Generating function for sequence $\frac{(-1)^{n+1} B_n(x)}{n}$

Question

I wonder what is generating function for sequence $\frac{(-1)^{n+1} B_n(x)}{n}$.

Or, in other words, what is

$$\sum_{n=1}^\infty \frac{(-1)^{n+1} B_n(x) t^n}{n}$$

Mathematica fails to provide answer.

$B_n(x)$ are Bernoulli polynomials.

Answer 1

If we start from $$\frac{e^{xz}}{e^z-1}=\frac{1}{z}+\sum_{k\geq 1}B_k(x)\frac{z^{k-1}}{k!}\tag{1} $$ we may replace $z$ with $-zt$ to get $$ \frac{e^{(1-x)tz}}{1-e^{tz}}+\frac{1}{tz}=\sum_{k\geq 1}(-1)^{k-1} B_k(x)\frac{z^{k-1} t^{k-1}}{k!}\tag{2} $$ and derive the wanted series by multiplying both sides by $e^{-z}$ and applying $\int_{0}^{+\infty}\left(\ldots\right)\,dz$: $$ \sum_{n\geq 1}B_n(x)\frac{(-1)^{n+1} t^{n}}{n} = t\int_{0}^{+\infty}\left(\frac{e^{(1-x)tz}}{1-e^{tz}}+\frac{1}{tz}\right)e^{-z}\,dz.\tag{3} $$ By replacing $z$ with $\frac{w}{t}$ we get: $$ \sum_{n\geq 1}B_n(x)\frac{(-1)^{n+1} t^{n}}{n} = \int_{0}^{+\infty}\left(\frac{e^{(1-x)w}}{1-e^{w}}+\frac{1}{w}\right)e^{-w/t}\,dz.\tag{4} $$ The RHS of $(4)$ is a generalization of an integral defining the Euler-Mascheroni constant $\gamma$.
By Frullani's theorem:

$$ \sum_{n\geq 1}B_n(x)\frac{(-1)^{n+1} t^{n}}{n} =\log(t)+\psi\left(x+\frac{1}{t}\right). \tag{5}$$