Value of an integral A function  $f$ is defined as
$$f(t)=t^{r-1}(2t-t^2)^{n-1}(2-2t)$$
where $r,n$ are natural numbers. Can we somehow find an a closed expression for the inegral $\int_0^1 f(t)\,dt$ or even an asymptotic expression as $n$ goes to infinity? One could substitution $u=2t-t^2$ but for the extra term $t^{r-1}$.
Mathematica yields the following value :
$$2^{-1 + 2 n + r} (B[1/2, -1 + n + r, n] - 2 B[1/2, n + r, n])$$
where $B(x,a,b)=\int_0^x u^{a-1}(1-u)^{b-1} du$ is the incomplete beta integral
 A: Partial solution: asymptotics at $n\to\infty$
$$I(n)=\int_0^1t^{r-1}(2t-t^2)^{n-1}(2-2t)dt=2^n\int_0^1t^{r-1}(1-t)e^{(n-1)\log(t(1-\frac{t}{2}))}dt$$
$\log(t(1-\frac{t}{2}))$ has maximum at $t=1$.
Near this pont
$\log(t(1-\frac{t}{2}))=\log{\frac{1}{2}}-(t-1)^2-\frac{1}{2}(t-1)^4+...$
Changing the variable $\,t\to{x}=1-t$ and expanding integration to $\infty$
$$I(n)=2^n{e}^{(n-1)\log{\frac{1}{2}}}\int_0^1(1-x)^{r-1}e^{-(n-1)\bigl(x^2+\frac{1}{2}x^4+...\bigr)}xdx=$$$$=2\int_0^1\bigl(1-(r-1)x+...)\bigr)e^{-(n-1)\bigl(x^2+\frac{1}{2}x^4+...\bigr)}xdx$$$$\sim2\int_0^\infty{e}^{-(n-1)x^2}xdx-2(r-1)\int_0^\infty{e}^{-(n-1)x^2}x^2dx+O\Bigl(\frac{1}{n^2}\Bigr)$$
$$\sim\frac{1}{n}\Bigl(1-\frac{\sqrt{\pi}(r-1)}{2\sqrt{n}}\Bigr)+O\Bigl(\frac{1}{n^2}\Bigr)$$
A: Since
$$f(t)=t^{r-1}(2t-t^2)^{n-1}(2-2t)\\
\qquad\qquad\qquad=2t^{r-1+n-1}(1-t)\sum_{i=0}^{n-1}(-1)^i2^{n-i}{n-1 \choose i}t^i$$
we get
$$\int_0^1f(t)\,\text dt=2\sum_{i=0}^{n-1} (-1)^i 2^{n-i}{n-1 \choose i} B(r+n+i-1,1)\\
\qquad =\sum_{i=0}^{n-1} (-1)^i 2^{n-i+1}{n-1 \choose i}\Big/(r+n+i-1)$$
