}}
where ( s k ) is the binomial coefficient .
The digamma function satisfies a reflection formula similar to that of the gamma function :
ψ
(
1
−
x
)
−
ψ
(
x
)
=
π
cot
π
x
{\displaystyle \psi (1-x)-\psi (x)=\pi \cot \pi x}
The digamma function satisfies the recurrence relation
ψ
(
x
+
1
)
=
ψ
(
x
)
+
1
x
.
{\displaystyle \psi (x+1)=\psi (x)+{\frac {1}{x}}.}
Thus, it can be said to "telescope" 1 / x , for one has
Δ
[
ψ
]
(
x
)
=
1
x
{\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
ψ
(
n
)
=
H
n
−
1
−
γ
{\displaystyle \psi (n)=H_{n-1}-\gamma }
where γ is the Euler–Mascheroni constant .
More generally, one has
ψ
(
x
+
1
)
=
−
γ
+
∑
k
=
1
∞
(
1
k
−
1
x
+
k
)
.
{\displaystyle \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
)
+
1
x
{\displaystyle 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:
ψ
(
x
+
N
)
−
ψ
(
x
)
=
∑
k
=
0
N
−
1
1
x
+
k
{\displaystyle \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
∑
r
=
1
m
ψ
(
r
m
)
=
−
m
(
γ
+
ln
m
)
,
{\displaystyle \sum _{r=1}^{m}\psi \left({\frac {r}{m}}\right)=-m(\gamma +\ln m),}
∑
r
=
1
m
ψ
(
r
m
)
⋅
exp
2
π
r
k
i
m
=
m
ln
(
1
−
exp
2
π
k
i
m
)
,
k
∈
Z
,
m
∈
N
,
k
≠
m
.
{\displaystyle \sum _{r=1}^{m}\psi \left({\frac {r}{m}}\right)\cdot \exp {\dfrac {2\pi rki}{m}}=m\ln \left(1-\exp {\frac {2\pi ki}{m}}\right),\qquad k\in \mathbb {Z} \ ,\ m\in \mathbb {N} \ ,\ k\neq m.}
∑
r
=
1
m
−
1
ψ
(
r
m
)
⋅
cos
2
π
r
k
m
=
m
ln
(
2
sin
k
π
m
)
+
γ
,
k
=
1
,
2
,
…
,
m
−
1
{\displaystyle \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 k=1,2,\ldots ,m-1}
∑
r
=
1
m
−
1
ψ
(
r
m
)
⋅
sin
2
π
r
k
m
=
π
2
(
2
k
−
m
)
,
k
=
1
,
2
,
…
,
m
−
1
{\displaystyle \sum _{r=1}^{m-1}\psi \left({\frac {r}{m}}\right)\cdot \sin {\frac {2\pi rk}{m}}={\frac {\pi }{2}}(2k-m),\qquad k=1,2,\ldots ,m-1}
are due to Gauss.[ 3] [ 4] More complicated formulas, such as
∑
r
=
0
m
−
1
ψ
(
2
r
+
1
2
m
)
⋅
cos
(
2
r
+
1
)
k
π
m
=
m
ln
(
tan
π
k
2
m
)
,
k
=
1
,
2
,
…
,
m
−
1
{\displaystyle \sum _{r=0}^{m-1}\psi \left({\frac {2r+1}{2m}}\right)\cdot \cos {\frac {(2r+1)k\pi }{m}}=m\ln \left(\tan {\frac {\pi k}{2m}}\right),\qquad k=1,2,\ldots ,m-1}
∑
r
=
0
m
−
1
ψ
(
2
r
+
1
2
m
)
⋅
sin
(
2
r
+
1
)
k
π
m
=
−
π
m
2
,
k
=
1
,
2
,
…
,
m
−
1
{\displaystyle \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 k=1,2,\ldots ,m-1}
∑
r
=
1
m
−
1
ψ
(
r
m
)
⋅
cot
π
r
m
=
−
π
(
m
−
1
)
(
m
−
2
)
6
{\displaystyle \sum _{r=1}^{m-1}\psi \left({\frac {r}{m}}\right)\cdot \cot {\frac {\pi r}{m}}=-{\frac {\pi (m-1)(m-2)}{6}}}
∑
r
=
1
m
−
1
ψ
(
r
m
)
⋅
r
m
=
−
γ
2
(
m
−
1
)
−
m
2
ln
m
−
π
2
∑
r
=
1
m
−
1
r
m
⋅
cot
π
r
m
{\displaystyle \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}{\frac {r}{m}}\cdot \cot {\frac {\pi r}{m}}}
∑
r
=
1
m
−
1
ψ
(
r
m
)
⋅
cos
(
2
l
+
1
)
π
r
m
=
−
π
m
∑
r
=
1
m
−
