= Solution
Let $d_n$ be the even <Fourier coefficient> of the initial displacement, with the normalization $\int_{-L}^L\cos^2(k_nx)dx=L$. Writing $s_n=\sin(n\pi/2)=(-1)^{(n-1)/2}$ gives
$$
\begin{aligned}
d_n&=\frac2L\int_0^Le^{-x/l}\cos(k_nx)\,dx\\
&=\boxed{\frac{2l}{L}\frac{1+k_nl\,s_ne^{-L/l}}{1+(k_nl)^2}}\qquad(n\text{ odd}).
\end{aligned}
$$
Zero initial velocity requires the real modal time factor
$$
T_n(t)=d_ne^{-\alpha t/2}\left[\cos(\omega_nt)+\frac{\alpha}{2\omega_n}\sin(\omega_nt)\right].
$$
One consistent choice in the question's complex representation is therefore
$$
\boxed{A_n=\frac{d_n}{2}\left(1-\frac{i\alpha}{2\omega_n}\right),\qquad B_n=\frac{d_n}{2}\left(1+\frac{i\alpha}{2\omega_n}\right).}
$$
For real $\omega_n$, $B_n=\overline{A_n}$; the real-part notation otherwise makes the complex coefficients nonunique. For imaginary $\omega_n$, the same formula is interpreted using <hyperbolic functions>. Its limit at $\omega_n=0$ is $d_ne^{-\alpha t/2}(1+\alpha t/2)$.
For $l/L\to0$ at fixed mode number,
$$
\boxed{d_n\simeq\frac{2l/L}{1+(n\pi l/2L)^2},}
$$
which can be substituted into $A_n,B_n$ above. The profile specified at $t=0$ is actually nonzero at $x=\pm L$. Thus for finite $l/L$ it cannot be the initial data of a solution continuous at the initial endpoint corners with fixed endpoints. The <Fourier series> reproduces it in the open interval, with zero endpoint values; it is an <L2 space> initial-data interpretation. The assumption $l\ll L$ makes this compatibility error exponentially small, rather than exactly zero.
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