$\int_a^{a + n \cdot h} (x-a)(x-(a+h))(x-(a+2h))\cdots(x-(a+n\cdot h))\, \text{d}x = ?$ 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: 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, $b_n=0$
A: We can actually simplify your equation even further by using a Rising factorial or Falling factorial, but the best way would be to use the equivalent Pochhammer Symbol:
$$(a)_b\mathop=^\text{def}\frac{Γ(a+b)}{Γ(a)}\implies (u)_{n+1}= \frac{Γ(u+n+1)}{Γ(u)} $$
Therefore your integral is nothing more than:
$$f(h)=h^{n+2} \int_{-n}^0 \frac{\Gamma\left[u + 1 + n\right]}{\Gamma\left[u\right]} \ \text{d} u \\
=h^{n+2}\int_{-n}^0 (u)_{n+1}du$$
You can then use a series representation to integrate easily.
Since your $n+1\in\Bbb N$, it makes the integration easier. Let’s use this  simple expansion which uses the Stirling Numbers of the First Kind $S_n^{(m)}$. We can interchange the sum and integral because it is a finite sum:
$$f(h)=h^{n+2} \int_{-n}^0 \frac{\Gamma\left[u + 1 + n\right]}{\Gamma\left[u\right]} \ \text{d} u \\
=h^{n+2}\int_{-n}^0 (u)_{n+1}du=h^{n+2}\int_{-n}^0 \sum_{k=0}^{n+1} (-1)^{k+n+1} S_{n+1}^{(k)} u^k du= h^{n+2}\sum_{k=0}^{n+1} (-1)^{k+n+1} S_{n+1}^{(k)} \frac{u^{k+1}}{k+1} \bigg|_{-n}^0= -h^{n+2}\sum_{k=0}^{n+1} (-1)^{k+n+1} S_{n+1}^{(k)} \frac{(-n)^{k+1}}{k+1}$$
Now let’s finish the result noting that $(-1)^{2\Bbb Z}=1$
$$-h^{n+2}\sum_{k=0}^{n+1} (-1)^{k+n+1} S_{n+1}^{(k)} \frac{(-n)^{k+1}}{k+1}= -h^{n+2}\sum_{k=0}^{n+1} (-1)^{n}(-1)^{k+1} (-1)^{k+1}S_{n+1}^{(k)} \frac{(n)^{k+1}}{k+1}=   -h^{n+2}\sum_{k=0}^{n+1} \frac{ (-1)^nS_{n+1}^{(k)}n^{k+1}}{k+1} $$
We can actually factor $n$ out even if the upper bound is $n$:
$$-h^{n+2}\sum_{k=0}^{n+1} \frac{ (-1)^nS_{n+1}^{(k)}n^{k+1}}{k+1}= -n(-1)^n h^{n+2}\sum_{k=0}^{n+1} \frac{ S_{n+1}^{(k)}n^k}{k+1}  $$
Therefore, your question’s gamma function formula has been evaluated assuming $n\in\Bbb N$:
$$f(h)=  n(-1)^{n+1} h^{n+2}\sum_{k=0}^{n+1} \frac{ S_{n+1}^{(k)}n^k}{k+1}  = -(-h)^{n+2}\sum_{k=1}^{n+1} \frac{ S_{n+1}^{(k-1)}n^k}k $$
Which is a closed form assuming finite $n$. Please correct me and give me feedback!
