I want to calculate the Fourier transform of the Fejer kernel $\mathcal{F}_{N}$ on the real line, which is given by $$ \mathcal{F}_N\left(x\right) = \begin{cases} N\left[\,{\frac{\sin\left(\,{\pi xN}\,\right)}{\pi xN}}\,\,\right]^{2}, & \text{if}\ x\neq 0 \\[2ex] N, & \text{if}\ x = 0 \end{cases} $$
I want to find $$ \hat{\mathcal{F}}_{N}\left(n\right) = N\int\left[\frac{\sin\left(\pi xN\right)\,\,} {\pi xN}\right]^{2} \mathrm{e}^{-2\pi\mathrm{i}xn}\,\,\,\mathrm{d}x\,, $$ but so far I can't think of any transformations to compute this integral. I would greatly appreciate any help.
Let us use the identity $$\frac{\sin ax}{a x} = \frac{1}{2a}\int_{-a}^{a} e^{i x\xi}\,\mathrm{d}\xi $$ with $a=\pi N$. Subsituting in the definition \begin{align} \hat{\mathcal{F}}_N(n) &= N\int_{-\infty}^{\infty} \left(\frac{\sin \pi x N}{\pi x N}\right)^2 e^{-2\pi i x n} \mathrm{d}x\\ &= N\int_{-\infty}^{\infty} \left(\frac{1}{2a}\int_{-a}^{a} e^{i x\xi_1}\,\mathrm{d}\xi_1\right) \left(\frac{1}{2a}\int_{-a}^{a} e^{i x\xi_2}\,\mathrm{d}\xi_2\right) e^{-2\pi i x n} \mathrm{d}x\\ &= \frac{N}{4a^2}\int_{-\infty}^{\infty} \mathrm{d}x \int_{-a}^{a}d\xi_1 \int_{-a}^{a}d\xi_2 e^{i x(\xi_1+\xi_2 - 2\pi n)}\,. \end{align} Rearranging the order of the integrals we may recognize the $\delta$ function, which may be generally written as $$ \delta(z-\alpha)=\frac{1}{2\pi} \int_{-\infty}^\infty e^{ik(z-\alpha)}\ dk =\frac{1}{2\pi} \int_{-\infty}^\infty e^{ik(z-\alpha)}\ dk\,, $$ with the identification $k\equiv x$ and $z-\alpha \equiv \xi_1+\xi_2-2\pi n$. Since $a=\pi N$, this leads to $$\hat{\mathcal{F}}_N(n) = \frac{1}{2\pi N} \int_{-\pi N}^{\pi N}d\xi_1 \int_{-\pi N}^{\pi N}d\xi_2 \,\delta(\xi_1+\xi_2 - 2\pi n)\,. $$ The integral is proportional to the length of the diagonal where $\xi_2=2\pi n - \xi_1$ within the square with edges at $\pm\pi N$. The result is zero if $|n|>N$ and nonzero otherwise. To make the calculation simpler we may shift the coordinates by changing the integration variable as $\xi_{1,2}=\xi'_{1,2}-N\pi$, $$\hat{\mathcal{F}}_N(n) = \frac{1}{2\pi N} \int_{0}^{2\pi N}d\xi'_1 \int_{0}^{2\pi N}d\xi'_2 \,\delta\left(\xi'_1+\xi'_2 - 2\pi \left(n+N\right)\right)\,. $$ This shows that the length of the side is $2\pi(n+N)$ if $0\leq 2\pi(n+N)\leq 2\pi N$ and decreases otherwise if $2\pi N\leq 2\pi(n+N)\leq 4\pi N$ \begin{equation} \hat{\mathcal{F}}_N(n) = \left\{ \begin{array}{cl} 0 & {\rm if~} |n|>N\\ \frac{n-N}{N} & {\rm if~} -N\leq n \leq 0\\ 1-\frac{n-N}{N} & {\rm if~} 0\leq n \leq N \end{array} \right. \end{equation} This may be simplified as \begin{equation} \hat{\mathcal{F}}_N(n) = \left\{ \begin{array}{cc} 0 & {\rm if~} |n| \geq N \\ 1-\frac{|n|}{N} & {\rm if~} |n| \leq N \end{array} \right.\,. \end{equation}
We may verify the result for $n=0$ and $n=\pm N$ by direct calculation \begin{align} \hat{\mathcal{F}}_N(0) &= N\int_{-\infty}^{\infty} \left(\frac{\sin \pi x N}{\pi x N}\right)^2 \mathrm{d}x = 1 \end{align} and \begin{align} \hat{\mathcal{F}}_N(\pm N) &= N\int_{-\infty}^{\infty} \left(\frac{\sin \pi x N}{\pi x N}\right)^2 \cos(\pm 2\pi x N) \mathrm{d}x \\ &= \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{\sin^2 \theta}{\theta^2} \cos(\pm 2\theta) \mathrm{d}\theta \\ &= \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{\sin^2\theta}{\theta^2} (1-2\sin^2\theta) \mathrm{d}\theta\\ &= \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{\sin^2\theta}{\theta^2} \mathrm{d}\theta -\frac{2}{\pi}\int_{-\infty}^{\infty} \frac{\sin^4\theta}{\theta^2} \mathrm{d}\theta = 1 - 1 = 0\,, \end{align} as required.