I wonder, can anyone describe an expression or formula of a transform that converts
$$\sum_{k=0}^\infty \frac{a_k x^k}{k!}$$
into
$$\sum_{k=0}^\infty \frac{a_k B_k(x)}{k!}$$
where $B_k(x)$ are Bernoulli polynomials.
For instance,
$$x^2\, \to\, x^2-x+1/6$$
$$e^x \,\to \, \frac{e^x}{e-1}$$
etc.
An operator performing the mapping is
$$B(\partial_x) = e^{b.\partial_x} =: \frac{\partial_x}{e^{\partial_x}-1},$$
with $\frac{\partial}{\partial x} = \partial_x$ and $(B.(0))^n=B_n(x)|_{x=0}= (b.)^n = b_n$, since the Bernoulli polynomials are an Appell sequence.
Edit (6/20/2017):
This operator is essentially the Todd operator. See the discussions on pg. 30 and Appendix B of "Permutohedra, associahedra, and beyond" by Postnikov of the Todd operator as a transform of the homogeneous volume polynomials for classes of polytopes into a generalized Ehrhart polynomial coding the number of lattice points in the polytopes.
(Edit 8/2018)
For some idea of the importance of this Todd operator in modern mathematics and physics, see New Models for Veneziano Amplitudes: Combinatorial, Symplectic and Supersymmetric Aspects by Kholodenko.
(Edit 9/2024)
A rep of the Bernoulli op that is more robust than
$$T_{x \to (x+b.)}= e^{b.\partial_x}=: \frac{\partial_x}{e^{\partial_x}-1}$$
is
$$T_{x \to (x+b.)} = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} T_{x \to x+j}.$$
Define the Bernoulli translation op $T_{x \to (x+b.)}$ as the op that umbrally translates the variable $x$ to $(x+b.) = B.(x)$ such that
$$T_{x \to (x+b.)} x^n = (x+b.)^n = \sum_{k=0}^n \binom{n}{k}(b.)^k x^{n-k}= \sum_{k=0}^n \binom{n}{k}b_k x^{n-k} = B_n(x)$$
where $b_n$ is the $n$-th Bernoulli number and $B_n(x)$ the $n$-th Bernoulli polynomial.
Then a diff op rep when acting on $x^n$ is
$$T_{x \to (x+b.)} = e^{b.\partial_x} =: \frac{\partial_x}{e^{\partial_x}-1}.$$
A more robust, extended rep of the Bernoulli op that allows action on $x^s$ where $s$ is complex can be realized via an operator compositional-inverse pair.
Define the finite diff op $\delta_x$ by
$$\delta_x x^s = (x+1)^s-x^s = (T_{x \to x+1}-1) x^s.$$
Then its diff op rep is (by analytic continuation)
$$\delta_x x^s = (e^{\partial_x}-1) x^s = (x+1)^s -x^s$$
and inverting
$$\partial_x = \ln(1+\delta_x),$$
so when acting on $x^n$
$$T_{x \to (x+b.)=B.(x)} = e^{b.\partial_x} = \frac{\partial_x}{e^{\partial_x}-1} = \frac{\partial_x}{\delta_x} = \frac{\ln(1+\delta_x)}{\delta_x}$$
$$ = \sum_{k \geq 0} (-1)^k \frac{1}{k+1} \delta_x^k = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} e^{j\partial_x}$$
$$ = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} T_{x \to x+j}.$$
Removing the intermediate steps gives the Bernoulli translation rep
