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The problem at hand is finding a partial fraction expansion of : $$(\gamma+\psi(1-z))^{n}$$ $$(\gamma\; \text{being the Euler–Mascheroni constant , and}\;\psi(\cdot )\;\text{being the digamma function })$$ For $n=2$, i used the Mittag-Leffler theorem, and with a bit of induction, i obtained: $$\left(\gamma+\psi(1-z) \right )^{2}=-\frac{\pi^{2}}{6}+\psi^{(1)}(1-z)+2\sum_{n=2}^{\infty}\frac{(\psi(n)+\gamma)z}{n(z-n)}$$ For $n=3$: $$\left(\gamma+\psi(1-z) \right )^{3}=-2\zeta(3)+\frac{\psi^{(2)}(1-z)}{2}+3\sum_{n=2}^{\infty}\frac{\psi(n)+\gamma}{(z-n)^{2}}$$ $$+3\sum_{n=1}^{\infty}\left(\psi^{(1)}(n)+(\psi(n)+\gamma)^{2}-\frac{\pi^{2}}{3} \right )\frac{z}{n(z-n)}$$ Where $\psi^{(k)}(\cdot)$ is the polygamma function of kth order My question:

1)For higher powers, what is the general form of $(\gamma+\psi(1-z))^{n}$ ?

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Just a quick note: You are referring to Mittag-Lefler summation, not the Mittag-Lefler Theorem on meromorphic functions. – pre-kidney Jul 14 '13 at 23:31
no, i'm referring to Mittag-Leffler theorem on partial fraction expansions of meromorphic functions of prescribed poles – Mohammad Al Jamal Jul 14 '13 at 23:48
up vote 1 down vote accepted

Using Cauchy's integral formula, and the fact that : $$\psi(1)+\gamma=0$$ we have: $$\left(\psi(1-s)+\gamma \right )^{n}=\frac{1}{2\pi i}\oint_{c}\left(\psi(1-z)+\gamma \right )^{n}\left[\frac{1}{z-s}-\frac{1}{z} \right ]$$ Where $\Re(s)<1$, and the contour $c$ encloses $s$, but none of the positive integers. We may choose the closed contour such that it consists of the segments: $I_{1}=[a-iT,a+iT]$,$I_{2}=[a+iT,-c+iT]$,$I_{3}=[-c+iT,-c-iT]$, $I_{4}=[-c-iT,a-iT]$. Furthermore, we may choose: $T\rightarrow \infty$,$\;\;c\rightarrow \infty$. In these limits, the integral is vanishingly small along the path $I_{2}\cup I_{2}\cup I_{3}$ . Thus, we are allowed to write the integral as: $$\left(\psi(1-s)+\gamma \right )^{n}=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\left(\psi(1-z)+\gamma \right )^{n}\left[\frac{1}{z-s}-\frac{1}{z} \right ]$$ Now, instead of closing the contour to the left, we close it to the right, picking up the residues at positive integers: $$\left(\psi(1-s)+\gamma \right )^{n}=-\sum_{k=1}^{\infty} \text{Res}_{z=k}\left[\left(\psi(1-z)+\gamma \right )^{n}\left[\frac{1}{z-s}-\frac{1}{z} \right ]\right]$$ $$=-\frac{1}{(n-1)!}\sum_{k=1}^{\infty}\lim_{z\rightarrow k}\frac{d^{n-1}}{dz^{n-1}}\left[(z-k)^{n}\left(\psi(1-z)+\gamma \right )^{n}\left[\frac{1}{z-s}-\frac{1}{z} \right ]\right]$$ The digamma function may be written as a Laurent expansion around each of its singularities: $$\gamma+\psi(1-z)=\frac{1}{z-k}+H_{k-1}+\sum_{m=1}^{\infty}\frac{a_{m,k}}{m!}(z-k)^{m}\;\;\;\;\left(\left | z-k \right |<1\right)$$$H_{k}$ being the kth harmonic number. The numbers $a_{m,k}$ are given by: $$a_{m,k}=\lim_{z\rightarrow k}\left[(-1)^{m}\psi^{(m)}(1-z)-\frac{(-1)^{m}m!}{(z-k)^{m+1}}\right]$$ $$=\frac{(-1)^{m+1}m!}{k^{m+1}}+(-1)^{m}\left(\psi^{(m)}(1)+m!H_{k}^{(m+1)}\right)$$ Where $H_{k}^{(m)}$ is the generalized harmonic number of order $k$ in power $m$. Using this representation, we have: $$(z-k)^{n}\left(\psi(1-z)+\gamma \right )^{n}=\left(1+H_{k-1}(z-k)+\sum_{m=2}^{\infty}\frac{ma_{m-1,k}}{m!}(z-k)^{m} \right )^{n}$$ $$=n!\left(\sum_{m=n}^{\infty}B_{m,n}(k)\frac{(z-k)^{(m-n)}}{m!} \right )$$ $B_{m,n}(k)$ being the incomplete Bell polynomials in 1, $H_{k-1}$, and the numbers $ma_{m-1,k}$. Thus: $$\left(\psi(1-s)+\gamma \right )^{n}=-n\sum_{k=1}^{\infty}\lim_{z\rightarrow k}\frac{d^{n-1}}{dz^{n-1}}\left[\left(\sum_{m=n}^{\infty}B_{m,n}(k)\frac{(z-k)^{(m-n)}}{m!} \right )\left(\frac{1}{z-s}-\frac{1}{z}\right)\right]$$ Now we compute the derivatives: $$\frac{d^{n-1}}{dz^{n-1}}\left[\left(\sum_{m=n}^{\infty}B_{m,n}(k)\frac{(z-k)^{(m-n)}}{m!} \right )\left(\frac{1}{z-s}-\frac{1}{z}\right)\right]_{z\rightarrow k}$$

$$=\sum_{i=0}^{n-1}\binom{n-1}{i}\frac{d^{i}}{dz^{i}}\left(\sum_{m=n}^{\infty}B_{m,n}(k)\frac{(z-k)^{(m-n)}}{m!} \right )\frac{d^{n-i-1}}{dz^{n-i-1}}\left(\frac{1}{z-s}-\frac{1}{z}\right)\left.\begin{matrix} \\ \\ \end{matrix}\right|_{z\rightarrow k}$$ $$=\sum_{i=0}^{n-1}\frac{(n-1)!}{(n+i)!}(-1)^{n-i-1} B_{n+i,n}(k)\left(\frac{1}{(k-s)^{n-i}}-\frac{1}{k^{n-i}}\right)$$ Thus, we obtain: $$\left(\psi(1-s)+\gamma \right )^{n}=\sum_{i=0}^{n-1}\frac{(-1)^{n-i}n!}{(n+i)!}\sum_{k=1}^{\infty} B_{n+i,n}(k)\left(\frac{1}{(k-s)^{n-i}}-\frac{1}{k^{n-i}}\right)$$

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