Solutions of the constant coefficient Helmholtz equation via the Fourier transform When $f$ is a rapidly decaying Schwartz function, 
$$
g(x) = \frac{1}{2\lambda} \int_{\mathbb{R}} \sin \left(2\lambda\left|x-y\right|\right) f(y)\ dy
$$
is an element of $C^\infty\left(\mathbb{R}\right)$ and a solution of the Helmholtz equation
$$
y''(t) + \lambda^2 y(t) = f(t).
\tag{1}
$$
Moreover, the Fourier transform of $g$ is the tempered distribution
which is the principal value of 
$$
\frac{\widehat{f}(\xi)}{\lambda^2-\xi^2}.
\tag{2}
$$
If we assume only that $f$ is in $L^1\left(\mathbb{R}\right)$, then $g$ is still an element of $C^\infty\left(\mathbb{R}\right)$ and it is still a solution of (1) (this follows easily from the dominated convergence theorem).  However, it is not clear to me that (2) is defined (the Fourier transform might not be Lipschitz continous, in which case the principal value integral might not exist).
I am not sure that (2) is not defined.  Is it the case that if $f$ is the Fourier transform of a $L^1\left(\mathbb{R}\right)$
function then 
$$
\lim_{\epsilon\to 0} \int_{|x|>\epsilon} \frac{f(x)}{x}\ dx
\tag{3}
$$
must exist?  It is easy to construct an example of a continuous function
$f(x)$ for which (3) does not exist -- for example,
$$
f(x) = \frac{\mbox{sign}(x)}{1+|\log(x)|}.
$$
I have not, however, been able to construct such a function and show that it is the image of an integrable function under the Fourier transform.
Assuming that (2) does not define a tempered distribution for all $f \in L^1\left(\mathbb{R}\right)$, is there some other way of interpreting (2)?  Clearly, one could obtain the Fourier transform of g as a limit of some tempered distributions.  For instance, let
$$
\phi(x) = \frac{1}{2\lambda i} \exp\left(\lambda i |x|\right)
$$
so that $g$ is the real part of $\phi*f$.  Also, let 
$$
\phi_t (x) = \frac{1}{2(\lambda+i t) i} \exp((\lambda+i t) i |x|).
$$
Since $\phi_t * f $ converges to $\phi *f$ in the space of tempered
distributions (in fact, it converges in $L^\infty\left(\mathbb{R}\right)$),
$$
\frac{\widehat{f}(\xi)}{(\lambda +i t)^2 - \xi^2}=\widehat{\phi_t} \widehat{f} = \widehat{\phi_t*f} \to \widehat{\phi *f }
\ \ \ \mbox{as} \ \ t \to 0
$$
in the space of tempered distributions.  And we can, of course, recover $\widehat{g}$ from $\widehat{\phi *f}$.
This is deeply unsatisfying to me for some reason --- perhaps because $g$ is defined as a convolution and I very much wish to interpret it as a product of Fourier transforms.
 A: Regarding (2), it is not true that if $f=\widehat{g}$, for some $g\in L^{1}(\mathbb{R})$, then
$$\lim_{\epsilon\downarrow 0}\int_{\left|x\right|\geq\epsilon}\dfrac{f(x)}{x}dx$$
exists. To see this, consider $g=\chi_{[0,1]}$ (i.e. the characteristic function of the unit interval). Then
$$\widehat{g}(\xi)=\int_{0}^{1}e^{-2\pi i y\xi}dy=-\dfrac{e^{-2\pi i\xi}-1}{2\pi i\xi}$$
Take $f:=\widehat{g}$, and fix $\epsilon>0$. Since $\left|e^{-2\pi i x}-1\right|\leq 2$ and $\left|\xi\right|^{-2}$ is integrable outside a neighborhood of the origin, it suffices to consider the integral over the annulus $1\geq\left|x\right|\geq\epsilon$. Observe that
$$\int_{1\geq\left|x\right|\geq\epsilon}\dfrac{f(x)}{x}dx=\dfrac{1}{2\pi i}\int_{\epsilon}^{1}\left[\dfrac{e^{-2\pi i x}-1}{\left|x\right|^{2}}-\dfrac{e^{2\pi ix}-1}{\left|x\right|^{2}}\right]dx=\dfrac{1}{\pi i}\int_{\epsilon}^{1}\dfrac{\cos(2\pi x)}{\left|x\right|^{2}}dx$$
The integral on the RHS blows up as $\epsilon\downarrow 0$, since $\cos(2\pi x)$ is bounded from below in a neighborhood of the origin and $\int_{\epsilon}^{1}\left|x\right|^{-2}dx=\epsilon^{-1}-1$.
