= Airy disk normalization and encircled energy
{c}
{title2=$E(U)=1-J_0(U)^2-J_1(U)^2$}
Peak normalization $P(0)=1$ does not make the <integral> of a <point spread function> equal to one. In the <paraxial approximation> to the <focal plane>, define $b=\pi D/(\lambda f)$ and $u=br$. The unit-<integral> <point spread function normalization> is
$$
p(r)=\frac{b^2}{4\pi}P(br),\qquad \int_{\mathbb R^2}p(r)\,d^2\mathbf r=1.
$$
Here $p$ has units of inverse <area>; multiplying by the detected <radiant flux> gives image-plane <irradiance>. Its <encircled energy>, meaning the fraction inside radius $R=U/b$, is
$$
E(U)=\int_0^{U/b}2\pi r p(r)\,dr=\frac12\int_0^UuP(u)\,du=1-J_0(U)^2-J_1(U)^2.
$$
This is an <integral> over <area>, so its radial weight is $u$. An unweighted <integral> of a line through the centre is a different quantity.
A short proof uses the <Bessel function of the first kind> identities $J_0'=-J_1$ and $J_1'=J_0-J_1/u$. They give
$$
\frac{d}{dU}\left(1-J_0(U)^2-J_1(U)^2\right)=\frac{2J_1(U)^2}{U}=\frac U2P(U).
$$
Both expressions for $E$ vanish at zero. Since these <Bessel functions of the first kind> tend to zero as $U\to\infty$, $E(\infty)=1$, establishing the normalization. At the first dark ring,
$$
E(j_{1,1})=1-J_0(j_{1,1})^2\simeq0.837784869.
$$
A circular detector region ending at the first zero therefore misses about $16.22\%$ of the ideal <radiant flux>, even though the surrounding rings have low peak <light intensity>. This matters when converting measured counts into total source brightness or choosing an image aperture for a <photodetector>.
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