1
r
⋅
sin
2
π
r
m
cos
2
π
r
m
−
cos
(
2
l
+
1
)
π
m
,
l
∈
Z
{\displaystyle \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 l\in \mathbb {Z} }
∑
r
=
1
m
−
1
ψ
(
r
m
)
⋅
sin
(
2
l
+
1
)
π
r
m
=
−
(
γ
+
ln
2
m
)
cot
(
2
l
+
1
)
π
2
m
+
sin
(
2
l
+
1
)
π
m
∑
r
=
1
m
−
1
ln
sin
π
r
m
cos
2
π
r
m
−
cos
(
2
l
+
1
)
π
m
,
l
∈
Z
{\displaystyle \sum _{r=1}^{m-1}\psi \left({\frac {r}{m}}\right)\cdot \sin {\dfrac {(2l+1)\pi r}{m}}=-(\gamma +\ln 2m)\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 l\in \mathbb {Z} }
∑
r
=
1
m
−
1
ψ
2
(
r
m
)
=
(
m
−
1
)
γ
2
+
m
(
2
γ
+
ln
4
m
)
ln
m
−
m
(
m
−
1
)
ln
2
2
+
π
2
(
m
2
−
3
m
+
2
)
12
+
m
∑
l
=
1
m
−
1
ln
2
sin
π
l
m
{\displaystyle \sum _{r=1}^{m-1}\psi ^{2}\left({\frac {r}{m}}\right)=(m-1)\gamma ^{2}+m(2\gamma +\ln 4m)\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 Blagouchine (2014)[ 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
ψ
(
r
m
)
=
−
γ
−
ln
(
2
m
)
−
π
2
cot
(
r
π
m
)
+
2
∑
n
=
1
⌊
m
−
1
2
⌋
cos
(
2
π
n
r
m
)
ln
sin
(
π
n
m
)
{\displaystyle \psi \left({\frac {r}{m}}\right)=-\gamma -\ln(2m)-{\frac {\pi }{2}}\cot \left({\frac {r\pi }{m}}\right)+2\sum _{n=1}^{\left\lfloor {\frac {m-1}{2}}\right\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]
∑
n
=
1
x
1
n
{\displaystyle \sum _{n=1}^{x}{\frac {1}{n}}}
the digamma function for x , also a real number, can be approximated by
ψ
(
x
)
=
ln
(
x
)
−
1
2
x
−
1
12
x
2
+
1
120
x
4
−
1
252
x
6
+
1
240
x
8
−
5
660
x
10
+
691
32760
x
12
−
1
12
x
14
+
O
(
1
x
16
)
{\displaystyle \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
ψ
(
x
)
=
ln
(
x
)
−
1
2
x
+
∑
n
=
1
∞
ζ
(
1
−
2
n
)
x
2
n
=
ln
(
x
)
−
1
2
x
−
∑
n
=
1
∞
B
2
n
2
n
x
2
n
{\displaystyle \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}}{2nx^{2n}}}}
where Bk is the k th Bernoulli number and ζ is the Riemann zeta function . Although the infinite sum does not converge for any 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
ψ
(
x
+
1
)
=
1
x
+
ψ
(
x
)
{\displaystyle \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).
As x goes to infinity, ψ (x ) gets arbitrarily close to both ln(x − 1/2) and ln x . Going down from x + 1 to x , ψ decreases by 1 / x , ln(x − 1/2) decreases by ln (x + 1/2) / (x − 1/2) , which is more than 1 / x , and ln x decreases by ln (1 + 1 / x) , which is less than 1 / x . From this we see that for any positive x greater than 1/2 ,
ψ
(
x
)
∈
(
ln
(
x
−
1
2
)
,
ln
x
)
{\displaystyle \psi (x)\in \left(\ln \left(x-{\tfrac {1}{2}}\right),\ln x\right)}
or, for any positive x ,
exp
ψ
(
x
)
∈
(
x
−
1
2
,
x
)
.
{\displaystyle \exp \psi (x)\in \left(x-{\tfrac {1}{2}},x\right).}
The exponential exp ψ (x ) is approximately x − 1/2 for large x , but gets closer to x at small x , approaching 0 at x = 0 .
