Regulating $\int_0^\infty \sin x \, \mathrm{d} x$
Here is a partial answer that seconds @Jack D'Aurizio's idea.
Let us focus on the class of locally integrable functions $f : [0,\infty) \to \Bbb{R}$ such that
- $f(0) = 1 = \lim_{x \to 0^+} f(x)$, and
- $ I(f, \epsilon) := \lim_{R\to\infty} \int_{0}^{R} f(\epsilon x) \sin x \, dx$ converges for all $\epsilon > 0$.
For appropriate choices of $f$, this covers many classical summability methods:
If $f(x) = \mathbf{1}_{[0,1]}(x)$, then $I(f,\epsilon) = \int_{0}^{1/\epsilon} \sin x \, dx$ corresponds to the ordinary definition of improper integrability.
If $f(x) = (1 - x)_+$, then $I(f, \epsilon) = \epsilon \int_{0}^{1/\epsilon} \int_{0}^{x} f(t) \, dtdx $ corredponds to the Cesaro summability.
If $f(x) = e^{-x}$, then $I(f, \epsilon) = \int_{0}^{\infty} e^{-\epsilon x} \sin x \, dx$ corresponds to the Abel summability.
It also covers all the examples you have discussed. We can simply choose $f(x)$ to be
$$ \frac{1}{1+x^2}, \quad \operatorname{sech}^2 x, \quad (1 + 2x^2)e^{-x^2}, \quad \cdots. $$
Since we know that 1 fails to converge while 2 and 3 converge, we may suspect that regularity of $f$ is one factor which determines how regularly $I(f,\epsilon)$ behaves. Toward this direction, let us make the following assumption:
Assumption 1. We have $f(x) = \int_{x}^{\infty} \rho(t) \, dt$ for some $\rho$ which is integrable on $[0, \infty)$.
Under this condition, let us show that $I(f, \epsilon)$ converges to $1$. Indeed, by the Fubini's theorem, we have
\begin{align*} \int_{0}^{R} f(\epsilon x) \sin x \, dx &= \int_{0}^{R} \left( \int_{\epsilon x}^{\infty} \rho(t) \, dt \right) \sin x \, dx \\ &= \int_{0}^{\infty} \left( \int_{0}^{(t/\epsilon) \wedge R} \sin x \, dx \right) \rho(t) \, dt \\ &= \int_{0}^{\infty} [ 1 - \cos ((t/\epsilon) \wedge R) ] \rho(t) \, dt. \end{align*}
Now we can apply the dominated convergence theorem to the last integral to find that $I(f, \epsilon)$ converges and is given by
$$ I(f, \epsilon) = \int_{0}^{\infty} ( 1 - \cos (t/\epsilon) ) \rho(t) \, dt. $$
Taking $\epsilon \to 0^+$, Riemann-Lebesgue lemma tells that the last integral converges to:
$$ \lim_{\epsilon \to 0^+} I(f, \epsilon) = \int_{0}^{\infty} \rho(t) \, dt = f(0^+) = 1. $$
This proof already hints us that this regularization captures the notion of the average of the antiderivative $\int_{0}^{x} \sin t \, dt = 1 - \cos x$. Indeed, with a bit of extra assumption, we can check that this regularization yields the integral of a nice function in Cesaro summability sense. Let us formalize this claim.
Assumption 2.
- $f$ satisfies Assumption 1 with $\rho : [0, \infty) \to \Bbb{R}$ which is of bounded variation and satisfies $\int_{[0,\infty)} x |d\rho(x)| < \infty$.
- $\varphi : [0,\infty) \to \Bbb{R}$ is locally integrable such that the antiderivative $\Phi(x) = \int_{0}^{x} \varphi(t) \, dt$ is bounded and has Cesaro-mean $\ell \in \Bbb{R}$, i.e., $\frac{1}{R} \int_{0}^{R} \Phi(x) \, dx \to \ell$ as $R \to \infty$.
Although this assumption seems technical, it loosely tells that $\rho$ does not oscillate heavily and that $\Phi$ is bounded and has an average.
