I have the equations
$$ \int\limits_0^\infty dk \ A(k)k \sinh(ka)\cos(kx)=0 \ \ ; \ \ 1<|x|<\infty \tag{1} $$ $$ \int\limits_0^\infty dk \ A(k) \cosh(ka) \cos(kx)=1 \ \ ; \ \ |x|<1 \tag{2} $$
Where $A(k)$ is unknown, $a$ is a real positive constant, and we expect $A=A(k,a)$. I think the general idea is to write
$$ \left[ A(k) \times \operatorname{convenient stuff}(k)\right]=\int\limits_0^1dx' \ g(x') \cos(kx') \tag{*} $$
Where $g$ is the new unknown, and (*) is chosen such that (1) is satisfied, then substitute this into (2) so that the integral over $k$ may be performed.
Question in brief: What substitution in (*) will allow me to make progress? Or is there a different 'trick' here altogether? Maybe an expansion on the right in something other than cosines?
The rest of this post is background, and my current attempt.
In a similar problem
$$ \int\limits_0^\infty dk \ A(k) k \sin(kx) =0 \ \ ; \ \ 1<|x|<\infty \tag{3} $$
$$ \int\limits_0^\infty dk \ A(k) \tanh(ka)\sin(kx)=1 \ \ ; \ \ |x|<1 \tag{4} $$
Setting
$$ A(k) k = \int\limits_0^1 dx' \ g(x') \sin(kx') $$
Allows one to make progress by substituting into (4), and even (eventually) obtain $g$ analytically. I have tried the similar substitution
$$ A(k) k \sinh(ak)=\int\limits_0^1 dx' g(x') \cos(kx') $$
So that (1) is satisfied, but substituting this into (2) is not fruitful (divergent integral over $k$).
It is possible that the Fourier expansion in $\cos$ on the RHS of (*) is not the way to go here. Maybe expanding in Bessel functions or something else; but I don't have the intuition to 'see' what is a good expansion. For example, in a (different) similar problem, a useful expansion is
$$ \left[ A(k) \times \operatorname{convenient stuff}(k)\right]=\int\limits_0^1dx' \ g(x') J_0(kx') $$
Where $J_0$ is a Bessel function of the first kind. The similar problems mentioned above are from: Mixed boundary value problems by Dean G. Duffy.
Current progress
By integrating (1) and (2) w.r.t. $x$ we obtain
$$ \int\limits_0^\infty dk \ A(k) \sinh(ka)\sin(kx)=0 \ \ ; \ \ 1<|x|<\infty \tag{1a} $$ $$ \int\limits_0^\infty dk \ A(k) k^{-1} \cosh(ka) \sin(kx)=x \ \ ; \ \ |x|<1 \tag{2a} $$
With the ansatz
$$ A(k) \cosh(ka) = \int\limits_0^1 dx' \ g(x') \sin(kx') \tag{**} $$
After substituting (**) into (2a), and some algebra, we find
$$ A(k) \cosh(ka) = J_1(k) $$
However, to justify this we must show it satisfies (1a)
$$ \int\limits_0^\infty dk \ \int\limits_0^1 dx' \ g(x') \sin(kx') \sin(kx) \tanh(ka) \stackrel{?}{=}0 \ \ ; \ \ 1<|x|<\infty \tag{1b} $$
Within the integral, we always have $x \neq x'$, so it would be nice to say (distributionally) this is true, but I'm not sure that it is. Alternative question: is (1b) true?
Update: Numerically integrating (1b) shows that it's not true.