Suppose one wants to evaluate the integral $$I = \int_0^\infty J_0(x) \, dx \ ,$$ where $J_0$ is the Bessel function of the first kind (od order zero). One seemingly simple way is to take well-known Lipschitz's integral, $$\int_0^\infty e^{-ax} J_0(bx) \, dx = \frac{1}{\sqrt{a^2 + b^2}} \quad (*)$$ valid for $a > 0$ and any real $b$, insert $b = 1$ and look at the limit $a \to 0^+$, which leads us to the correct result $I = 1$.
Side note: one way to prove (*) is to take integral representation of the function $J_0$ and exchange order of integration (cf. Watson: "Treatise on the Theory of Bessel Functions", page 384).
Now, the problem is how to prove that $$\lim_{a \to 0^+} \int_0^\infty e^{-ax} J_0(x) \, dx = \int_0^\infty \lim_{a \to 0^+} e^{-ax} J_0(x) \, dx \ ?$$ Monotone convergence theorem does not seem to be useful as $J_0$ is not a non-negative (nor non-positive) function, and dominated convergence theorem does not seem to be useful as $|J_0|$ is not integrable (cf. On the absolute integrability of Bessel functions).
Is this procedure, exchange of limit and integral above, even meaningful?
Note that by dominated convegence, we have that $$\lim_{a\to 0}\int_0^c e^{-ax}J_0(x)dx=\int_0^c J_0(x)dx.$$
Thus it suffices to show that $$\lim_{c\to \infty}\lim_{a\to 0}\int_c^{\infty}e^{-ax}J_0(x)dx=0.$$
As you have observed, there seems to be no cheap way to do this, but it isn't that bad employing the asymptotic for $J_0(x)$ are $x\to \infty$. We can write $$J_0(x)=\sqrt{\frac{2\pi}{x}}\cos(x-\pi/4)+\frac{1}{x^{3/2}}g(x),$$ where $g(x)$ a bounded function on $[1,\infty)$, such that say $|g(x)|\le C$.
Using this we see that
$$\int_c^{\infty}e^{-ax}J_0(x)dx=\int_c^{\infty}e^{-ax}\sqrt{\frac{2\pi}{x}}\cos(x-\pi/4)dx+\int_c^{\infty}\frac{e^{-ax}}{x^{3/2}}g(x)dx.$$
For the second integral, note that $$|\int_c^{\infty}\frac{e^{-ax}}{x^{3/2}}g(x)dx|\le C|\int_c^{\infty}\frac{1}{x^{3/2}}dx|\le \frac{C}{2c^{1/2}},$$ so that $$\lim_{c\to \infty}\lim_{a\to 0}\int_c^{\infty}\frac{e^{-ax}}{x^{3/2}}g(x)dx=0.$$
To bound the first integral, we can use integration by parts: $$\int_c^{\infty}e^{-ax}\sqrt{\frac{2\pi}{x}}\cos(x-\pi/4)dx=-\int_c^{\infty}\frac{\sqrt{2\pi}}{2x^{3/2}}\int_x^{\infty}e^{-ay}\cos(y-\pi/4)dy+\sqrt{\frac{2\pi}{c}}\int_c^{\infty}e^{-ay}\cos(y-\pi/4)dy.$$
Now we can compute that $$\int_{c}^\infty e^{-ax}cos(x-\pi/4)dx=e^{-ac}\frac{(1+a)cos(c)-(1-a)sin(c)}{\sqrt{2}(1+a^2)}.$$ We see from the right that for such that for $a<1$, $$|\int_c^{\infty}e^{-ay}\cos(y-\pi/4)dy|\le 4.$$
Thus
$$|\sqrt{\frac{2\pi}{c}}\int_c^{\infty}e^{-ay}\cos(y-\pi/4)dy|\le \sqrt{4\frac{2\pi}{c}}.$$ $$|\int_c^{\infty}\frac{\sqrt{2\pi}}{2x^{3/2}}\int_x^{\infty}e^{-ay}\cos(y-\pi/4)dy|\le-\int_c^{\infty}\frac{4\sqrt{2\pi}}{2x^{3/2}}\le \frac{4\sqrt{2\pi}}{c^{1/2}}.$$
Both of these expressions vanish taking interactive limits, so we see that $$\lim_{c\to \infty}\lim_{a\to 0}\int_c^{\infty}e^{-ax}\sqrt{\frac{2\pi}{x}}\cos(x-\pi/4)dx=0.$$
This completes the proof.