As in the previous proof, we find that
$$ \int_{0}^{R} f(\epsilon x) \varphi(x) \, dx = \int_{0}^{\infty} \Phi((t/\epsilon)\wedge R) \rho(t) \, dt \xrightarrow[R\to\infty]{} \int_{0}^{\infty} \Phi(t/\epsilon) \rho(t) \, dt. $$
Now from the assumption that $\rho$ is of bounded variation, we easily find that $\rho(t) \to 0$ as $t \to \infty$. Thus by the Fubini's theorem again,
\begin{align*} \int_{0}^{\infty} \Phi(t/\epsilon) \rho(t) \, dt &= \int_{0}^{\infty} \Phi(t/\epsilon) \left( - \int_{(t,\infty)} d\rho(u) \right) \, dt \\ &= \int_{(0,\infty)} \left( \int_{0}^{u} \Phi(t/\epsilon) \, dt \right) \, d\rho(u) \\ &= \int_{(0,\infty)} \left( \frac{\epsilon}{u} \int_{0}^{u/\epsilon} \Phi(t') \, dt' \right) u \, d\rho(u) \end{align*}
where $t' = t/\epsilon$. Then by the assumption, we can invoke the dominated convergence theorem to find that the last integral converges to:
$$ \lim_{\epsilon \to 0^+} \int_{(0,\infty)} \left( \frac{\epsilon}{u} \int_{0}^{u/\epsilon} \Phi(t') \, dt' \right) u \, d\rho(u) = \int_{(0,\infty)} \ell u \, d\rho(u) = \ell. $$
Putting together, we have
$$ \lim_{\epsilon \to 0^+} \int_{0}^{\infty} f(\epsilon x) \varphi(x) \, dx = \ell = \lim_{R\to\infty} \frac{1}{R}\int_{0}^{R} \int_{0}^{x} \varphi(t) \, dtdx. \tag{*}$$
Let $f(\epsilon x)$ a function which decays to zero (to be safe, i think it is sufficent if it can be also fastly oscillating) as $x\rightarrow \infty$ with $f(0)=1$. Furthermore uit should not contain any singular points. Then we have
$$ F(\epsilon)=\int_0^{\infty}\sin(x)f(x\epsilon)\, \mathrm dx= \frac{1}{\epsilon}\int_0^{\infty}\sin\left({\frac{x}{\epsilon}}\right)f(x)\, \mathrm dx=\frac{1}{\epsilon}\Im\int_0^{\infty}\exp\left({\frac{i x}{\epsilon}}\right)f(x) \, \mathrm dx $$
if $\epsilon$ goes to zero this is a classic example for a integral which might be tackled by the method of stationary phase and the dominating contribution will be from the boundary point at $x=0$ (see formula 2.22 in the given reference)
$$ F(\epsilon)\underset{\epsilon \to 0}{\sim}\frac{1}{\epsilon}\Im\left(\epsilon\frac{f(0)}{-i} \right)+\mathcal{O}(\epsilon)=1+\mathcal{O}(\epsilon) $$
This is because heuristically the heaviy oscillations tend to cancel out each other (but to get them you need an additional small parameter $\epsilon$)
Suppose $f\in C^1([0,\infty),$ $\lim_{x\to \infty} f(x) = 0,$ and both $f,f' \in L^1([0,\infty)).$ Then
$$\tag 1 \lim_{\epsilon\to 0} \int_0^\infty f(\epsilon x) \sin x\, dx = f(0).$$
Proof: Integrating by parts shows the integral equals
$$f(\epsilon x)(-\cos x)\big |_0^\infty + \int_0^\infty \epsilon f'(\epsilon x) \cos x\, dx = f(0) + \int_0^\infty f'(y) \cos (y/\epsilon)\, dy.$$
As $\epsilon \to 0,$ the last integral $\to 0$ by the Riemann-Lebesgue lemma. This gives $(1).$
Remarks: 1. In the cases $f(x) = e^{-x}, 1/(1+x^2), \mathrm {sech}^2 x, (1+2x^2)e^{-x^2},$ $(1)$ gives the results you mentioned (although with $\sqrt \epsilon$ in place of $\epsilon$ for the second and fourth functions).
- The hypothesis $\lim_{x\to \infty} f(x) = 0$ is not really needed, because it's implied by the other hypotheses. This is because $f'\in L^1$ implies $f$ is uniformly continuous, and a uniformly continuous function in $L^1([0,\infty))$ must vanish at $\infty.$ I included the $f(x)\to 0$ hypothesis to keep the proof simple.
$(1)$ can be generalized: Keep the hypotheses on $f$ as above, and assume $G\in C^1([0,\infty)),$ with both $G,G'$ bounded. Then
$$\tag 2 \lim_{\epsilon\to 0} \int_0^\infty f(\epsilon x) G'(x)\, dx = -G(0)f(0).$$
For example, taking $G(x) = - \cos x$ gives $(1).$ The proof of $(2)$ is very much like that of $(1);$ we need an analogue of Riemann-Lebesgue, but that's straightforward.