I ran across an abandon post from 2013 where the OP has no work shown but just a problem statement. The OP was last seen May 2013 so I doubt they will be returning to edit their post with relevant work. However, the problem seems interesting so I am posting it with my own work. The original post can be found here . On the post, @DilipSarwate comments saying write cosine in exponential form and use the Binomial theorem.
We are trying to find the Fourier transform of $$ f(x) = \chi\cos^n(\pi x) $$ where $n\in\mathbb{N}$, $x\in\mathbb{R}$, and $$ \chi = \begin{cases} 1, & -1/2 <x< 1/2\\ 0, & x\not\in (-1/2, 1/2) \end{cases} $$ Using the following defintion, $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{i\omega x}dx$, we have \begin{align} \frac{1}{\sqrt{2\pi}}\int_{-1/2}^{1/2}\cos^n(\pi x)e^{i\omega x}dx &= \frac{1}{\sqrt{2\pi}}\int_{-1/2}^{1/2}\Bigl[\sum_{k=0}^n\frac{1}{2^n}\binom{n}{k}e^{i\pi x(n-k)}e^{-i\pi xk}\Bigr]e^{i\omega x}dx\\ &= \frac{1}{\sqrt{2\pi}}\int_{-1/2}^{1/2}\sum_{k=0}^n\frac{1}{2^n}\binom{n}{k}\exp\bigl[i(\pi n - 2\pi k + \omega)x\bigr]dx \end{align} Can we swap the sum and integral here? If so, we have
$$ \frac{1}{\sqrt{2\pi}}\sum_{k=0}^n\frac{1}{2^n}\binom{n}{k}\int_{-1/2}^{1/2}\exp\bigl[i(\pi n - 2\pi k + \omega)x\bigr]dx = \frac{1}{\sqrt{2\pi}}\sum_{k=0}^n\frac{1}{2^n}\binom{n}{k}\frac{2\sin\bigl[1/2(\pi n - 2\pi k + \omega)\bigr]}{\pi n - 2\pi k + \omega} $$ From Mathematica, I know that sum evaluates to $$ \sum = -\frac{ n! \sin \bigl(\frac{1}{2} (\pi n+\omega )\bigr) \Gamma \bigl(-\frac{\pi n+\omega }{2 \pi }\bigr)}{2^n\pi \Gamma \Bigl[(\frac{1}{2} \bigl(n-\frac{\omega }{\pi }+2\bigr)\Bigr]} $$ but I don't know how to evaluate it analytically. This whole part was under the assumption the sum and integral can be interchanged.
Since $n$ is finite, there is just a finite sum. Therefore it is allowed to interchange the summation and integration.
To calculate the summation, we start with this, without having interchanged summation and integration:
\begin{align} \sum&=\int_{-1/2}^{1/2}\sum_{k=0}^n\frac{1}{2^n}\binom{n}{k}\exp\bigl[i(\pi n - 2\pi k + \omega)x\bigr]dx\\ &=\int_{-1/2}^{1/2}\frac{1}{2^n}\left[\sum_{k=0}^n\binom{n}{k}e^{-2i\pi kx}\right]e^{i(n\pi+ \omega)x}dx\\ &=\frac{1}{2^n}\int_{-1/2}^{1/2}(1+e^{-2i\pi x})^ne^{i(n\pi+ \omega)x}dx\equiv \frac{1}{2^n}I(n,n\pi+\omega)\\ &=\frac{1}{2^n}\int_{-1/2}^{1/2}(1+e^{-2i\pi x})^n\left[\frac{e^{i(n\pi+ \omega)x}}{i(n\pi+\omega)}\right]'dx\\ &=\frac{1}{2^n}\left[(1+e^{-2i\pi x})^n\frac{e^{i(n\pi+ \omega)x}}{i(n\pi+\omega)}\right]_{-1/2}^{1/2}+\frac{1}{2^n}\int_{-1/2}^{1/2}2in\pi e^{-2i\pi x}(1+e^{-2i\pi x})^{n-1}\left[\frac{e^{i(n\pi+ \omega)x}}{i(n\pi+\omega)}\right]dx\\ &=\frac{1}{2^n}\frac{2\pi n}{n\pi+\omega}\int_{-1/2}^{1/2}(1+e^{-2i\pi x})^{n-1}e^{i(n\pi+\omega-2\pi)x}dx\\ &=\frac{1}{2^n}\frac{2\pi n}{n\pi+\omega}I(n-1,(n-2)\pi+\omega) \end{align}
Repeating this $n$ times, it follows: \begin{align} I(n,n\pi+\omega)&=n!\prod_{k=0}^{n-1}\frac{1}{\frac{\omega}{2\pi}+\frac{n}{2}-k}I(0,\omega-n\pi)\\ &=n!\prod_{k=0}^{n-1}\frac{-1}{k-\frac{\omega}{2\pi}-\frac{n}{2}}\int_{-1/2}^{1/2}e^{i(\omega-n\pi)x}dx\\ &=n!(-1)^{n}\prod_{k=0}^{n-1}\frac{1}{k-\frac{\omega}{2\pi}-\frac{n}{2}}\left[\frac{e^{i(\omega-n\pi)x}}{2\pi i(\frac{\omega}{2\pi}-\frac{n}{2})}\right]_{-1/2}^{1/2}\\ &=n!(-1)^{n+1}\prod_{k=0}^{n}\frac{1}{k-\frac{\omega}{2\pi}-\frac{n}{2}}\left[\frac{2i\sin\left(\frac{1}{2}(\omega-n\pi)\right)}{2\pi i}\right]\\ &=n!(-1)^{n+1}\frac{\Gamma(-\frac{\omega}{2\pi}-\frac{n}{2})}{\pi\Gamma(n+1-\frac{\omega}{2\pi}-\frac{n}{2})}\sin\left(\frac{1}{2}(\omega-n\pi)\right)\\ &=n!(-1)^{n+1}\frac{\Gamma(-\frac{\omega}{2\pi}-\frac{n}{2})}{\pi\Gamma(1-\frac{\omega}{2\pi}+\frac{n}{2})}\sin\left(\frac{1}{2}(\omega-n\pi)\right)\\ &=n!(-1)^{n+1}\frac{\Gamma(-\frac{\omega}{2\pi}-\frac{n}{2})}{\pi\Gamma(1-\frac{\omega}{2\pi}+\frac{n}{2})}(-1)^n\sin\left(\frac{1}{2}(\omega+n\pi)\right)\\ \end{align}
Conclusion: $$\sum=-\frac{n!\Gamma(-\frac{\omega}{2\pi}-\frac{n}{2})}{\pi\Gamma(1-\frac{\omega}{2\pi}+\frac{n}{2})}\sin\left((\frac{1}{2}(\omega+n\pi)\right)$$