I am investigating the following type of functions
\begin{equation}I(\alpha) = \int_{0}^{\pi}f(t)\cos(\alpha t)\,\mathrm{d}t\,.\end{equation}
where $f(t)$ is a real-valued function with non-negative values, no symmetry and no periodicity.My question is the following: what are the conditions/properties that $f(t)$ needs to fulfill, so that $I(\alpha)$ is periodic, say with period length $L$?
In general this means that $I(\alpha) = I(\alpha + L)\,\forall\alpha$. Inserting this into the above equation and accounting for $\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$ results in
\begin{equation}I(\alpha) \overset{!}{=} I(\alpha+L) = \int_{0}^{\pi}f(t)\cos\left([\alpha+L] t\right)\,\mathrm{d}t = \int_{0}^{\pi}f(t)\cos\left(\alpha t\right)\cos\left(L t\right)\,\mathrm{d}t - \int_{0}^{\pi}f(t)\sin\left(\alpha t\right)\sin\left(L t\right)\,\mathrm{d}t\end{equation}
Further, acounting for $1-\cos(x) = 2\sin^{2}\left(x/2\right)$ and with\begin{equation}I(\alpha) - \int_{0}^{\pi}f(t)\cos\left(\alpha t\right)\cos\left(L t\right)\,\mathrm{d}t = \int_{0}^{\pi}f(t)\cos(\alpha t)\left[1-\cos\left(L t\right)\right]\,\mathrm{d}t = 2\int_{0}^{\pi}f(t)\cos(\alpha t)\sin^{2}\left(Lt/2\right)\,\mathrm{d}t\end{equation}
results in
\begin{equation}\begin{split}&2\int_{0}^{\pi}f(t)\cos(\alpha t)\sin^{2}\left(Lt/2\right)\,\mathrm{d}t = - \int_{0}^{\pi}f(t)\sin\left(\alpha t\right)\sin\left(L t\right)\,\mathrm{d}t\\&\Leftrightarrow \int_{0}^{\pi}2f(t)\cos(\alpha t)\sin^{2}\left(Lt/2\right) + f(t)\sin\left(\alpha t\right)\sin\left(L t\right) \,\mathrm{d}t = 0\end{split}\end{equation}
Which would imply that
\begin{equation}\begin{split}2f(t)\cos(\alpha t)\sin^{2}\left(Lt/2\right) + f(t)\sin\left(\alpha t\right)\sin\left(L t\right) = f(t)\left[2\cos(\alpha t)\sin^{2}\left(Lt/2\right) + \sin\left(\alpha t\right)\sin\left(L t\right)\right] = 0\end{split}\end{equation}
Since I don't consider the trivial case ($f$ is not a zero-only function) this would mean that finding conditions/properties such that $I(\alpha) \overset{!}{=} I(\alpha+L)$ would be independent of the function $f$ itself. This is obviously not what I want. Unfortunately, at this point I am stuck. Do you have any suggestions how I can resolve this contradcition and how to continue in order to obtain a set of conditions/properties that $f(t)$ needs to fulfill so that $I(\alpha) = I(\alpha + L)\,\forall\alpha$? Thank you!