Finding a closed form for $I_n=\int_0^1 \prod_{l=1}^n\left[x^2-\frac{l^2}{n^2}\right]dx$ I'm now dealing with this integral and trying to give a closed form to it: 
$$I_n=\int_0^1 \prod_{l=1}^n\left[x^2-\dfrac{l^2}{n^2}\right]dx$$
Where the first few values are given: 
$$I_1=\int_0^1 (x^2-1) dx=-\dfrac23$$
$$I_2=\int_0^1 \left(x^2-\dfrac14\right)(x^2-1)dx=\dfrac1{30}$$
$$I_3=\int_0^1 \left(x^2-\frac19\right)\left(x^2-\frac49\right)(x^2-1)dx=-\dfrac{136}{8505}$$
I've tried to expand the polynomial, but I could merely give a general expression for the first few and the last few coefficients, and they become complicated very fast. I have searched the Internet; did I miss something that I couldn't find them because I don't know the Theorem's name? 
 A: $$I_n=\frac{1}{n^{2n}}\int_0^1\prod_{l=1}^{n}[(xn)^2-l^2]dx\qquad x=\frac{y}{n}$$
$$I_n=\frac{1}{n^{2n+1}}\int_0^n\prod_{l=1}^{n}[y^2-l^2]dy$$
We can expand the product, without knowing the coefficients for now as following;
$$\prod_{l=1}^{n}[y^2-l^2]=\sum_{j=0}^n(-1)^{n-j}y^{2j}t(n,j)$$
$$I_n=\frac{1}{n^{2n+1}}\sum_{j=0}^n(-1)^{n-j}t(n,j)\int_0^ny^{2j}dy$$
$$I_n=\frac{1}{n^{2n+1}}\sum_{j=0}^n\frac{(-1)^{n-j}t(n,j)n^{2j+1}}{2j+1}$$
$$I_n=\frac{(-1)^n}{n^{2n}}\sum_{j=0}^n\frac{(-1)^{j}t(n,j)n^{2j}}{2j+1}$$
Now back to the coefficients, ive found it on OEIS A008955
they are called the central factorial numbers. I think this is the closest we can get for a closed form.
A: You can relate the product to the rising factorials $x^\overline{n} = x(x+1)\dotsb (x+n-1) = \sum_{k=0}^n \begin{bmatrix}n\\k\end{bmatrix} x^k$ where the coefficients are Stirling's numbers of the first kind.
\begin{align}
I_n
&= \int_0^1 \prod_{l=1}^n \left(x^2 - \frac{l^2}{n^2}\right)\;dx
= \int_0^1 \prod_{l-1 =k=0}^{n-1} \frac{xn - k + 1}{n}\cdot \prod_{k=0}^{n-1} \frac{xn + k + 1}{n}\;dx\\
&= \frac{(-1)^n }{n^{2n}}\int_0^1 (-xn - 1)^{\overline{n}}(xn +1)^{\overline{n}}\;dx\\
%
&= \frac{(-1)^n }{n^{2n}}\int_0^1 \sum_{i=0}^n\sum_{j=0}^n \begin{bmatrix}n\\i\end{bmatrix} 
 \begin{bmatrix}n\\j\end{bmatrix} (-xn - 1)^i (xn + 1)^j\;dx\\
%
&=  \frac{(-1)^n }{n^{2n}} \sum_{i=0}^n\sum_{j=0}^n (-1)^i \begin{bmatrix}n\\i\end{bmatrix} 
 \begin{bmatrix}n\\j\end{bmatrix} \int_0^1 (xn + 1)^{i+j} \;dx\\
%
&=  \frac{(-1)^n }{n^{2n}} \sum_{i=0}^n\sum_{j=0}^n (-1)^i \begin{bmatrix}n\\i\end{bmatrix} 
 \begin{bmatrix}n\\j\end{bmatrix} \frac{(n+1)^{i+j+1}-1}{n(i+j+1)}\\
%
&= \sum_{i=0}^n\sum_{j=0}^n  (-1)^{i+n}\begin{bmatrix}n\\i\end{bmatrix} 
 \begin{bmatrix}n\\j\end{bmatrix} \frac{(n+1)^{i+j+1}-1}{n^{2n+1}(i+j+1)}\\
%
\end{align}
