# Polygamma function

In mathematics, the polygamma function of order m is a meromorphic function on the complex numbers defined as the (m + 1)th derivative of the logarithm of the gamma function:

$\psi ^{(m)}(z):={\frac {d^{m}}{dz^{m}}}\psi (z)={\frac {d^{m+1}}{dz^{m+1}}}\ln \Gamma (z).$ Thus

$\psi ^{(0)}(z)=\psi (z)={\frac {\Gamma '(z)}{\Gamma (z)}}$ holds where ψ(z) is the digamma function and Γ(z) is the gamma function. They are holomorphic on \ −0. At all the nonpositive integers these polygamma functions have a pole of order m + 1. The function ψ(1)(z) is sometimes called the trigamma function.

## Integral representation

When m > 0 and Re z > 0, the polygamma function equals

{\begin{aligned}\psi ^{(m)}(z)&=(-1)^{m+1}\int _{0}^{\infty }{\frac {t^{m}e^{-zt}}{1-e^{-t}}}\,dt\\&=-\int _{0}^{1}{\frac {t^{z-1}}{1-t}}(\ln t)^{m}\,dt.\end{aligned}}

This expresses the polygamma function as the Laplace transform of $(-1)^{m+1}t^{m}/(1-e^{-t})$ . It follows from Bernstein's theorem on monotone functions that, for m > 0 and x real and non-negative, $(-1)^{m+1}\psi ^{(m)}(x)$  is a completely monotone function.

Setting m = 0 in the above formula does not give an integral representation of the digamma function. The digamma function has an integral representation, due to Gauss, which is similar to the m = 0 case above but which has an extra term $e^{-t}/t$ .

## Recurrence relation

It satisfies the recurrence relation

$\psi ^{(m)}(z+1)=\psi ^{(m)}(z)+{\frac {(-1)^{m}\,m!}{z^{m+1}}}$

which – considered for positive integer argument – leads to a presentation of the sum of reciprocals of the powers of the natural numbers:

${\frac {\psi ^{(m)}(n)}{(-1)^{m+1}\,m!}}=\zeta (1+m)-\sum _{k=1}^{n-1}{\frac {1}{k^{m+1}}}=\sum _{k=n}^{\infty }{\frac {1}{k^{m+1}}}\qquad m\geq 1$

and

$\psi ^{(0)}(n)=-\gamma \ +\sum _{k=1}^{n-1}{\frac {1}{k}}$

for all n. Like the log-gamma function, the polygamma functions can be generalized from the domain uniquely to positive real numbers only due to their recurrence relation and one given function-value, say ψ(m)(1), except in the case m = 0 where the additional condition of strict monotonicity on + is still needed. This is a trivial consequence of the Bohr–Mollerup theorem for the gamma function where strictly logarithmic convexity on + is demanded additionally. The case m = 0 must be treated differently because ψ(0) is not normalizable at infinity (the sum of the reciprocals doesn't converge).

## Reflection relation

$(-1)^{m}\psi ^{(m)}(1-z)-\psi ^{(m)}(z)=\pi {\frac {d^{m}}{dz^{m}}}\cot {(\pi z)}=\pi ^{m+1}{\frac {P_{m}(\cos(\pi z))}{\sin ^{m+1}(\pi z)}}$

where Pm is alternately an odd or even polynomial of degree |m − 1| with integer coefficients and leading coefficient (−1)m⌈2m − 1. They obey the recursion equation

{\begin{aligned}P_{0}(x)&=x\\P_{m+1}(x)&=-\left((m+1)xP_{m}(x)+\left(1-x^{2}\right)P'_{m}(x)\right).\end{aligned}}

## Multiplication theorem

The multiplication theorem gives

$k^{m+1}\psi ^{(m)}(kz)=\sum _{n=0}^{k-1}\psi ^{(m)}\left(z+{\frac {n}{k}}\right)\qquad m\geq 1$

and

$k\psi ^{(0)}(kz)=k\log(k)+\sum _{n=0}^{k-1}\psi ^{(0)}\left(z+{\frac {n}{k}}\right)$

for the digamma function.

## Series representation

The polygamma function has the series representation

$\psi ^{(m)}(z)=(-1)^{m+1}\,m!\sum _{k=0}^{\infty }{\frac {1}{(z+k)^{m+1}}}$

which holds for m > 0 and any complex z not equal to a negative integer. This representation can be written more compactly in terms of the Hurwitz zeta function as

$\psi ^{(m)}(z)=(-1)^{m+1}\,m!\,\zeta (m+1,z).$

Alternately, the Hurwitz zeta can be understood to generalize the polygamma to arbitrary, non-integer order.

