$\int_a^{a + n \cdot h} (x-a)(x-(a+h))(x-(a+2h))\cdots(x-(a+n\cdot h))\, \text{d}x = ?$ I would like to dissolve this expression:
\begin{align}
f(h) = \int_a^{a + n \cdot h} (x-a)(x-(a+h))(x-(a+2h))\cdots(x-(a+nh)) \ \text{d} x
\end{align}
I have a guess by working out the first two expressions that the following appears.
\begin{align}
f(h) = -b_n \cdot h^{n+2}
\end{align}
And I noticed that for even $n$ the integral disappears. But the $b_n$ that appear seem very chaotic. I have listed them:
\begin{align}
b_n = \left(\frac{1}{6},0,\frac{9}{10},0,\frac{1375}{84},0,\frac{57281}{90},0,\frac{1891755}{44},\ldots\right)
\end{align}
I was also able to express this product term in gamma functions. But now I don't know if this makes it so much easier to calculate the expression.
\begin{align}
f(h) &= h^{n+1}\int_a^{a+nh} \frac{\Gamma\left[\frac{a - x}{h} + 1 + n\right]}{\Gamma\left[\frac{a-x}{h}\right]} \ \text{d} x \\
&= -h^{n+2} \int_0^{-n} \frac{\Gamma\left[u + 1 + n\right]}{\Gamma\left[u\right]} \ \text{d} u \\
&= h^{n+2} \int_{-n}^0 \frac{\Gamma\left[u + 1 + n\right]}{\Gamma\left[u\right]} \ \text{d} u \\
\end{align}

A closed form can be given in terms of the  Gregory co-efficients. Note that we can write $$b_n=\int_{0}^{n}x\left(x-1\right)\left(x-2\right)...\left(x-n\right)dx$$
We have, $$\int_{ }^{ }x\left(x-1\right)\left(x-2\right)...\left(x-n\right)dx=\psi_{n+2}\left(x\right)(n+1)!\tag{1}$$
Where, $\psi_n(x)$ is the $n$-th Bernoulli polynomial of the second kind and hence,
$$b_{n}=\left(\psi_{n+2}\left(n\right)-\psi_{n+2}\left(0\right)\right)(n+1)!\tag{2}$$.
Some things to consider: the linked article states the following properties of the polynomials,
$$\begin{align}\psi_{n+2}(0)&=G_{n+2}=(-1)^{n+1}|G_{n+2}|\\\psi_{n+2}(n)&=-|G_{n+2}|\end{align}$$
Where $G_n$ are the Gregory coefficients, which appears in the expansion of $x/\ln(1+x)$ for $|x|<1$,
$$\begin{align}\frac{x}{\ln(1+x)}&=1+G_1x+G_2x^2+G_3x^3+ \cdots\\&=1+\frac{1}{2}x-\frac{1}{12}x^{2}+\frac{1}{24}x^{3}-....\end{align}$$
Substituting this in $(2)$,
$$b_n=\left(\left(-1\right)^{n}-1\right)\left(n+1\right)!\left|G_{n+2}\right|\tag{3}$$
We can see from $(3)$ that for $n$ being even, $b_n=0$
