I am trying to prove the following relation in (3) where $\alpha,\beta,\gamma,\delta,\omega \in \mathbb{R}$. Given the integral
$$ I=\frac{1}{2}\int_0^\alpha dx \left( \beta \big(y'(x)\big)^2+\epsilon y^2\right), \quad \quad \quad (1) $$ using a Fourier Mode decomposition $$ y(\omega)=\int_0^\alpha dx e^{i\omega x} y(x),\quad \quad \quad (2) $$ show that the integral becomes $$ I=\frac{1}{2\alpha} \sum_\omega \big( \beta \omega^2+\epsilon\big)\big( y^2(\omega)\big) \quad \quad \quad (3) $$ where $\omega=\pm 2\pi n/\alpha, n=0,\pm 1,\pm 2,...$. I am having trouble proving (3). I tried just plugging (2) into (1), but didn't get anywhere after $$ I=\frac{1}{2}\int_0^\alpha dx \left(\beta\big(-i\omega\big)^2+\epsilon y^2(\omega) \right)=\frac{1}{2}\int_0^\alpha dx \left(\beta \omega^2+\epsilon y^2(\omega) \right) $$ It seems this integral becomes a sum along with a factor of $\alpha^{-1}$, but how?
You need to use the inverse Fourier transform, and please use different symbols for the function and its Fourier transform. If $$ Y(ω)=\int_0^α e^{i ω x} \, y(x) \,dx,\tag{2} $$ then $$ y(x)=\frac1{2π}\int_{-∞}^∞e^{-i ω x} \,Y(ω)\,dω\tag4 $$ and $$ y'(x)=\frac1{2π}\int_{-∞}^∞(-iω)e^{-i ω x} \,Y(ω)\,dω\tag5 $$