For x < 1 , we can calculate limits based on the fact that between 1 and 2, ψ (x ) ∈ [−γ , 1 − γ ] , so
ψ
(
x
)
∈
(
−
1
x
−
γ
,
1
−
1
x
−
γ
)
,
x
∈
(
0
,
1
)
{\displaystyle \psi (x)\in \left(-{\frac {1}{x}}-\gamma ,1-{\frac {1}{x}}-\gamma \right),\quad x\in (0,1)}
or
exp
ψ
(
x
)
∈
(
exp
(
−
1
x
−
γ
)
,
e
exp
(
−
1
x
−
γ
)
)
.
{\displaystyle \exp \psi (x)\in \left(\exp \left(-{\frac {1}{x}}-\gamma \right),e\exp \left(-{\frac {1}{x}}-\gamma \right)\right).}
From the above asymptotic series for ψ , one can derive an asymptotic series for exp(−ψ (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.
1
exp
ψ
(
x
)
=
1
x
+
1
2
⋅
x
2
+
5
4
⋅
3
!
⋅
x
3
+
3
2
⋅
4
!
⋅
x
4
+
47
48
⋅
5
!
⋅
x
5
−
5
16
⋅
6
!
⋅
x
6
+
⋯
{\displaystyle {\frac {1}{\exp \psi (x)}}={\frac {1}{x}}+{\frac {1}{2\cdot x^{2}}}+{\frac {5}{4\cdot 3!\cdot x^{3}}}+{\frac {3}{2\cdot 4!\cdot x^{4}}}+{\frac {47}{48\cdot 5!\cdot x^{5}}}-{\frac {5}{16\cdot 6!\cdot x^{6}}}+\cdots }
This can be considered a Taylor expansion of exp(−ψ (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
ψ
(
x
+
1
2
)
=
x
+
1
4
!
⋅
x
−
37
8
⋅
6
!
⋅
x
3
+
10313
72
⋅
8
!
⋅
x
5
−
5509121
384
⋅
10
!
⋅
x
7
+
O
(
1
x
9
)
for
x
>
1
{\displaystyle \exp \psi \left(x+{\tfrac {1}{2}}\right)=x+{\frac {1}{4!\cdot x}}-{\frac {37}{8\cdot 6!\cdot x^{3}}}+{\frac {10313}{72\cdot 8!\cdot x^{5}}}-{\frac {5509121}{384\cdot 10!\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:
ψ
(
1
)
=
−
γ
ψ
(
1
2
)
=
−
2
ln
2
−
γ
ψ
(
1
3
)
=
−
π
2
3
−
3
ln
3
2
−
γ
ψ
(
1
4
)
=
−
π
2
−
3
ln
2
−
γ
ψ
(
1
6
)
=
−
π
3
2
−
2
ln
2
−
3
ln
3
2
−
γ
ψ
(
1
8
)
=
−
π
2
−
4
ln
2
−
π
+
ln
(
2
+
2
)
−
ln
(
2
−
2
)
2
−
γ
.
{\displaystyle {\begin{aligned}\psi (1)&=-\gamma \\\psi \left({\tfrac {1}{2}}\right)&=-2\ln {2}-\gamma \\\psi \left({\tfrac {1}{3}}\right)&=-{\frac {\pi }{2{\sqrt {3}}}}-{\frac {3\ln {3}}{2}}-\gamma \\\psi \left({\tfrac {1}{4}}\right)&=-{\frac {\pi }{2}}-3\ln {2}-\gamma \\\psi \left({\tfrac {1}{6}}\right)&=-{\frac {\pi {\sqrt {3}}}{2}}-2\ln {2}-{\frac {3\ln {3}}{2}}-\gamma \\\psi \left({\tfrac {1}{8}}\right)&=-{\frac {\pi }{2}}-4\ln {2}-{\frac {\pi +\ln \left(2+{\sqrt {2}}\right)-\ln \left(2-{\sqrt {2}}\right)}{\sqrt {2}}}-\gamma .\end{aligned}}}
Moreover, by the series representation, it can easily be deduced that at the imaginary unit ,
Re
ψ
(
i
)
=
−
γ
−
∑
n
=
0
∞
n
−
1
n
3
+
n
2
+
n
+
1
,
Im
ψ
(
i
)
=
∑
n
=
0
∞
1
n
2
+
1
=
1
2
+
π
2
coth
π
.