One more series may be permitted for the polygamma functions. As given by Schlömilch,

${\frac {1}{\Gamma (z)}}=ze^{\gamma z}\prod _{n=1}^{\infty }\left(1+{\frac {z}{n}}\right)e^{-{\frac {z}{n}}}.$

This is a result of the Weierstrass factorization theorem. Thus, the gamma function may now be defined as:

$\Gamma (z)={\frac {e^{-\gamma z}}{z}}\prod _{n=1}^{\infty }\left(1+{\frac {z}{n}}\right)^{-1}e^{\frac {z}{n}}.$

Now, the natural logarithm of the gamma function is easily representable:

$\ln \Gamma (z)=-\gamma z-\ln(z)+\sum _{n=1}^{\infty }\left({\frac {z}{n}}-\ln \left(1+{\frac {z}{n}}\right)\right).$

Finally, we arrive at a summation representation for the polygamma function:

$\psi ^{(n)}(z)={\frac {d^{n+1}}{dz^{n+1}}}\ln \Gamma (z)=-\gamma \delta _{n0}-{\frac {(-1)^{n}n!}{z^{n+1}}}+\sum _{k=1}^{\infty }\left({\frac {1}{k}}\delta _{n0}-{\frac {(-1)^{n}n!}{(k+z)^{n+1}}}\right)$

Where δn0 is the Kronecker delta.

Also the Lerch transcendent

$\Phi (-1,m+1,z)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(z+k)^{m+1}}}$

can be denoted in terms of polygamma function

$\Phi (-1,m+1,z)={\frac {1}{(-2)^{m+1}m!}}\left(\psi ^{(m)}\left({\frac {z}{2}}\right)-\psi ^{(m)}\left({\frac {z+1}{2}}\right)\right)$

## Taylor series

The Taylor series at z = 1 is

$\psi ^{(m)}(z+1)=\sum _{k=0}^{\infty }(-1)^{m+k+1}{\frac {(m+k)!}{k!}}\zeta (m+k+1)z^{k}\qquad m\geq 1$

and

$\psi ^{(0)}(z+1)=-\gamma +\sum _{k=1}^{\infty }(-1)^{k+1}\zeta (k+1)z^{k}$

which converges for |z| < 1. Here, ζ is the Riemann zeta function. This series is easily derived from the corresponding Taylor series for the Hurwitz zeta function. This series may be used to derive a number of rational zeta series.

## Asymptotic expansion

These non-converging series can be used to get quickly an approximation value with a certain numeric at-least-precision for large arguments:

$\psi ^{(m)}(z)\sim (-1)^{m+1}\sum _{k=0}^{\infty }{\frac {(k+m-1)!}{k!}}{\frac {B_{k}}{z^{k+m}}}\qquad m\geq 1$

and

$\psi ^{(0)}(z)\sim \ln(z)-\sum _{k=1}^{\infty }{\frac {B_{k}}{kz^{k}}}$

where we have chosen B1 = 1/2, i.e. the Bernoulli numbers of the second kind.

## Inequalities

The hyperbolic cotangent satisfies the inequality

${\frac {t}{2}}\operatorname {coth} {\frac {t}{2}}\geq 1,$

and this implies that the function

${\frac {t^{m}}{1-e^{-t}}}-\left(t^{m-1}+{\frac {t^{m}}{2}}\right)$

is non-negative for all $m\geq 1$  and $t\geq 0$ . It follows that the Laplace transform of this function is completely monotone. By the integral representation above, we conclude that

$(-1)^{m+1}\psi ^{(m)}(x)-\left({\frac {(m-1)!}{x^{m}}}+{\frac {m!}{2x^{m+1}}}\right)$

is completely monotone. The convexity inequality $e^{t}\geq 1+t$  implies that

$\left(t^{m-1}+t^{m}\right)-{\frac {t^{m}}{1-e^{-t}}}$

is non-negative for all $m\geq 1$  and $t\geq 0$ , so a similar Laplace transformation argument yields the complete monotonicity of

$\left({\frac {(m-1)!}{x^{m}}}+{\frac {m!}{x^{m+1}}}\right)-(-1)^{m+1}\psi ^{(m)}(x).$

Therefore, for all m ≥ 1 and x > 0,

${\frac {(m-1)!}{x^{m}}}+{\frac {m!}{2x^{m+1}}}\leq (-1)^{m+1}\psi ^{(m)}(x)\leq {\frac {(m-1)!}{x^{m}}}+{\frac {m!}{x^{m+1}}}.$