# Mittag-Leffler polynomials

In mathematics, the Mittag-Leffler polynomials are the polynomials g$_n$(x) or M$_n$(x) studied by Mittag-Leffler (1891).
M$_n$(x) is a special case of the Meixner polynomial M$_n$(x;b,c) at b = 0, c = -1.

## Definition and examples
### Generating functions

The Mittag-Leffler polynomials are defined respectively by the generating functions
    $\displaystyle \sum _{n=0}^{\infty }g_{n}(x)t^{n}:={\frac {1}{2}}{\Bigl{\frac {1+t}{1-t}}{\Bigr )}^{x}$ and
    $\displaystyle \sum _{n=0}^{\infty }M_{n}(x){\frac {t^{n}}{n!}}:={\Bigl{\frac {1+t}{1-t}}{\Bigr )}^{x}=(1+t)^{x}(1-t)^{-x}=\exp(2x{\text{ artanh }}t).$
They also have the bivariate generating function
    $\displaystyle \sum _{n=1}^{\infty }\sum _{m=1}^{\infty }g_{n}(m)x^{m}y^{n}={\frac {xy}{(1-x)(1-x-y-xy)}}.$

### Examples

The first few polynomials are given in the following table. The coefficients of the numerators of the $g_{n}(x)$ can be found in the OEIS, though without any references, and the coefficients of the $M_{n}(x)$ are in the OEIS as well.
    ng$_n$(x)M$_n$(x) 0${\frac {1}{2}}$$1$ 1$x$$2x$ 2$x^{2}$$4x^{2}$ 3${\frac {1}{3}}(x+2x^{3})$$8x^{3}+4x$ 4${\frac {1}{3}}(2x^{2}+x^{4})$$16x^{4}+32x^{2}$ 5${\frac {1}{15}}(3x+10x^{3}+2x^{5})$$32x^{5}+160x^{3}+48x$ 6${\frac {1}{45}}(23x^{2}+20x^{4}+2x^{6})$$64x^{6}+640x^{4}+736x^{2}$ 7${\frac {1}{315}}(45x+196x^{3}+70x^{5}+4x^{7})$$128x^{7}+2240x^{5}+6272x^{3}+1440x$ 8${\frac {1}{315}}(132x^{2}+154x^{4}+28x^{6}+x^{8})$$256x^{8}+7168x^{6}+39424x^{4}+33792x^{2}$ 9${\frac {1}{2835}}(315x+1636x^{3}+798x^{5}+84x^{7}+2x^{9})$$512x^{9}+21504x^{7}+204288x^{5}+418816x^{3}+80640x$ 10${\frac {1}{14175}}(5067x^{2}+7180x^{4}+1806x^{6}+120x^{8}+2x^{10})$$1024x^{10}+61440x^{8}+924672x^{6}+3676160x^{4}+2594304x^{2}$

## Properties

The polynomials are related by $M_{n}(x)=2\cdot {n!}\,g_{n}(x)$ and we have $g_{n}(1)=1$ for $n\geqslant 1$. Also $g_{2k}({\frac {1}{2}})=g_{2k+1}({\frac {1}{2}})={\frac {1}{2}}{\frac {(2k-1)!!}{(2k)!!}}={\frac {1}{2}}\cdot {\frac {1\cdot 3\cdots (2k-1)}{2\cdot 4\cdots (2k)}}$.

### Explicit formulas

Explicit formulas are
    $g_{n}(x)=\sum _{k=1}^{n}2^{k-1}{\binom {n-1}{n-k}}{\binom {x}{k}}=\sum _{k=0}^{n-1}2^{k}{\binom {n-1}{k}}{\binom {x}{k+1}}$
    $g_{n}(x)=\sum _{k=0}^{n-1}{\binom {n-1}{k}}{\binom {k+x}{n}}$
    $g_{n}(m)={\frac {1}{2}}\sum _{k=0}^{m}{\binom {m}{k}}{\binom {n-1+m-k}{m-1}}={\frac {1}{2}}\sum _{k=0}^{\min(n,m)}{\frac {m}{n+m-k}}{\binom {n+m-k}{k,n-k,m-k}}$
(the last one immediately shows $ng_{n}(m)=mg_{m}(n)$, a kind of reflection formula), and
    $M_{n}(x)=(n-1)!\sum _{k=1}^{n}k2^{k}{\binom {n}{k}}{\binom {x}{k}}$, which can be also written as
    $M_{n}(x)=\sum _{k=1}^{n}2^{k}{\binom {n}{k}}(n-1)_{n-k}(x)_{k}$, where $(x)_{n}=n!{\binom {x}{n}}=x(x-1)\cdots (x-n+1)$ denotes the falling factorial.
In terms of the Gaussian hypergeometric function, we have
    $g_{n}(x)=x\!\cdot_{2}\!F_{1}(1-n,1-x;2;2).$

### Reflection formula

As stated above, for $m,n\in \mathbb {N}$, we have the reflection formula $ng_{n}(m)=mg_{m}(n)$.

