Question:
Are there any useful interpretations or "applications" of the formula $$\intop_0^\infty e^{-ax}\frac{\sin x}{x}\,dx=\frac{\pi}{2}-\arctan(a)\qquad \forall\;a\in \mathbb{R}, $$ in which the left hand side is not defined in a usual way for $a<0$?
Explanation / Motivation:
Using basic techniques, one can show that $$\intop_0^\infty e^{-ax}\frac{\sin x}{x}\,dx=\frac{\pi}{2}-\arctan(a)\qquad \forall\;a\geq 0, $$ where the left hand side is well-defined as an improper Riemann integral. From this observation, it follows, for example, that the Dirichlet integral has the value $\intop_0^\infty \frac{\sin x}{x}\,dx=\frac{\pi}{2}$.
Now, the analytic function $a\mapsto \frac{\pi}{2}-\arctan(a)$ is defined on all of $\mathbb{R}$, while the improper Riemann integral exists only for $a\geq 0$. We thus may view the identity in my question above as a "forbidden analytic extension", in the same spirit as zeta function regularization leads to the funny formula $$ 1+2+3+\ldots=-\frac{1}{12}. $$ Apparently the latter does have some useful "applications" or "interpretations" in physics, although it looks like plain nonsense at first sight. Thus, maybe the same is true for the formula in my question above.
Note that the arctangent fulfills a functional equation ($\arctan(1/a)=\text{sgn}(a)\frac{\pi}{2}+\arctan a$), a property it has in common with the Riemann zeta function. Using the functional equation, we can re-write the formula from the question as
$$ \intop_0^\infty e^{-x/a}\frac{\sin x}{x}\,dx=\arctan a,\qquad a\in \mathbb{R} $$ with the convention $e^{-x/0}\equiv0$ for $x>0$, where again the left-hand side is undefined for $a<0$.
