Find the function $f(x)=\sum_{n=1}^{\infty}\frac{H_{n-1}(-x)^n}{n!}$

Question

I want to find the function defines by : $$f(x)=\sum_{n=1}^{\infty}\frac{H_{n-1}(-x)^n}{n!}$$

Where $H_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ is the Harmonic series.

My work

We have the relation with the digamma function :

$$\psi(x+n)-\psi(x)=\sum_{k=0}^{n-1}\frac{1}{x+k}$$

So setting $x=1$ we have :

$$\psi(n)-\psi(1)=H_{n-1}$$

So we have :

$$f(x)=\sum_{n=1}^{\infty}\frac{(\psi(n)-\psi(1))(-x)^n}{n!}$$

With the help of WA I get :

$$f(x)=-x\operatorname{F}_1^{(1,0,0)}(1,2-x)$$

Where $F_1(a;b;z)$ denotes the confluent hypergeometric function of the first kind

Now applying the Ramanujan master theorem we get :

$$\int_{0}^{\infty}x^{s-1}f(x)dx=\Gamma(s)(\psi(-s)-\psi(1))$$

My question :

Can someone add some steps to get the result of Wolfram alpha ?

Any helps is greatly appreciated

Thanks a lot for all your contributions .

Answer 1

We have \begin{align*} \frac{d}{{da}}{}_1F_1 (a;2; - x) = \frac{d}{{da}}\sum\limits_{n = 0}^\infty {\frac{{(a)_n }}{{(2)_n }}\frac{{( - x)^n }}{{n!}}} = \sum\limits_{n = 0}^\infty {\frac{{(a)_n (\psi (n + a) - \psi (a))}}{{(2)_n }}\frac{{( - x)^n }}{{n!}}} \\ = \sum\limits_{n = 0}^\infty {\frac{{(a)_n (\psi (n + a) - \psi (a))}}{{(n + 1)!}}\frac{{( - x)^n }}{{n!}}} . \end{align*} Thus, \begin{align*} - x\left[ {\frac{d}{{da}}{}_1F_1 (a;2; - x)} \right]_{a = 1} & = - x\sum\limits_{n = 0}^\infty {\frac{{(1)_n (\psi (n + 1) - \psi (1))}}{{(n + 1)!}}\frac{{( - x)^n }}{{n!}}} \\ & = \sum\limits_{n = 0}^\infty {(\psi (n + 1) - \psi (1))\frac{{( - x)^{n + 1} }}{{(n + 1)!}}} = \sum\limits_{n = 1}^\infty {(\psi (n) - \psi (1))\frac{{( - x)^n }}{{n!}}} . \end{align*}