$$T_{x \to (x+b.)} = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} T_{x \to x+j}$$
with action
$$T_{x \to (x+b.)}x^s = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} (x+j)^s = -s\zeta(-s+1,x),$$
essentially the Helmut Hasse formula for the Hurwitz zeta function $\zeta(s,x)$. Taking this as the interpretation of
$$T_{x \to (x+b.)}x^s = (x+b.)^s = (B.(x))^s = B_s(x)$$
gives the Bernoulli function
$$B_s(x) = -s\zeta(-s+1,x) = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} (x+j)^s = T_{x \to (x+b.)}x^s,$$
and as
$$T_{x \to (x+b.)} \frac{(x+1)^{n+1}-x^{n+1}}{n+1} = e^{b.\partial_x} \frac{(x+1)^{n+1}-x^{n+1}}{n+1}$$
$$ = \frac{(b.+x+1)^{n+1}-(b.+x)^{n+1}}{n+1}$$
$$ = \frac{(B.(x+1))^{n+1}-(B.(x))^{n+1}}{n+1} = \frac{B_{n+1}(x+1)-B_{n+1}(x)}{n+1} $$
$$= \frac{\partial_x}{\delta_x}\delta_x \frac{ x^{n+1}}{n+1} = \partial_x \frac{ x^{n+1}}{n+1} = x^n$$
so does
$$T_{x \to (x+b.)} \frac{(x+1)^{s+1}-x^{s+1}}{s+1} = \frac{B_{s+1}(x+1)-B_{s+1}(x)}{s+1}$$
$$ = - \zeta(-s,x+1) -(-\zeta(-s,x)) = x^s = \partial_x \frac{x^{s+1}}{s+1}.$$
Note the action of the inverse op
$$ \frac{\delta_x}{\partial_x}x^n = \frac{e^{\partial_x}-1}{\partial_x}x^n = \sum_{k \geq 0} \frac{1}{k+1} \frac{\partial_x^k}{k!}x^n = \sum_{k \geq 0} \binom{n}{k} \frac{1}{k+1} x^{n-k}= \int_0^1 (x+t)^ndt $$
$$= \int_{x}^{x+1}t^n dt = \frac{(x+1)^{n+1}-x^{n+1}}{n+1} $$
has a rep as a sliding average, which, with $\hat{b}_k = \frac{1}{k+1}$, gives the action
$$T_{ x \to (x+\hat{b}.) = \hat{B}.(x)}x^s = \int_{x}^{x+1}t^s dt = \int_0^1 (x+t)^sdt = \frac{(x+1)^{s+1}-x^{s+1}}{s+1} = \hat{B}_s(x),$$
with a log function in the limiting case $s \to -1$.
Thus the umbral inversion relation for the Bernoulli polynomials
$$\hat{B}_n(B.(x)) =x^n$$
is generalized to
$$\hat{B}_s(B.(x)) =x^s.$$
This is consistent with Borel summation and analytic continuation of the action of $e^{b.\partial_x}$ and $e^{\hat{b}.\partial_x}$ on the Euler hybrid Mellin-Laplace rep of $x^s$ (and the more broadly valid Hankel contour rep of this integral), noting that
$$e^{b.\partial_x}e^{\hat{b}.\partial_x} = e^{(b.+\hat{b}.)\partial_x} = 1.$$
For example, the divergent series
$$e^{b.\partial_x}x^{-s} = e^{b.\partial_x}\int_0^{\infty} e^{-xt} \frac{t^{s-1}}{(s-1)!}dt$$
is Borel summed via
$$ \int_0^{\infty}e^{b.\partial_x} e^{-xt} \frac{t^{s-1}}{(s-1)!}dt = \int_0^{\infty}\frac{-t}{e^{-t}-1}e^{-xt} \frac{t^{s-1}}{(s-1)!}dt = B_{-s}(x)$$
$$ = \int_0^{\infty} e^{-(x+b.)t} \frac{t^{s-1}}{(s-1)!}dt = \int_0^{\infty} e^{-B.(x)t} \frac{t^{s-1}}{(s-1)!}dt$$
$$=: \frac{1}{(B.(x))^s} =: (B.(x))^{-s} := B_{-s}(x),$$
an example of Riemann-Ramanujan modified Mellin transform interpolation / analytic continuation. Then the various actions are reduced to actions on the Laplace kernel $e^{-xt}$.
