= Solution
Take the negative-exponential <Half-range Fourier transform> and set
$$
\omega(k)=k^2-i\alpha k,\qquad\widehat q_0(k)=\frac1{a+ik},\qquad
G_j(k,t)=\int_0^t e^{\omega(k)s}\partial_x^jq(0,s)\,ds.
$$
The <advection-diffusion equation> has the <local relation> with flux $X=-q_x-(ik+\alpha)q$. Integration over the half-line gives the <half-line drift global relation>
$$
e^{\omega t}\widehat q(k,t)=\widehat q_0(k)-G_1(k,t)-(ik+\alpha)G_0(k,t),
\qquad\operatorname{Im}k\leq0.
$$
For the upper decay domain
$$
D_+=\{k:\operatorname{Im}k>0,\ \operatorname{Re}(k^2-i\alpha k)<0\},
$$
write $k=u+iv$. It is the region above $v=(\alpha+\sqrt{\alpha^2+4u^2})/2$, in particular above $v=\alpha$. Its boundary is oriented from upper-left infinity through $i\alpha$ to upper-right infinity, with the domain on the left. <Fourier inversion> and deformation of the boundary terms give
$$
q(x,t)=\frac1{2\pi}\int_{\mathbb R}e^{ikx-\omega t}\widehat q_0(k)\,dk
-\frac1{2\pi}\int_{\partial D_+}e^{ikx-\omega t}\bigl[G_1+(ik+\alpha)G_0\bigr]dk.
$$
Use the <dispersion symmetry elimination of a boundary trace> $\nu(k)=i\alpha-k$. It preserves $\omega$ and maps $D_+$ into the lower half-plane. Because $i\nu+\alpha=-ik$, the <global relation> at $\nu$ implies
$$
G_1=\widehat q_0(\nu)+ikG_0-e^{\omega t}\widehat q(\nu,t).
$$
The last term contributes an <holomorphic> upper-domain <integral> multiplied by $e^{ikx}$, so it vanishes upon closing the <contour> for $x>0$. The solution representation with only known data is consequently
$$
\boxed{q(x,t)=\frac1{2\pi}\int_{\mathbb R}\frac{e^{ikx-\omega t}}{a+ik}\,dk
-\frac1{2\pi}\int_{\partial D_+}e^{ikx-\omega t}\left[\frac1{a-\alpha-ik}+(2ik+\alpha)G_0(k,t)\right]dk.}
$$
For the boundary value $\cos t$, direct integration gives
$$
G_0(k,t)=\frac{e^{\omega t}(\omega\cos t+\sin t)-\omega}{\omega^2+1}.
$$
This quotient is an <entire function> of $\omega$ when its apparent singularities at $\omega=\pm i$ are filled by their limits. Those points are not genuine forcing poles of the finite-time transform.
To obtain an exponentially decaying spectral <integrand>, choose $0<\theta<\pi/4$ and the <exponential-decay contour for half-line drift diffusion>
$$
\Gamma:\quad k=i\alpha+r e^{i(\pi-\theta)}\quad(r:\infty\to0),\qquad
k=i\alpha+r e^{i\theta}\quad(r:0\to\infty).
$$
Deform $\partial D_+$ to $\Gamma$. No pole is crossed: the initial-data denominator in that boundary <integral> has its only pole at $k=i(\alpha-a)$, which lies below $i\alpha$ since $a>0$. The apparent singularities of the forcing transform are removable. In the swept outer sectors, $e^{-\omega t}G_0=\int_0^t e^{-\omega(t-s)}\cos s\,ds$ has the appropriate bounded growth for the deformation. The <contour> formula becomes
$$
\boxed{\begin{aligned}
q(x,t)={}&\frac1{2\pi}\int_{\mathbb R}\frac{e^{ikx-\omega t}}{a+ik}\,dk\\
&-\frac1{2\pi}\int_\Gamma e^{ikx}\left[
\frac{e^{-\omega t}}{a-\alpha-ik}
+(2ik+\alpha)\frac{\omega\cos t+\sin t-\omega e^{-\omega t}}{\omega^2+1}
\right]dk.
