# Closed form for $\sum_{k=0}^{\infty}\frac{\Gamma(k+z)\psi(k+1)}{k!}x^{k}$

I remember seeing a closed form for the series : $$\sum_{k=0}^{\infty}\frac{\Gamma(k+z)\psi(k+1)}{k!}x^{k}\;\;\;\; \left | x \right |<1$$ But i don't seem to recall what it was . I tried replacing the gamma function with its Euler's integral , and then changing the order of summation and integration, using the fact that : $$\sum_{k=0}^{\infty}\frac{\psi(k+1)}{k!}y^{k}=e^{y}\left(\log y +\Gamma(0,y\right)$$ Thus : $$\sum_{k=0}^{\infty}\frac{\Gamma(k+z)\psi(k+1)}{k!}x^{k}=\int_{0}^{\infty}y^{z-1}e^{(x-1)y}\left[(\log(xy)+\Gamma(0,xy)) \right ]dy$$ $$=x^{-z}\int_{0}^{\infty}\omega^{z-1}e^{\left(1-\frac{1}{x} \right )\omega}\left[\log \omega +\Gamma(0,\omega) \right ]d\omega$$ Where $\Gamma(\cdot,\cdot)$ is the incomplete gamma function. But i don't know how to do the integral !

• Is $z$ and integer or do you want an expression for $z$ in the reals? Jan 29 '17 at 8:13
• no actually. $z$ is a complex parameter. $x$ is real. Jan 29 '17 at 8:17

Lemma 1. By the extended binomial theorem, $$\sum_{k\geq 0}\frac{\Gamma(k+z)}{k!}\,x^k =\frac{\Gamma(z)}{(1-x)^z}.\tag{1}$$ $\phantom{}$ Lemma 2. The series definition of the digamma function ensures $$\begin{eqnarray*} \psi(k+1) = -\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+k}\right) &=& -\gamma+\sum_{n\geq 1}\int_{0}^{+\infty}e^{-nt}(1-e^{-kt})\,dt\\&=&-\gamma+\int_{0}^{+\infty}\frac{1-e^{-kt}}{e^{t}-1}\,dt\\&=&-\gamma+\int_{0}^{1}\frac{u^k-1}{u-1}\,du.\end{eqnarray*}\tag{2}$$
By Lemma 1 and Lemma 2, $$\sum_{k\geq 0}\frac{\Gamma(k+z)\,\psi(k+1)}{k!}\,x^k = -\frac{\gamma\,\Gamma(z)}{(1-x)^z}+\Gamma(z)\int_{0}^{1}\frac{1}{u-1}\left(\frac{1}{(1-ux)^z}-\frac{1}{(1-x)^z}\right)\,du$$ and by applying integration by parts $$\sum_{k\geq 0}\frac{\Gamma(k+z)\,\psi(k+1)}{k!}\,x^k = -\frac{\gamma\,\Gamma(z)}{(1-x)^z}-\Gamma(z+1)\int_{0}^{1}\frac{x \log(1-u)}{(1-u x)^{z+1}}\,du.\tag{3}$$ Now the last integral can be seen as the derivative of a hypergeometric function (by substituting $u\mapsto(1-u)$ then exploiting $\log(u)=\left.\frac{d}{d\alpha}u^{\alpha}\right|_{\alpha=0^+}$) or expanded as a series by exploiting $\log(1-u)=-\sum_{m\geq 1}\frac{u^m}{m}$.