I'm solving a PDE by using the method of separation of variables, and this led me to a Sturm-Liouville of the form $$ \dfrac{d^2f}{dx^2} + \lambda f = 0 \\[15pt] \left\{\begin{array}{l} \dfrac{df}{dx}(0) - \alpha f(0) = 0 \\[5pt] \dfrac{df}{dx}(\pi) + \alpha f(\pi) = 0 \end{array}\right.\ ,\quad \alpha > 0 $$
If I study the case where $\lambda = 0$ and where $\lambda < 0$, both of them results in $f$ being trivial, so no interest here. However, if $\lambda > 0$ then I get that infinite $\lambda_n$ exists (though, they are difficult to calculate exactly) such that $f$ is not trivial, and then I get a set of solutions with elements $$ f_n(x) = A_n\cos(\sqrt{\lambda_n}x) + B_n\sin(\sqrt{\lambda_n}x)\ . $$
My question: how can I apply the orthogonality property of eigenfunctions to express some initial condition as a generalized Fourier series?
I know that if $\phi_n$ and $\phi_m$ are the eigenfunctions related to eigenvalues $\lambda_n$ and $\lambda_m$, with $n \neq m$, then $$ \int_0^\pi \phi_n(x)\phi_m(x)\ dx = 0\ , $$ which usually allow us to get an expression for the coefficients of the Fourier series. However, I don't know if the eigenfunctions are the sines and cosines separately, or the $f_n$ combination itself. If I take them separately, then I guess that $$ \int_0^\pi \cos(\sqrt{\lambda_n}x)\sin(\sqrt{\lambda_m}x)\ dx = 0\ ,\quad\forall n,m $$ but what about $$ \int_0^\pi \cos(\sqrt{\lambda_n}x)\cos(\sqrt{\lambda_m}x)\ dx $$ and $$ \int_0^\pi \sin(\sqrt{\lambda_n}x)\sin(\sqrt{\lambda_m}x)\ dx\ , $$ which probably lead to different values if $n = m$ or $n \neq m$. I don't know how to correctly get a Fourier series from this if I don't have an explicit expression for $\lambda_n$.
Consider a solution $f$ of $f''+\lambda f=0$ with $$ f'(0)-\alpha f(0) = 0 \\ f'(0)+\alpha f(0) = 1. $$ The second condition is permitted without loss of generalization, just to provide normalization. This amounts to $$ f'(0) = 1/2,\;\; f(0)=1/2\alpha. $$ This determines a solution of the form $$ f_{\lambda}(x) = A\cos(\sqrt{\lambda}x)+B\sin(\sqrt{\lambda}x) \\ A=1/2\alpha,\;\; B\sqrt{\lambda}=1/2 \\ \implies f_{\lambda}(x) = \frac{1}{2\alpha}\cos(\sqrt{\lambda}x)+\frac{1}{2\sqrt{\lambda}}\sin(\sqrt{\lambda}x). $$ Notice that $f_{\lambda}(x)$ is an entire function of $\lambda$ with a limiting value of $\frac{1}{2\alpha}+\frac{x}{2}$ at $\lambda=0$, which is the correct solution for the problem where $\lambda=0$. The second condition $f'(0)+\alpha f(0)=1$ determines the eigenvalues to be the zeros of the entire function of $\lambda$ given by $$ 0 = f_{\lambda}'(\pi)+\alpha f_{\lambda}(\pi)= \\ = -\frac{\sqrt{\lambda}}{2\alpha}\sin(\sqrt{\lambda}\pi)+\frac{1}{2}\cos(\sqrt{\lambda}\pi)+\frac{1}{2}\cos(\sqrt{\lambda}\pi)+\frac{\alpha}{2\sqrt{\lambda}}\sin(\sqrt{\lambda}\pi) \\ = \frac{1}{2}(-\frac{\sqrt{\lambda}}{\alpha}+\frac{\alpha}{\sqrt{\lambda}})\sin(\sqrt{\lambda}\pi)+\cos(\sqrt{\lambda}\pi) \\ =\frac{1}{2}(\alpha-\frac{\lambda}{\alpha})\frac{\sin(\sqrt{\lambda}\pi)}{\sqrt{\lambda}}+\cos(\sqrt{\lambda}\pi). $$ There are solutions $\lambda_1 < \lambda_2 < \lambda_3 < \cdots$, and it is the eigenfunctions $\{ f_{\lambda_n} \}$ that are mutually orthogonal, not the component functions $\cos$ and $\sin$. These eigenfunctions $\{ f_{\lambda_n} \}$ can be used to expand a function in a series: $$ f = \sum_{n=1}^{\infty}\frac{\langle f,f_{\lambda_n}\rangle}{\langle f_{\lambda_n},f_{\lambda_n}\rangle} f_{\lambda_n}. $$