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?
Best Answer
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.