### Recursion formulas

The polynomials $M_{n}(x)$ can be defined recursively by
    $M_{n}(x)=2xM_{n-1}(x)+(n-1)(n-2)M_{n-2}(x)$, starting with $M_{-1}(x)=0$ and $M_{0}(x)=1$.
Another recursion formula, which produces an odd one from the preceding even ones and vice versa, is
    $M_{n+1}(x)=2x\sum _{k=0}^{\lfloor n/2\rfloor }{\frac {n!}{(n-2k)!}}M_{n-2k}(x)$, again starting with $M_{0}(x)=1$.
As for the $g_{n}(x)$, we have several different recursion formulas:
    $\displaystyle (1)\quad g_{n}(x+1)-g_{n-1}(x+1)=g_{n}(x)+g_{n-1}(x)$
    $\displaystyle (2)\quad (n+1)g_{n+1}(x)-(n-1)g_{n-1}(x)=2xg_{n}(x)$
    $(3)\quad x{\Biglg_{n}(x+1)-g_{n}(x-1){\Bigr )}=2ng_{n}(x)$
    $(4)\quad g_{n+1}(m)=g_{n}(m)+2\sum _{k=1}^{m-1}g_{n}(k)=g_{n}(1)+g_{n}(2)+\cdots +g_{n}(m)+g_{n}(m-1)+\cdots +g_{n}(1)$
Concerning recursion formula (3), the polynomial $g_{n}(x)$ is the unique polynomial solution of the difference equation $x(f(x+1)-f(x-1))=2nf(x)$, normalized so that $f(1)=1$. Further note that (2) and (3) are dual to each other in the sense that for $x\in \mathbb {N}$, we can apply the reflection formula to one of the identities and then swap $x$ and $n$ to obtain the other one. (As the $g_{n}(x)$ are polynomials, the validity extends from natural to all real values of $x$.)

### Initial values

The table of the initial values of $g_{n}(m)$ (these values are also called the "figurate numbers for the n-dimensional cross polytopes" in the OEIS) may illustrate the recursion formula (1), which can be taken to mean that each entry is the sum of the three neighboring entries: to its left, above and above left, e.g. $g_{5}(3)=51=33+8+10$. It also illustrates the reflection formula $ng_{n}(m)=mg_{m}(n)$ with respect to the main diagonal, e.g. $3\cdot 44=4\cdot 33$.
    nm 12345678910 1 1111111111 2 24681012141618 3 391933517399129 4 4164496180304476 5 52585225501985 6 6361464561182 7 749231833 8 864344 9 981 10 10

### Orthogonality relations

For $m,n\in \mathbb {N}$ the following orthogonality relation holds:
    $\int _{-\infty }^{\infty }{\frac {g_{n}(-iy)g_{m}(iy)}{y\sinh \pi y}}dy={\frac {1}{2n}}\delta _{mn}.$
(Note that this is not a complex integral. As each $g_{n}$ is an even or an odd polynomial, the imaginary arguments just produce alternating signs for their coefficients. Moreover, if $m$ and $n$ have different parity, the integral vanishes trivially.)

### Binomial identity

Being a Sheffer sequence of binomial type, the Mittag-Leffler polynomials $M_{n}(x)$ also satisfy the binomial identity
    $M_{n}(x+y)=\sum _{k=0}^{n}{\binom {n}{k}}M_{k}(x)M_{n-k}(y)$.

