Reading this paper (page $3$, theorem $2$) I got confused.
First is defined the integral $I_n$ as:
\begin{align} I_n = (-1)^n \frac{ \pi^{2n} }{(2n)!} \int_0^1 t^{2n}(1-t)^{2n} \sin \pi t \: dt . \end{align}
Now, $I_n$ consists of terms of the shape:
\begin{align} \int_0^1 t^{m} \sin \pi t \: dt , \: m \leq 2n. \end{align}
Integrate by parts the integral above:
\begin{align} \int_0^1 t^{m} \sin \pi t \: dt = \frac{1}{\pi} - \frac{m(m-1)}{\pi^2} \int_0^1 t^{m-2} \sin \pi t \: dt. \end{align}
The following conclusion I don't understand: "Hence $I_n = \pi^{-1}A_n(\pi^{-2}) $, where $A_n$ is an $(n-1)$st degree polynomial with integral coefficients."
I understand the above as being, $I_n = \pi^{-1}( a_{n-1}\pi^{-2(n-1)} + a_{n-2}\pi^{-2(n-2)} + \cdots + a_0).$ But this is wrong.
Can someone give a detailed explanation, please?
I think that there is typo in the paper; it should be "Hence $I_n = \pi^{-1}A_n(\pi^{-2})$ where $A_n$ is an $\color{red}{n}$ degree polynomial with integral coefficients."
I prefered to write it as $$I_n=\frac 1 \pi \sum_{k=0}^n \frac {a_k} {\pi^{2k}}$$
Using what is given and computing, we have $$\left( \begin{array}{cc} n & \pi I_n \\ 1 & 2-\frac{24}{\pi ^2} \\ 2 & 2-\frac{360}{\pi ^2} +\frac{3360}{\pi ^4}\\ 3 & 2-\frac{1680}{\pi ^2}+\frac{151200}{\pi ^4}-\frac{1330560}{\pi ^6} \\ 4 & 2-\frac{5040}{\pi ^2}+\frac{1663200}{\pi ^4}-\frac{121080960}{\pi ^6}+\frac{1037836800}{\pi ^8}\\ 5 & 2-\frac{11880}{\pi ^2}+\frac{10090080}{\pi ^4}-\frac{2421619200}{\pi ^6}+\frac{158789030400}{\pi ^8}-\frac{1340885145600}{\pi ^{10}} \\ 6 &2 -\frac{24024}{\pi ^2}+\frac{43243200}{\pi ^4}-\frac{24700515840}{\pi^6}+\frac{5028319296000}{\pi ^8}-\frac{309744468633600}{\pi ^{10}}+\frac{2590590101299200}{\pi ^{12}} \end{array} \right)$$
In fact, given by a CAS, there is a general expression in terms of hypergeometric function $$I_n=(-1)^n\frac{ \pi ^{2 n+\frac{3}{2}} }{2^{4 n+2}\,\Gamma \left(2 n+\frac{3}{2}\right)}\, _2F_3\left(n+1,n+\frac{3}{2};\frac{3}{2},2 n+\frac{3}{2},2 n+2;-\frac{\pi ^2}{4}\right)$$