Example related to Milnor's "On polylogarithms, Hurwitz zeta functions, and the Kubert identities":
On pgs. 21 & 22 of Milnor's paper are the relations
$$\frac{e^{\partial_x}-1}{\partial_x}\gamma_1(x) = \int_x^{x+1} \gamma_1(t)dt = \int_{x}^{x+1}\ln\left(\frac{\left(t-1\right)!}{\sqrt{2\ \pi}}\right)dt = x\ln\left(x\right)-x .$$
Milnor states that the inversion with the orthodox Bernoulli op
$$\frac{\partial_x}{e^{\partial_x}-1}(x\ln(x)-x)= e^{b.\partial_x}(x\ln(x)-x)$$
yields a divergent asymptotic series for $\gamma_1(x)$ (which is actually used in some proofs of the Harer-Zagier formula).
However, the more robust rep I present yields for $x>0$
$$\sum_{k=0}^{\infty}\frac{1}{k+1\ }\ \sum_{j=0}^{k}\left(-1\right)^{j}\ \frac{k!}{j!\left(k-j\right)!}\ \left(\left(x+j\right)\ln\left(x+j\right)-\left(x+j\right)\right) = \ln\left(\frac{\left(x-1\right)!}{\sqrt{2\ \pi}}\right)$$
$$= \gamma_1(x).$$
(Cut and paste the LaTex into Desmos online with the upper limit of the sum reduced to $20$ for a numerical check.)
Another way, somewhat related with the above answers, is the $p$-adic Volkenborn integral. You can find this, for example, in Schikhof's or in Alain Robert's books on $p$-adic calculus, or Henri Cohen vol. 2 of his books on number theory. This approach is useful because of the relation of Bernoulli numbers and L-functions: one can easily define good and elementary $p$-adic zeta functions using the Volkenborn integral (actually, this was Kubota and Leopoldt's original approach).
Let $\mathbb{Z}_p$ be the ring of $p$-adic integers and let $\mathbb{C}_p$ be the topological completion of an algebraic closure of the field of fractions $\mathbb{Q}_p$ of $\mathbb{Z}_p$ (a nice and large field for doing $p$-adic analysis). Let $f:\mathbb{Z}_p\to\mathbb{C}_p$ be an analytic function, that is, $f$ is of the form
$$f(x)=\sum_{n\ge0}a_n\frac{x^n}{n!},\qquad a_n\in\mathbb{C}_p,\quad \frac{a_n}{n!}\to0.$$
(We suppose $f$ analytic for simplicity and because of what you are asking). Then the Volkenborn integral of $f$ is defined by the following $p$-adic limit:
$$\int_{\mathbb{Z}_p}f(t)dt=\lim_{m\to\infty}p^{-m}\sum_{k=0}^{p^m-1}f(k).$$
Then, one has the following relation with Bernoulli numbers and polynomials:
$$\int_{\mathbb{Z}_p}t^ndt=B_n$$ and $$\int_{\mathbb{Z}_p}(x+t)^ndt=B_n(x).$$
This Volkenborn integral is a continuous linear operator on a Banach space of functions (see the books mentioned above). Hence, with $f$ as above, one obtains:
$$\int_{\mathbb{Z}_p}f(t)dt=\sum_{n\ge0}a_n\frac{B_n}{n!}$$ and $$\int_{\mathbb{Z}_p}f(x+t)dt=\sum_{n\ge0}a_n\frac{B_n(x)}{n!}.$$
Hope this helps.
Note: This integral is a special case of "$p$-adic distributions", which are one of the main tools that are now used to define $p$-adic zeta functions attached to arthmetic objects. See, for example, Washington or Lang books on cyclotomic fields for a nice introduction.
PS: For a nice "general zeta functions" interpretation of your question, see Lemma 2.4 in this article by Friedman and Pereira https://arxiv.org/abs/1105.2603 It was published in the IJNT, but the arxiv version is the same as the published version.
The transfert operator describe an expression or formula of a transform that $\sum_{k=0}^\infty \frac{a_k x^k}{k!}$ into $ \sum_{k=0}^\infty \frac{a_k B_k(x)}{k!}$ is the p-adic operator such that :The eigenvalues of the p-adic transfer operator are the Bernoulli polynomials,and are associated with the eigenvalues $p^{-n}$ , Try to check this paper by LINAS VEPŠTAS, page 8. Theorem with proof show that
