Convolution theorem for a function that doesn't vanish at infinity I'm studying a paper of Rapp and Kassal (1968), in which a second order differential function that doesn't vanish at $\pm \infty$ is solved. I will put here the parts of the paper that matter:
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I think that there's a typo at $(26)$: the "$\lim_{t\to+\infty}$" shouldn't appear there.
The differential equation that I'm trying to solve is $(24)$ which has the intial condition $(25)$. It seems that the authors of the paper applied a Fourier transform to $(24)$ and used the convolution theorem to get the solution $(26)$.
Let $y(t)=\tilde{y}(t)-\tilde{y}_0$. Let also $h(t)=-\frac{1}{\mu}\frac{\gamma A''}{L}\text{sech}^2\left(\frac{v_o t}{2L}\right)$ and $\omega=\sqrt{\frac{f}{\mu}}$. 
I will define the Fourier transform of a function $f(t)$ as:
$$F(\xi)=\mathscr{F}[f(t)]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}f(t)e^{-i\xi t}\,dt$$
Applying the Fourier trasform to $(24)$, we would get:
$$\frac{1}{\sqrt{2\pi}}\left[\dot{y}(t)+i\xi y(t)\right]e^{-i\xi t}\bigg\rvert_{t=-\infty}^{t=+\infty}-\xi^2Y(\xi)+w^2Y(\xi)=H(\xi)\Leftrightarrow$$
$$\Leftrightarrow Y(\xi)=-\frac{1}{\sqrt{2\pi}}\left[\dot{y}(t)+i\xi y(t)\right]e^{-i\xi t}\bigg\rvert_{t=-\infty}^{t=+\infty}G(\xi)+ H(\xi)G(\xi) $$
where $G(\xi)=\frac{1}{-\xi^2+\omega^2}$.
And then, by using the convolution theorem we would have:
$\tilde{y}(t)-\tilde{y}_0=$
$-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\mathscr{F}^{-1}\left\{\frac{1}{\sqrt{2\pi}}\left[\dot{y}(t)+i\xi y(t)\right]e^{-i\xi t}\bigg\rvert_{t=-\infty}^{t=+\infty}\right\}g(t-s)\,ds+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}h(s)g(t-s)\,ds$
My main problem is how to deal with the $\mathscr{F}^{-1}\left\{\frac{1}{\sqrt{2\pi}}\left[\dot{y}(t)+i\xi y(t)\right]e^{-i\xi t}\bigg\rvert_{t=-\infty}^{t=+\infty}\right\}$ term. It should only depend on $s$. But for that, I would need to remove the dependency on $t$, which I don't know if it is even possible (since the operand function isn't convergent). Is there any other way to achive $(26)$? My attempt seems not to be a good one.
[Note: I actually tried to find the procedure to get the relation $(26)$ from $[11]$ reference, but I wasn't successful.]
 A: Consider the function, related to the variation of constant method,
$$
g_t(s)=\cos(ω(t-s))(y(s)-\bar y)+\frac{\sin(ω(t-s))}{ω}y'(s).
$$
Then
$$
g_t'(s)=\frac{\sin(ω(t-s))}ω[ω^2(y(s)-\bar y)+y''(s)]=\frac{\sin(ω(t-s))}{ωμ}f(s)
$$
Now $g_t(t)=y(t)-\bar y$ so that by integration
$$
y(t)-\bar y-g_t(a)=\frac{1}{ωμ}\int_a^t\sin(ω(t-s))f(s)\,ds
$$
or
$$
y(t)-\bar y=\cos(ω(t-a))(y(a)-\bar y)+\frac{\sin(ω(t-a))}{ω}y'(a)+\frac{1}{ωμ}\int_a^t\sin(ω(t-s))f(s)\,ds
$$
Now if $a$ and $t$ are both close to $-\infty$, then the right side forcing function is very small, its contribution goes to zero. The integral from $-\infty$ to $a$ is just as small, as the function $f(s)$ in the integrand is essentially a falling exponential, looking in direction $s\to -\infty$. This means that the bounded oscillatory behavior demanded by
$$
\lim_{t\to-\infty}\Bigl[y(t)-\bar y -B_i\sin(ωt+\delta_i)\Bigr]=0
$$
has to come from the first terms,
$$
\cos(ω(t-a))(y(a)-\bar y)+\frac{\sin(ω(t-a))}{ω}y'(a)\approx B_i\sin(ωt+δ_i)
$$
which then gives the solution without limit
$$
y(t)-\bar y=B_i\sin(ωt+δ_i)+\frac{1}{ωμ}\int_{-\infty}^t\sin(ω(t-s))f(s)\,ds
$$

I do not think that you should read the limits in the cited image as the standard function limits. Here this symbolism seems to stand for "behaves asymptotically like". As the last solution formula is exact for all $t$, it is also asymptotically true for $t\to+\infty$.
