Integration $\frac{1}{2\pi}\int_{-\pi}^{\pi}(x-a)^ke^{-i\omega x}dx, \ \ \ \ a\in\mathbb R$. Give a compact form for the solution of integral:
$$\frac{1}{2\pi}\int_{-\pi}^{\pi}(x-a)^ke^{-i\omega x}dx, \ \ \ \ a\in\mathbb R,k\in\mathbb N$$
any suggestions please?
 A: You may just integrate by parts
$$\frac{1}{2\pi}\int_{-\pi}^{\pi}(x-a)^k e^{-i\omega x}dx =\left.-(x-a)^k\frac{e^{-i \omega x}}{2\pi i\omega }\right|_{-\pi}^{\pi} +\frac{k}{2\pi i\omega}\int_{-\pi}^{\pi}(x-a)^{k-1} e^{-i\omega x}dx$$ thus, with an obvious notation, you have
$$
I_k(a,\omega) =\frac{(-1)^k }{2\pi i\omega }\left((a+\pi)^k e^{i \pi \omega }- (a-\pi)^k e^{-i \pi \omega }\right)+\frac{k}{i\omega}I_{k-1}(a,\omega) 
$$ multiplying by $\displaystyle  \frac{(i\omega)^k}{k!}$,
$$
\frac{(i\omega)^k}{k!}I_k(a,\omega) - \frac{(i\omega)^{k-1}}{(k-1)!}I_{k-1}(a,\omega)=\frac{(-1)^k (i\omega)^{k}}{2\pi i \omega \:k! }\left((a+\pi)^k e^{i \pi \omega }- (a-\pi)^k e^{-i \pi \omega }\right) 
$$ telescoping by summing from $k=1$ to $k=n$,
$$
\frac{(i\omega)^n}{n!}I_n(a,\omega) - \frac{\sin (\pi  \omega)}{\pi  \omega }=\frac{1}{2i\pi \omega}\sum_{k=1}^{n}\frac{(-1)^k (i\omega)^{k}}{k! }\left((a+\pi)^k e^{i \pi \omega }- (a-\pi)^k e^{-i \pi \omega }\right) 
$$ giving 

$$
\frac{1}{2\pi}\int_{-\pi}^{\pi}(x-a)^n e^{-i\omega x}dx=\frac{n!}{2\pi i \omega}\sum_{k=0}^{n}\frac{(-1)^k (i\omega)^{k-n}}{k! }\left((a+\pi)^k e^{i \pi \omega }- (a-\pi)^k e^{-i \pi \omega }\right) 
$$

with $a\in\mathbb C, \, \omega \in\mathbb C, \,n\in\mathbb N$.
A: Considering the problem of the antiderivtive , change variable $x=a+iy$ $$\int(x-a)^ke^{-i\omega x}dx=i^{k+1}e^{-ia\omega}\int y^k e^{\omega y}dy=-y^{k+1}i^{k+1}e^{-ia\omega} E_{-k}(-y \omega )$$ where appears the elliptic integral.
