From Beta Function
$$\int_0^{1} (1-x)^{a} \cdot x^{b} \ du = \dfrac{1}{a+b+1} \cdot \dfrac{1}{\binom{a+b}{a}}$$
So
$$\int_0^{1} (1-x)^{n-a} \cdot x^{a} \ du = \dfrac{1}{n+1} \cdot \dfrac{1}{\binom{n}{a}}$$
But now I search for a closed formula for $I$, for $\alpha \in \left[0, \ 2\pi\right]$
$$I(n, \ a, \ \alpha) = \int_0^{1} (1-x)^{n-a} \cdot x^{a} \cdot \exp\left(i\alpha x\right) \ du $$
I thought about expanding the $\exp$ by series and then integrating
$$\begin{align}I(n, \ a, \ \alpha) & = \int_{0}^{1} \left(1-x\right)^{n-a} \cdot x^{a} \left(\sum_{k=0}^{\infty} \dfrac{\left(i\alpha x\right)^{k}}{k!}\right) \ dx \\ & = \sum_{k=0}^{\infty}\dfrac{(i\alpha)^{k}}{k!} \cdot \int_{0}^{1}\left(1-x\right)^{n-a} \cdot x^{a+k} \ dx \\ & = \sum_{k=0}^{\infty}\dfrac{(i\alpha)^{k}}{k!}\cdot \dfrac{1}{n+k+1} \cdot \dfrac{1}{\binom{n+k}{a}}\end{align}$$
But now I don't know how to simplify it and take out the infinity sum.
Motivation:
Let $\mathbf{C}(x)$ be a bezier curve defined on $\left[0, \ 1\right]$ of $(n+1)$ points and (unknown) control points $\mathbf{P}$:
$$\mathbf{C}(x) = \sum_{j=0}^{n} \binom{n}{j} (1-x)^{n-j} \cdot x^j \cdot \mathbf{P}_j$$
Let $\mathbf{u}$ and $\mathbf{v}$ be the two axes of an ellipse, then I want to compute $\mathbf{P}$ to fit a $\dfrac{\alpha}{2\pi}$ part of this ellipse
$$\mathbf{C}(x) \approx \mathbf{u} \cdot \cos \alpha x + \mathbf{v} \cdot \sin \alpha x$$
To fit, I use least squares on these two equations: $$\mathbf{P}_{j} = P_{ju} \cdot \dfrac{ \mathbf{u}}{u^2} + P_{jv} \cdot \dfrac{ \mathbf{v}}{v^2}$$ $$\sum_{j=0}^{n} \binom{n}{j} (1-x)^{n-j} x^{j} \cdot \underbrace{\langle \mathbf{P}_j, \ \mathbf{u}\rangle}_{P_{ju}} = \langle \mathbf{C}(x), \ \mathbf{u}\rangle \approx \langle \mathbf{u}, \ \mathbf{u}\rangle \cdot \cos \alpha x = u^2 \cdot \cos \alpha x$$ $$\sum_{j=0}^{n} \binom{n}{j} (1-x)^{n-j} x^{j} \cdot \underbrace{\langle \mathbf{P}_j, \ \mathbf{v}\rangle}_{P_{jv}} = \langle \mathbf{C}(x), \ \mathbf{v}\rangle \approx \langle \mathbf{v}, \ \mathbf{v}\rangle \cdot \sin \alpha x = v^2 \cdot \sin \alpha x$$
Then minimizing $J(P_{ju})$, which results in a linear system
$$J(P_{ju}) = \int_{0}^{1} \left(\binom{n}{j} (1-x)^{n-j} x^{j} \cdot P_{ju} - u^2 \cdot \cos \alpha x\right)^2 \ du$$ $$\nabla J = \mathbf{0} \Rightarrow \sum_{k=0}^{n} A_{jk} \cdot P_{ku} = B_{j} $$ $$\begin{align}A_{jk} & = \int_0^1 \ \binom{n}{j}\binom{n}{k}(1-x)^{2n-j-k} x^{j+k} \ dx = \dfrac{\binom{n}{j}\binom{n}{k}}{\binom{2n}{j+k}} \\ B_{j} & = u^2 \binom{n}{j} \int_{0}^{1}(1-x)^{n-j} x^{j} \cdot \cos \alpha x \ dx \end{align}$$
So, the right-hand-side is the real part of the integral of this question, while to compute $P_{vj}$, it's the imaginary part.