i'm working on a project and i need to prove those are true (sorry for bad notation it's my first time here and sorry for bad English i'm Brazilian so it's not my native language.)
if n is even
$$\int_0^n \prod_{j = 0}^n \frac{t-j}{(t-i)}dt= \int_0^n\prod_{j = 0}^n \frac{t-j}{(t+i-n)}dt$$
After some manipulation this is the result:
$$\int_0^n \frac{2i-n}{(t-i)(t+i-n)}* \prod_{j = 0}^n (t-j)dt= 0$$
if n is odd
$$\int_0^n\prod_{j = 0}^n \frac{t-j}{(t-i)}dt= -\int_0^n\prod_{j = 0}^n \frac{t-j}{(t+i-n)}dt$$n
After some manipulation this is the result:
$$\int_0^n \frac{2t-n}{(t-i)(t+i-n)}* \prod_{j = 0}^n (t-j) dt= 0$$
$n ≥ i ≥ 0$ and i is a positive integer, n ≥ 0 and $n$ is a positive integer.
I've tested this in many ways and i "know" that those are true but i can't prove it.
The big problem is that there is no way to integrate the product as "n", I always need to define n before and by doing so i lose my generalization.
Here are my ideas, i know two things from my tests:
When i test for random i and n the result of the integral is symmetric in n/2. However, i can't generalize and test for n (because i cant integrate the the product with n).
The result of the integrals have roots in t=0 and t=n, that's why the definite integral equals 0. However, i can't prove that the polynomial that this integral results have these roots, mainly because i cant find this polynomial with regards to n because again i need to define n to be even in order to be able to integrate.
Note: the denominators are just there to cancel the times where j=i and j= (i-n), you can just add the restrictions to the productory like this:
$$\int_0^n \prod_{j = 0, j≠i}^n \frac{t-j}{1}dt= \int_0^n\prod_{j = 0, j≠(n-i)}^n \frac{t-j}{1}dt$$
Let me provide an example of why this works:
$$\int_0^n \frac{2i-n}{(t-i)(t+i-n)}* \prod_{j = 0}^n (t-j)dt= 0$$
n=2, i=0
$$\int_0^2 \frac{2*0-2}{(t-0)(t+0-2)}* \prod_{j = 0}^2 (t-j)dt= 0$$
$$\int_0^2 \frac{-2}{(t-0)(t-2)}*(t-0)(t-1)(t-2)dt= 0$$
$$\int_0^2 \frac{-2(t-1)}{1}dt= 0$$
$$2*2-(2^2)= 0$$
Inputing 0 also equals 0
n=3, i=0
$$\int_0^3 \frac{2*t-3}{(t-0)(t+0-3)}* \prod_{j = 0}^3 (t-j)dt= 0$$
$$\int_0^3 \frac{2t-3}{(t-0)(t-3)}*(t-0)(t-1)(t-2)(t-3)dt= 0$$
$$\int_0^3 \frac{(2t-3)(t-1)(t-2)}{1}dt= 0$$
$$\frac{(3^4)}{2}-3*3^3+\frac{(13*3^2)}{2}-6*3= 0$$
Inputing 0 also equals 0
I've just recently came back working on math projects and i'm a little rusty. I'm sure there is a [symmetry, odd, even] property that i'm forgetting that makes this easy. After many tests i think it's the way to solve this.
PS: Googling about products i discovered i'm dealing with a kind of descending factorial here.
The key is to substitute $t\rightarrow n-t$, $$ \int_0^n\frac{\prod_{j=0}^n(t-j)}{t-i} dt = \int_0^n \frac{\prod_{j=0}^n(n-t-j)}{n-t-i}dt, $$ then reverse the order of the product. That is, let $j = n$ be the first term, then $n-1$, then $n-2$, etc. The numerators will then be $-t,\,1-t\,2-t\,...$ So the product of the numerators can equivalently be written as $\prod_{j=0}^n(j - t)$. Thus we have $$ \int_0^n\frac{\prod_{j=0}^n(t-j)}{t-i} dt = \int_0^n \frac{\prod_{j=0}^n(j-t)}{n-t-i}dt = \int_0^n\frac{\prod_{j=0}^n(-1)(t-j)}{(-1)(t+i-n)} dt=(-1)^n\int_0^n\frac{\prod_{j=0}^n(t-j)}{t+i-n} dt $$ Since $(-1)^n = 1$ when $n$ is even and $(-1)^n = -1$ when $n$ is odd, this recovers your result.