### Integral representations

Based on the representation as a hypergeometric function, there are several ways of representing $g_{n}(z)$ for $|z|<1$ directly as integrals, some of them being even valid for complex $z$, e.g.
    $(26)\qquad g_{n}(z)={\frac {\sin(\pi z)}{2\pi }}\int _{-1}^{1}t^{n-1}{\Bigl{\frac {1+t}{1-t}}{\Bigr )}^{z}dt$
    $(27)\qquad g_{n}(z)={\frac {\sin(\pi z)}{2\pi }}\int _{-\infty }^{\infty }e^{uz}{\frac {(\tanh {\frac {u}{2}})^{n}}{\sinh u}}du$
    $(32)\qquad g_{n}(z)={\frac {1}{\pi }}\int _{0}^{\pi }\cot ^{z}({\frac {u}{2}})\cos({\frac {\pi z}{2}})\cos(nu)du$
    $(33)\qquad g_{n}(z)={\frac {1}{\pi }}\int _{0}^{\pi }\cot ^{z}({\frac {u}{2}})\sin({\frac {\pi z}{2}})\sin(nu)du$
    $(34)\qquad g_{n}(z)={\frac {1}{2\pi }}\int _{0}^{2\pi }(1+e^{it})^{z}(2+e^{it})^{n-1}e^{-int}dt$.

