Mittag-Leffler polynomials

In mathematics, the Mittag-Leffler polynomials are the polynomials gn(x) or Mn(x) studied by.

Mn(x) is a special case of the Meixner polynomial Mn(x;b,c) at b = 0, c = -1.

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
 * $$ \displaystyle \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} g_n(m)x^my^n =\frac{xy}{(1-x)(1-x-y-xy)}. $$

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.
 * {| class="wikitable"

!n !! gn(x) !! Mn(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 || $${\frac{1}{15}} (3x+10x^3+2x^5)$$ || $$32 x^5 + 160 x^3 + 48 x$$
 * 6 || $${\frac{1}{45}} (23x^2+20x^4+2x^6)$$ || $$64 x^6 + 640 x^4 + 736 x^2$$
 * 7 || $${\frac{1}{315}} (45 x + 196 x^3 + 70 x^5 + 4 x^7)$$ || $$128 x^7 + 2240 x^5 + 6272 x^3 + 1440 x$$
 * 8 || $${\frac{1}{315}} (132 x^2 + 154 x^4 + 28 x^6 + x^8)$$ || $$256 x^8 + 7168 x^6 + 39424 x^4 + 33792 x^2$$
 * 9 || $${\frac{1}{2835}} (315 x + 1636 x^3 + 798 x^5 + 84 x^7 + 2 x^9)$$ || $$512 x^9 + 21504 x^7 + 204288 x^5 + 418816 x^3 + 80640 x$$
 * 10 || $${\frac{1}{14175}} (5067 x^2 + 7180 x^4 + 1806 x^6 + 120 x^8 + 2 x^{10})$$ || $$1024 x^{10} + 61440 x^8 + 924672 x^6 + 3676160 x^4 + 2594304 x^2$$
 * }
 * 6 || $${\frac{1}{45}} (23x^2+20x^4+2x^6)$$ || $$64 x^6 + 640 x^4 + 736 x^2$$
 * 7 || $${\frac{1}{315}} (45 x + 196 x^3 + 70 x^5 + 4 x^7)$$ || $$128 x^7 + 2240 x^5 + 6272 x^3 + 1440 x$$
 * 8 || $${\frac{1}{315}} (132 x^2 + 154 x^4 + 28 x^6 + x^8)$$ || $$256 x^8 + 7168 x^6 + 39424 x^4 + 33792 x^2$$
 * 9 || $${\frac{1}{2835}} (315 x + 1636 x^3 + 798 x^5 + 84 x^7 + 2 x^9)$$ || $$512 x^9 + 21504 x^7 + 204288 x^5 + 418816 x^3 + 80640 x$$
 * 10 || $${\frac{1}{14175}} (5067 x^2 + 7180 x^4 + 1806 x^6 + 120 x^8 + 2 x^{10})$$ || $$1024 x^{10} + 61440 x^8 + 924672 x^6 + 3676160 x^4 + 2594304 x^2$$
 * }
 * 9 || $${\frac{1}{2835}} (315 x + 1636 x^3 + 798 x^5 + 84 x^7 + 2 x^9)$$ || $$512 x^9 + 21504 x^7 + 204288 x^5 + 418816 x^3 + 80640 x$$
 * 10 || $${\frac{1}{14175}} (5067 x^2 + 7180 x^4 + 1806 x^6 + 120 x^8 + 2 x^{10})$$ || $$1024 x^{10} + 61440 x^8 + 924672 x^6 + 3676160 x^4 + 2594304 x^2$$
 * }
 * 10 || $${\frac{1}{14175}} (5067 x^2 + 7180 x^4 + 1806 x^6 + 120 x^8 + 2 x^{10})$$ || $$1024 x^{10} + 61440 x^8 + 924672 x^6 + 3676160 x^4 + 2594304 x^2$$
 * }

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 $$g_{2k}(\frac12)=g_{2k+1}(\frac12)=\frac12\frac{(2k-1)!!}{(2k)!!}=\frac12\cdot \frac{1\cdot 3\cdots (2k-1) }{2\cdot 4\cdots (2k)} $$.