{\displaystyle {\begin{aligned}\operatorname {Re} \psi (i)&=-\gamma -\sum _{n=0}^{\infty }{\frac {n-1}{n^{3}+n^{2}+n+1}},\\[8px]\operatorname {Im} \psi (i)&=\sum _{n=0}^{\infty }{\frac {1}{n^{2}+1}}\\&={\frac {1}{2}}+{\frac {\pi }{2}}\coth \pi .\end{aligned}}}
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 = 1.461632 144 968 ... . All others occur single between the poles on the negative axis:
x
1
=
−
0.504
083
008...
,
x
2
=
−
1.573
498
473...
,
x
3
=
−
2.610
720
868...
,
x
4
=
−
3.635
293
366...
,
⋮
{\displaystyle {\begin{aligned}x_{1}&=-0.504\,083\,008...,\\x_{2}&=-1.573\,498\,473...,\\x_{3}&=-2.610\,720\,868...,\\x_{4}&=-3.635\,293\,366...,\\&\qquad \vdots \end{aligned}}}
Already in 1881, Charles Hermite observed[citation needed ] that
x
n
=
−
n
+
1
ln
n
+
o
(
1
(
ln
n
)
2
)
{\displaystyle x_{n}=-n+{\frac {1}{\ln n}}+o\left({\frac {1}{(\ln n)^{2}}}\right)}
holds asymptotically. A better approximation of the location of the roots is given by
x
n
≈
−
n
+
1
π
arctan
(
π
ln
n
)
n
≥
2
{\displaystyle x_{n}\approx -n+{\frac {1}{\pi }}\arctan \left({\frac {\pi }{\ln n}}\right)\qquad n\geq 2}
and using a further term it becomes still better
x
n
≈
−
n
+
1
π
arctan
(
π
ln
n
+
1
8
n
)
n
≥
1
{\displaystyle x_{n}\approx -n+{\frac {1}{\pi }}\arctan \left({\frac {\pi }{\ln n+{\frac {1}{8n}}}}\right)\qquad n\geq 1}
which both spring off the reflection formula via
0
=
ψ
(
1
−
x
n
)
=
ψ
(
x
n
)
+
π
tan
π
x
n
{\displaystyle 0=\psi (1-x_{n})=\psi (x_{n})+{\frac {\pi }{\tan \pi x_{n}}}}
and substituting ψ (xn ) by its not convergent asymptotic expansion. The correct second term of this expansion is 1 / 2n , where the given one works good to approximate roots with small n .
Regarding the zeros, the following infinite sum identities were recently proved by István Mező [ 9]
∑
n
=
0
∞
1
x
n
2
=
γ
2
+
π
2
2
,
∑
n
=
0
∞
1
x
n
4
=
γ
4
+
π
4
9
+
2
3
γ
2
π
2
+
4
γ
ζ
(
3
)
.
{\displaystyle {\begin{aligned}\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}}+{\frac {2}{3}}\gamma ^{2}\pi ^{2}+4\gamma \zeta (3).\end{aligned}}}
Here γ is the Euler–Mascheroni constant .
Regularization
The digamma function appears in the regularization of divergent integrals
∫
0
∞
d
x
x
+
a
,
{\displaystyle \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
∑
n
=
0
∞
1
n
+
a
=
−
ψ
(
a
)
.
{\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n+a}}=-\psi (a).}
See also
References
^ a b
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.
^ Weisstein, Eric W. "Digamma function" . MathWorld .
^ R. Campbell. Les intégrales eulériennes et leurs applications , Dunod, Paris, 1966.
^ H.M. Srivastava and J. Choi. Series Associated with the Zeta and Related Functions , Kluwer Academic Publishers, the Netherlands, 2001.
^ Blagouchine, Iaroslav V. (2014). "A theorem for the closed-form evaluation of the first generalized Stieltjes constant at rational arguments and some related summations" . Journal of Number Theory . 148 . Elsevier: 537–592. arXiv :1401.3724 . doi :10.1016/j.jnt.2014.08.009 .
^ Bernardo, José M. (1976). "Algorithm AS 103 psi(digamma function) computation" (PDF) . Applied Statistics . 25 : 315–317.
^ Beal, Matthew J. (2003). Variational Algorithms for Approximate Bayesian Inference (PDF) (PhD thesis). The Gatsby Computational Neuroscience Unit, University College London. pp. 265–266.
^ If it converged to a function f (y ) then ln(f (y ) / y ) would have the same Maclaurin series as ln(1 / y ) − φ (1 / y ) . But this does not converge because the series given earlier for φ (x ) does not converge.
^ Mező, István (2014). "A note on the zeros of the Digamma function and the derivative of the log-Barnes function". arXiv :1409.2971v1 .
External links