### Closed forms of integral families

There are several families of integrals with closed-form expressions in terms of zeta values where the coefficients of the Mittag-Leffler polynomials occur as coefficients. All those integrals can be written in a form containing either a factor $\tan ^{\pm n}$ or $\tanh ^{\pm n}$, and the degree of the Mittag-Leffler polynomial varies with $n$. One way to work out those integrals is to obtain for them the corresponding recursion formulas as for the Mittag-Leffler polynomials using integration by parts.
1. For instance, define for $n\geqslant m\geqslant 2$
    $I(n,m):=\int _{0}^{1}{\dfrac {{\text{artanh}}^{n}x}{x^{m}}}dx=\int _{0}^{1}\log ^{n/2}{\Bigl{\dfrac {1+x}{1-x}}{\Bigr )}{\dfrac {dx}{x^{m}}}=\int _{0}^{\infty }z^{n}{\dfrac {\coth ^{m-2}z}{\sinh ^{2}z}}dz.$
These integrals have the closed form
    $(1)\quad I(n,m)={\frac {n!}{2^{n-1}}}\zeta ^{n+1}~g_{m-1}({\frac {1}{\zeta }})$
in umbral notation, meaning that after expanding the polynomial in $\zeta$, each power $\zeta ^{k}$ has to be replaced by the zeta value $\zeta (k)$. E.g. from $g_{6}(x)={\frac {1}{45}}(23x^{2}+20x^{4}+2x^{6})\$ we get $\ I(n,7)={\frac {n!}{2^{n-1}}}{\frac {23~\zeta (n-1)+20~\zeta (n-3)+2~\zeta (n-5)}{45}}\$ for $n\geqslant 7$.
2. Likewise take for $n\geqslant m\geqslant 2$
    $J(n,m):=\int _{1}^{\infty }{\dfrac {{\text{arcoth}}^{n}x}{x^{m}}}dx=\int _{1}^{\infty }\log ^{n/2}{\Bigl{\dfrac {x+1}{x-1}}{\Bigr )}{\dfrac {dx}{x^{m}}}=\int _{0}^{\infty }z^{n}{\dfrac {\tanh ^{m-2}z}{\cosh ^{2}z}}dz.$
In umbral notation, where after expanding, $\eta ^{k}$ has to be replaced by the Dirichlet eta function $\eta (k):=\left(1-2^{1-k}\right)\zeta (k)$, those have the closed form
    $(2)\quad J(n,m)={\frac {n!}{2^{n-1}}}\eta ^{n+1}~g_{m-1}({\frac {1}{\eta }})$.
3. The following holds for $n\geqslant m$ with the same umbral notation for $\zeta$ and $\eta$, and completing by continuity $\eta (1):=\ln 2$.
    $(3)\quad \int \limits _{0}^{\pi /2}{\frac {x^{n}}{\tan ^{m}x}}dx=\cos {\Bigl{\frac {m}{2}}\pi {\Bigr )}{\frac {(\pi /2)^{n+1}}{n+1}}+\cos {\Bigl{\frac {m-n-1}{2}}\pi {\Bigr )}{\frac {n!~m}{2^{n}}}\zeta ^{n+2}g_{m}({\frac {1}{\zeta }})+\sum \limits _{v=0}^{n}\cos {\Bigl{\frac {m-v-1}{2}}\pi {\Bigr )}{\frac {n!~m~\pi ^{n-v}}{(n-v)!~2^{n}}}\eta ^{n+2}g_{m}({\frac {1}{\eta }}).$
Note that for $n\geqslant m\geqslant 2$, this also yields a closed form for the integrals
    $\int \limits _{0}^{\infty }{\frac {\arctan ^{n}x}{x^{m}}}dx=\int \limits _{0}^{\pi /2}{\frac {x^{n}}{\tan ^{m}x}}dx+\int \limits _{0}^{\pi /2}{\frac {x^{n}}{\tan ^{m-2}x}}dx.$
4. For $n\geqslant m\geqslant 2$, define $\quad K(n,m):=\int \limits _{0}^{\infty }{\dfrac {\tanh ^{n}(x)}{x^{m}}}dx$.
If $n+m$ is even and we define $h_{k}:=(-1)^{\frac {k-1}{2}}{\frac {(k-1)!(2^{k}-1)\zeta (k)}{2^{k-1}\pi ^{k-1}}}$, we have in umbral notation, i.e. replacing $h^{k}$ by $h_{k}$,
    $(4)\quad K(n,m):=\int \limits _{0}^{\infty }{\dfrac {\tanh ^{n}(x)}{x^{m}}}dx={\dfrac {n\cdot 2^{m-1}}{(m-1)!}}(-h)^{m-1}g_{n}(h).$
Note that only odd zeta values (odd $k$) occur here (unless the denominators are cast as even zeta values), e.g.
    $K(5,3)=-{\frac {2}{3}}(3h_{3}+10h_{5}+2h_{7})=-7{\frac {\zeta (3)}{\pi ^{2}}}+310{\frac {\zeta (5)}{\pi ^{4}}}-1905{\frac {\zeta (7)}{\pi ^{6}}},$
    $K(6,2)={\frac {4}{15}}(23h_{3}+20h_{5}+2h_{7}),\quad K(6,4)={\frac {4}{45}}(23h_{5}+20h_{7}+2h_{9}).$
5. If $n+m$ is odd, the same integral is much more involved to evaluate, including the initial one $\int \limits _{0}^{\infty }{\dfrac {\tanh ^{3}(x)}{x^{2}}}dx$. Yet it turns out that the pattern subsists if we define $s_{k}:=\eta '(-k)=2^{k+1}\zeta (-k)\ln 2-(2^{k+1}-1)\zeta '(-k)$, equivalently $s_{k}={\frac {\zeta (-k)}{\zeta '(-k)}}\eta (-k)+\zeta (-k)\eta (1)-\eta (-k)\eta (1)$. Then $K(n,m)$ has the following closed form in umbral notation, replacing $s^{k}$ by $s_{k}$:
    $(5)\quad K(n,m)=\int \limits _{0}^{\infty }{\dfrac {\tanh ^{n}(x)}{x^{m}}}dx={\frac {n\cdot 2^{m}}{(m-1)!}}(-s)^{m-2}g_{n}(s)$, e.g.
    $K(5,4)={\frac {8}{9}}(3s_{3}+10s_{5}+2s_{7}),\quad K(6,3)=-{\frac {8}{15}}(23s_{3}+20s_{5}+2s_{7}),\quad K(6,5)=-{\frac {8}{45}}(23s_{5}+20s_{7}+2s_{9}).$
Note that by virtue of the logarithmic derivative ${\frac {\zeta '}{\zeta }}(s)+{\frac {\zeta '}{\zeta }}(1-s)=\log \pi -{\frac {1}{2}}{\frac {\Gamma '}{\Gamma }}\left({\frac {s}{2}}\right)-{\frac {1}{2}}{\frac {\Gamma '}{\Gamma }}\left({\frac {1-s}{2}}\right)$ of Riemann's functional equation, taken after applying Euler's reflection formula, these expressions in terms of the $s_{k}$ can be written in terms of ${\frac {\zeta '(2j)}{\zeta (2j)}}$, e.g.
    $K(5,4)={\frac {8}{9}}(3s_{3}+10s_{5}+2s_{7})={\frac {1}{9}}\left\{{\frac {1643}{420}}-{\frac {16}{315}}\ln 2+3{\frac {\zeta '(4)}{\zeta (4)}}-20{\frac {\zeta '(6)}{\zeta (6)}}+17{\frac {\zeta '(8)}{\zeta (8)}}\right\}.$
6. For $n<m$, the same integral $K(n,m)$ diverges because the integrand behaves like $x^{n-m}$ for $x\searrow 0$. But the difference of two such integrals with corresponding degree differences is well-defined and exhibits very similar patterns, e.g.
    $(6)\quad K(n-1,n)-K(n,n+1)=\int \limits _{0}^{\infty }\left({\dfrac {\tanh ^{n-1}(x)}{x^{n}}}-{\dfrac {\tanh ^{n}(x)}{x^{n+1}}}\right)dx=-{\frac {1}{n}}+{\frac {(n+1)\cdot 2^{n}}{(n-1)!}}s^{n-2}g_{n}(s)$.