Explicit formulas
Explicit formulas are
 * $$ g_n(x) = \sum_ {k = 1}^{n} 2^{k-1}\binom{n-1}{n-k}\binom xk = \sum_ {k = 0}^{n-1} 2^{k}\binom{n-1}{k}\binom x{k+1} $$
 * $$ g_n(x) = \sum_{k = 0}^{n-1} \binom{n-1}k\binom{k+x}n $$
 * $$ g_n(m) = \frac12\sum_{k = 0}^m \binom mk\binom{n-1+m-k}{m-1}=\frac12\sum_{k = 0}^{\min(n,m)} \frac m{n+m-k}\binom{n+m-k}{k,n-k,m-k}$$

(the last one immediately shows $$ng_n(m)=mg_m(n) $$, a kind of reflection formula), and
 * $$ M_n(x)=(n-1)!\sum_ {k = 1}^{n}k2^k\binom nk \binom xk $$, which can be also written as
 * $$ M_n(x)=\sum_ {k = 1}^{n}2^k\binom nk(n-1)_{n-k}(x)_k$$, where $$(x)_n = n!\binom xn = x(x-1)\cdots(x-n+1)$$ 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:


 * $$ \displaystyle (1)\quad g_n(x + 1) - g_{n-1}(x + 1)= g_n(x) + g_{n-1}(x) $$


 * $$ \displaystyle (2)\quad (n + 1)g_{n+1}(x) - (n - 1)g_{n-1}(x) = 2xg_n(x) $$


 * $$ (3)\quad x\Bigl(g_n(x+1)-g_n(x-1)\Bigr) = 2ng_n(x)$$


 * $$ (4)\quad g_{n+1}(m)= g_{n}(m)+2\sum_{k=1}^{m-1}g_{n}(k)=g_{n}(1)+g_{n}(2)+\cdots+g_{n}(m)+g_{n}(m-1) +\cdots+g_{n}(1) $$

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. $$3\cdot44=4\cdot33 $$.
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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
 * $$ M_n(x+y)=\sum_{k=0}^n\binom nk M_k(x)M_{n-k}(y)$$.

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.


 * $$(26)\qquad g_n(z) = \frac{\sin(\pi z)}{2\pi}\int _{-1}^1 t^{n-1} \Bigl(\frac{1+t}{1-t}\Bigr)^z dt$$


 * $$(27)\qquad g_n(z) = \frac{\sin(\pi z)}{2\pi} \int_{-\infty}^{\infty} e^{uz}\frac{(\tanh \frac u2)^n}{\sinh u} du$$


 * $$(32)\qquad g_n(z) = \frac1\pi\int _0^\pi \cot^z (\frac u2) \cos (\frac{\pi z}2) \cos (nu)du$$


 * $$(33)\qquad g_n(z) = \frac1\pi\int _0^\pi \cot^z (\frac u2) \sin (\frac{\pi z}2) \sin (nu)du$$


 * $$(34)\qquad g_n(z) = \frac1{2\pi}\int _0^{2\pi} (1+e^{it})^z (2+e^{it})^{n-1} e^{-int}dt$$.

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}^nx}{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^2z} dz.$$ These integrals have the closed form
 * $$(1)\quad I(n,m)=\frac{n!}{2^{n-1}}\zeta^{n+1}~g_ {m-1}(\frac1{\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}^nx}{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^2z} 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
 * $$ (2)\quad J(n,m)=\frac{n!}{2^{n-1}} \eta^{n+1}~g_ {m-1}(\frac1{\eta} )$$.

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(\frac1{\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(\frac1{\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 $$s_k:=\eta'(-k)=2^{k+1}\zeta(-k)\ln2-(2^{k+1}-1)\zeta'(-k)$$, 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 $$s_k$$:
 * $$ (5)\quad K(n,m)=\int\limits_0^\infty\dfrac{\tanh^n(x)}{x^m}dx=\frac{n\cdot2^{m}}{(m-1)!}(-s)^{m-2}g_n(s)$$, 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.
 * $$K(5,4)=\frac{8}{9}(3s_3+10s_5+2s_7)=\frac 19\left\{ \frac{1643}{420}-\frac{16 }{315}\ln2+3\frac{\zeta'(4) }{\zeta(4) }-20\frac{\zeta'(6) }{\zeta(6) }+17\frac{\zeta'(8) }{\zeta(8) }\right\}.$$

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.
 * $$ (6)\quad K(n-1,n)-K(n,n+1)=\int\limits_0^\infty\left(\dfrac{\tanh^{n-1}(x)}{x^{n}}-\dfrac{\tanh^{n}(x)}{x^{n+1}}\right)dx= -\frac 1n + \frac{ (n+1)\cdot2^{n}}{(n-1)!}s^{n-2}g_n(s) $$.