\end{aligned}}
$$
Evaluate the second quotient as a removable expression at $\omega=\pm i$, rather than splitting it into artificial residue terms. On either ray,
$$
|e^{ikx}|=e^{-\alpha x-rx\sin\theta},\qquad
\operatorname{Re}\omega=r^2\cos2\theta-\alpha r\sin\theta.
$$
Since $\cos2\theta>0$, the transient factors have Gaussian decay as $r\to\infty$. The nontransient forcing term is $O(r^{-1})e^{-rx\sin\theta}$ and thus also decays exponentially for every $x>0$. The real-axis <integral> has the Gaussian factor $e^{-k^2t}$. These estimates provide the requested improvement over integration along $\operatorname{Re}\omega=0$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-71-contours.png]
{title=Upper cubic-dispersion contour and exponentially decaying drift-diffusion contour, with their orientations}
The first panel shows the cubic domain from Question 1. The second uses $\alpha=1$ and $\theta=\pi/8$; its marked initial-data pole corresponds to $a=0.6$ and lies below both boundary <contours>.
An independent real-kernel form also makes the initial and boundary traces explicit. Apply the <Dirichlet gauge transform for constant drift>, $u=e^{\alpha x/2}q$, and then multiply by $e^{\alpha^2t/4}$ to obtain the <heat equation>. The odd-reflected <heat kernel> for initial data and its <Dirichlet boundary-forcing heat-kernel formula> yield
$$
\begin{aligned}
q(x,t)={}&\frac{e^{-\alpha x/2-\alpha^2t/4}}{\sqrt{4\pi t}}
\int_0^\infty\left[e^{-(x-y)^2/(4t)}-e^{-(x+y)^2/(4t)}\right]e^{(\alpha/2-a)y}\,dy\\
&+e^{-\alpha x/2}\int_0^t\frac{x\,e^{-x^2/(4\tau)-\alpha^2\tau/4}}{2\sqrt\pi\,\tau^{3/2}}\cos(t-\tau)\,d\tau.
\end{aligned}
$$
The first kernel tends to the initial delta kernel on $x>0$ as $t\downarrow0$ and vanishes at $x=0$. For the second kernel, the substitution $v=x/(2\sqrt\tau)$ gives the unit-mass approximate boundary delta as $x\downarrow0$, so its trace is $\cos t$; it vanishes initially for fixed $x>0$. Differentiating the kernels for $x,t>0$ verifies the <advection-diffusion equation>. The <integrals> converge for every positive $a,\alpha$, even when the exponentially weighted initial datum grows, because the spatial Gaussian controls that growth. Thus this gives the same decaying solution without introducing an additional restriction on $a$.
For a general decaying initial datum, an unweighted <Fourier sine transform> does not directly diagonalize the drift operator. Define
$$
S(k,t)=\int_0^\infty\sin(kx)q(x,t)\,dx,
\qquad C(k,t)=\int_0^\infty\cos(kx)q(x,t)\,dx.
$$
Two <integrations by parts> for $q_{xx}$ and one for $q_x$ give
$$
\boxed{S_t=-k^2S-\alpha kC+k\cos t.}
$$
The cosine transform remains unknown, so this is not a closed sine-transform evolution. In contrast, the <weighted sine transform for half-line drift diffusion> of $u=e^{\alpha x/2}q$ is closed:
$$
\widetilde S_t=-\left(k^2+\frac{\alpha^2}{4}\right)\widetilde S+k\cos t,
$$
$$
\widetilde S(k,t)=e^{-(k^2+\alpha^2/4)t}\int_0^\infty\sin(ky)e^{\alpha y/2}q_0(y)\,dy
+k\int_0^t e^{-(k^2+\alpha^2/4)(t-s)}\cos s\,ds.
$$
When the weighted initial transform exists, <Fourier sine inversion> gives $q(x,t)=e^{-\alpha x/2}(2/\pi)\int_0^\infty\sin(kx)\widetilde S(k,t)\,dk$. This method requires suitable decay of $e^{\alpha x/2}q_0$, not merely decay of $q_0$. For the exponential datum it is an ordinary integrable sine transform only when $a>\alpha/2$. At $a\leq\alpha/2$ one would need a generalized or regularized transform instead; the <contour> and real-kernel formulas above already work without that restriction. Therefore \b[the ordinary sine transform of $q$ is not a closed solution method; a weighted sine transform works under the stated extra weighted-decay condition].
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