Evaluation of the series $S(\omega)=\sum\limits_{k=0}^\infty (-1)^k {\alpha \choose k}\cos(k\omega)$ I had a problem evaluating the series
\begin{equation}
S(\omega)=\sum_{k=0}^\infty (-1)^k {\alpha \choose k}\cos(k\omega),\quad 0<\alpha<2,\quad \omega\in(-\pi,\pi)
\end{equation}
where
\begin{equation}
{\alpha \choose k}=\frac{\Gamma(1+\alpha)}{\Gamma(\alpha-k+1)\Gamma(k+1)}
\end{equation}
is the binomial coefficient generalized to non-integer.
Seems it is a bit like Fourier series. However the coefficients are strange. I have drawn the curve of $S(\omega)$ vs. $\omega$ and through visualization I thought that $S(\omega)$ may be a well behaved function which has a simpler form.
Can you help me find a simple, equivalent expression to the above series? If that doesn't exist, is there an approximation to the sum?
Any answer would be appreciated. 
P.S.:
Some answers given are concerned with complex numbers $(1-e^{i\omega})$. As far as I know, $(1-e^{i\omega})^\alpha$ is a multi-valued function. It is not convenient for evaluation in Matlab. 
Is there an equivalent function that only involves real numbers?
 A: Assume that $\omega\ne0$. 
Then $1-\mathrm e^{\mathrm i\omega}\ne0$ and users @MichaelHardy and @i707107 explained in comments why 
$$
S(\omega)
=
\Re\exp(\alpha\cdot\mathrm{Log}(1-\mathrm e^{\mathrm i\omega})),
$$ 
under the condition that 
$$
\left|\alpha\cdot\arg(1-\mathrm e^{\mathrm i\omega})\right|
\lt
\pi,
$$ 
where $\arg$ denotes the principal argument, with values in $(-\pi,\pi]$, and $\mathrm{Log}$ denotes the principal branch of the complex logarithm, defined on $\mathbb C\setminus\mathbb R_-$ by the identity 
$$
\mathrm{Log}(r\mathrm e^{\mathrm it})
=
\ln(r)+\mathrm i\mathrm t,
\quad 
r\gt0,
\quad 
|t|\lt\pi.
$$
Thus, the main task is to identify $z=1-\mathrm e^{\mathrm i\omega}$ as $z=r\mathrm e^{\mathrm it}$ with $r\gt0$ and $|t|\lt\pi$, and to check the argument condition. 
To do so, note that 
$$
z
=
\mathrm e^{\mathrm i\omega/2}\cdot(\mathrm e^{-\mathrm i\omega/2}-\mathrm e^{\mathrm i\omega/2})
=
-2\mathrm i\,\sin(\omega/2)\,\mathrm e^{\mathrm i\omega/2},
$$ 
hence 
$$
r
=
2\,|\sin(\omega/2)|,
\qquad
\mathrm e^{\mathrm it}
=
-\mathrm i\,\mathrm{sgn}(\omega)\,\mathrm e^{\mathrm i\omega/2}
=
\mathrm e^{\mathrm i(\omega-\mathrm{sgn}(\omega)\pi)/2}.
$$ 
This proves that $t=\frac12(\omega-\mathrm{sgn}(\omega)\pi)$ since this number is in the interval $(-\pi/\pi)$. 
Furthermore, separating the cases $\omega\gt0$ and $\omega\lt0$, one can see that $|t|\lt\pi/2$. 
Since $|\alpha|\lt2$, the argument condition holds and 
$$
S(\omega)
=
r^\alpha\,\Re\mathrm e^{\mathrm i\alpha t}
=
r^\alpha\,\cos(\alpha t),
$$
that is,

$$
S(\omega)
=
2^\alpha\,\left|\sin\left(\tfrac12\omega\right)\right|^\alpha\,\cos\left(\tfrac12\alpha(\omega-\mathrm{sgn}(\omega)\pi)\right).
$$ 

Exercise: 
Extend this formula to the case $\omega=0$. 
Check that this defines an even function on $(-\pi,\pi)$. 
When $\alpha=1$, check that the RHS is $2\sin^2(\omega/2)$ for every $\omega$ in $(-\pi,\pi)$ and explain why it ought to.
A: Hint
If you replace $\cos(k\omega)$ by $\dfrac{{e^{ik\omega}+e^{-ik\omega}}}{2}$ or if you compute $$S(\omega)=\sum_{k=0}^\infty (-1)^k {\alpha \choose k}\cos(k\omega)$$ as being the real part of $$T(\omega)=\sum_{k=0}^\infty (-1)^k {\alpha \choose k}e^{i k \omega}$$ using the generalised binomial theorem, you end with $$S(\omega)=\sum_{k=0}^\infty (-1)^k {\alpha \choose k}\cos(k\omega)=\frac{1}{2} \left(\left(1-e^{i \omega}\right)^\alpha+\left(1-e^{-i \omega}\right)^\alpha\right)$$ which a real valued function.
A: Use the identity 

$$\cos(k\omega) = \frac{e^{ik\omega}+e^{-ik\omega}}{2}$$

and the Newton's generalised binomial theorem as

$$ S_1 = \sum_{k=0}^{\infty}(-1)^k {\alpha\choose k}e^{ik\omega} = (1-e^{i\omega})^{\alpha}. $$

You do the rest of the problem.  
A: Thank you very much for the answer given by @Did.
In this reply, I'll try to answer the exercise he proposes.


*

*The formula 
\begin{equation}
S(\omega) = 2^\alpha|\sin(\omega/2)|^\alpha \cos\left[\frac{\alpha}{2}(\omega-\text{sgn}(\omega)\pi)\right]
\end{equation}
can be extended by defining $\text{sgn}(0) = \frac{1}{\alpha}$ such that $\omega=0$ is also included.

*The formula is even because
\begin{equation}
\cos\left[\frac{\alpha}{2}(\omega-\text{sgn}(\omega)\pi)\right] = \cos(\alpha\omega/2)\cos(\alpha\text{sgn}(\omega)\pi/2)+\sin(\alpha\omega/2)\sin(\alpha\text{sgn}(\omega)\pi/2)
\end{equation}
and
\begin{equation}
\begin{split}
&\cos(\alpha\text{sgn}(\omega)\pi/2)=\cos(\alpha\pi/2) ,\quad \omega>0\\
&\cos(\alpha\text{sgn}(\omega)\pi/2)=\cos(\alpha\pi/2) ,\quad \omega<0\\
&\sin(\alpha\text{sgn}(\omega)\pi/2)=\sin(\alpha\pi/2) ,\quad \omega>0\\
&\sin(\alpha\text{sgn}(\omega)\pi/2)=-\sin(\alpha\pi/2) ,\quad \omega<0\\
\end{split}
\end{equation}

*When $\alpha=1$, the formula becomes
\begin{equation}
S(\omega)=2|\sin(\omega/2)| \cos\left[(\omega-\text{sgn}(\omega)\pi)/2\right]
\end{equation}
and since
\begin{equation}
\begin{split}
&\cos\left[(\omega-\text{sgn}(\omega)\pi)/2\right] = \cos(\omega/2-\pi/2)=\sin(\omega/2),\quad \omega>0\\
&\cos\left[(\omega-\text{sgn}(\omega)\pi)/2\right] = \cos(\omega/2+\pi/2)=-\sin(\omega/2),\quad \omega<0\\
\end{split}
\end{equation}
We conclude
\begin{equation}
S(\omega)=2\sin^2(\omega/2)
\end{equation}
