Formula for integration bounds of recursively defined polynomial sequence We can recursively define a sequence of polynomials by 
$$P_0(x) := 1$$
and then with the definite integral
$$P_n(x) := \int_{c_n}^x P_{n-1}(t) ~\mathrm dt$$
where the $c_n$ are to be chosen so that
$$\int_0^1 P_n(t)~\mathrm dt = 0$$
so for $n = 1$ we have $P_1(x) = x - c_1$ so $c_1 = 1/2$. However already for $n=2$ it becomes difficult since $\int_{c_2}^x (t-1/2)~\mathrm dt =  t^2/2 - t/2- c_2^2/2 + c_2/2$ and to have $\int_0^1 P_2(t)~\mathrm dt = 0$ there are two solutions for $c_2$ namely 
$$c_2 = 1/2 \pm \sqrt 3/6.$$
So $P_2(x) = x^2/2 - x/2 + 1/12$. For $n=3$ we get $c_3 = 0$ and $P_3(x) = x^3/6 - x^2/4 + x/12$ however for $n=4$ we have $c_4=1/2 - 1/2\sqrt (1 - 2/15\sqrt 30)$ (and 3 other solutions) for $P_4(x)=x^4/24 - x^3/12 + x^2/24 - 1/720$. For $n=5$ again $c_5=0$ (however there are another 4 solutions). Is there some general Formula for the $c_n$ so we could have a shortcut for calculating the coefficients of $P_n(x)$? Also with an eye to fractional calculus it would be nice to know if for the Riemann-Liouville integral
$$\frac1{\Gamma(\alpha)} \int_c^x P(t) (x-t)^{\alpha-1}~\mathrm dt$$
and
$$\int_0^1 P(t)~\mathrm dt = 0$$
if there even is a solution for this - let alone if it would fit in the previously defined sequence. The dream result would of course be not only to have a formula for $c_n$ but a function $c(n)$ that continuously defines such a polynomial sequence. But that seems very remote since even in simple cases with for example $\alpha = 1/2$ and $c>0$ the integral produces complex values...
 A: I can't comment, but your problem looks awfully like the method for generating the Bernoulli polynomials $B_n(x)$, except that instead of the relation $\frac{\mathrm d}{\mathrm dx}P_n(x)=P_{n-1}(x)$ that you have, the Bernoulli polynomials have the relation $\frac{\mathrm d}{\mathrm dx}B_n(x)=nB_{n-1}(x)$. That might be something to start with...

As Fabian seems to have settled the relationship of your polynomials to the more conventional Bernoulli polynomials, I'll address the question of generalizing the Bernoulli polynomials to arbitrary index; one way to this is the use of the exponential generating function
$$\frac{t\exp(zt)}{\exp(t)-1}=\sum_{n=0}^\infty \frac{B_n(z)}{n!}t^n$$
along with the Cauchy differentiation formula
$$f^{(n)}(a)=\frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z-a)^{n+1}}\mathrm dz;$$
i.e., consider the contour integral
$$B_\alpha(z)=\frac{\Gamma(\alpha+1)}{2\pi i}\oint_\gamma \frac{t\exp(zt)}{\exp(t)-1} \frac{\mathrm dt}{t^{\alpha+1}},$$
where $\gamma$ is an anticlockwise contour that doesn't cross the negative real axis, and is within the circle with radius $2\pi$ (due to the restricted radius of convergence of the EGF). I'd say more, but it seems there is prior work by Butzer et al., and thus I'll just have to ask you to see that paper.
A: Don't know about the $c$'s. But finding the polynomials $P_n(t)$ is not so hard. As gorilla already noted
$$\frac{d}{dx} P_n(t) = P_{n-1}(t). \qquad  (1)$$
This determines the polynomials $P_n(t)$ 
together with the normalization
$$\int_0^1 dt\, P_n(t) = 0 \qquad n\neq0 \qquad (2)$$
and the initial condition $P_0(t)=1$ uniquely (all the different $c$'s in you post which are the solutions of some polynomial equation lead to the same  $P_n$!).
Every recursion step adds a degree to the polynomial and therefore we can parameterize the $n$-th polynomial as
$$ P_n(t) = \sum_{j=0}^n \frac{a_{n-j}}{j!} t^j = \frac{a_0}{n!} t^n + \frac{a_1}{(n-1)!} t^{n-1}+ \dots\;.$$
With this ansatz Eq. (1) is automatically fulfilled. Equation (2) leads to the condition
$$ \sum_{j=0}^n \frac{a_{n-j}}{(j+1)!} =0$$
with the solution
$$a_n = -\sum_{j=1}^n \frac{a_{n-j}}{(j+1)!}.$$
This defines a recursion relation for the coefficients $a_n$ (with $a_0=1$ as initial condition).
The recursion relation has the solution
$a_n = \frac{B_n}{n!}$
with $B_n$ the $n$-th Bernoulli number. The polynomial $P_n$ thus are given by
$$P_n(t) = \sum_{j=0}^n \frac{B_{n-j}}{j!(n-j)!} t^j = \frac{B_n(t)}{n!},$$
with $B_n(t)$ the Bernoulli polynomials.
