= Solution
Write the endpoint condition in its linear <Robin boundary condition> form $y'(1)=y(1)/\alpha$. For a positive <eigenvalue> $\lambda=q^2$, the left <Dirichlet boundary condition> gives $y=C\sin(qx)$, and the right condition gives
$$
q\cos q=\frac1\alpha\sin q,\qquad\boxed{\tan q=\alpha q.}
$$
For each positive root $q_n$, the squared <norm> is
$$
N_n=\int_0^1\sin^2(q_nx)\,dx=\frac12-\frac{\sin(2q_n)}{4q_n}.
$$
Consequently the sine coefficients in any complete expansion are
$$
\boxed{a_n=\frac{\int_0^1f(x)\sin(q_nx)\,dx}{N_n}=\frac{2\int_0^1f(x)\sin(q_nx)\,dx}{1-\sin(2q_n)/(2q_n)}.}
$$
When $\alpha>1$, these are all modes and the <Sturm-Liouville eigenfunction expansion> gives exactly the stated sine series, with the usual convergence interpretation. Positivity of all <eigenvalues> follows from <integration by parts> and the <Cauchy-Schwarz inequality>:
$$
\int_0^1y(-y'')dx=\int_0^1(y')^2dx-\frac{y(1)^2}{\alpha}\geq\left(1-\frac1\alpha\right)\int_0^1(y')^2dx>0
$$
for a nonzero admissible $y$, since $y(1)=\int_0^1y'\,dx$.
However, the original PDF states only $\alpha>0$, and with that range its sine-only completeness claim is false. The missing modes are as follows. At $\alpha=1$, $\lambda=0$ admits $y=x$, so the correct expansion is
$$
\boxed{f(x)=3x\int_0^1t f(t)\,dt+\sum_{n\geq1}a_n\sin(q_nx),\qquad\alpha=1.}
$$
For $0<\alpha<1$, a negative <eigenvalue> $\lambda=-\kappa^2$ admits $y=\sinh(\kappa x)$, with
$$
\boxed{\tanh\kappa=\alpha\kappa,\qquad\kappa>0.}
$$
There is exactly one positive root: $\tanh\kappa/\kappa$ decreases strictly from $1$ to zero. For example, its <derivative> has the sign of $\kappa\operatorname{sech}^2\kappa-\tanh\kappa<0$, because $\sinh\kappa\cosh\kappa>\kappa$. Thus the corrected expansion is
$$
\boxed{f(x)=a_-\sinh(\kappa x)+\sum_{n\geq1}a_n\sin(q_nx),\quad a_-=\frac{\int_0^1f(t)\sinh(\kappa t)\,dt}{\sinh(2\kappa)/(4\kappa)-1/2},\quad0<\alpha<1.}
$$
These are the <nonpositive modes in a Dirichlet-Robin interval expansion>. As an explicit counterexample to the printed claim, at $\alpha=1$ take $f(x)=x$: it satisfies both <boundary conditions>, and each positive root obeys
$$
\int_0^1x\sin(q_nx)\,dx=-\frac{\cos q_n}{q_n}+\frac{\sin q_n}{q_n^2}=0
$$
because $\tan q_n=q_n$. Thus all $a_n$ vanish despite $f\ne0$. Likewise $f=\sinh(\kappa x)$ is a counterexample for $\alpha<1$, by <orthogonality of Sturm-Liouville eigenfunctions> with distinct <eigenvalues>. The formulas above give the full solution without silently assuming away the omitted modes.
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