Huygens-Fresnel principle The math derivation of Huygens principle in 2D, applied to a plane wavefront propagation, gives the following field complex amplitude at a distance "d" of the initial wavefront (the unit for distances is the wavelength):
integral
Numerical integration gives the expected value (pure delay operator):
pure delay
But Mathematica(r) fails to derive a close form solution to the above integrale.
Does anybody know a way to derive it?
 A: Given the discussion in the comments above, I now realize that the equality between these two expressions is an approximate one expected to hold only when $d$ is large. In the large-$d$ limit, this integral can be evaluated using the stationary phase approximation.
We start by making the change of variables $x = u\, d$, which causes the integral to become
\begin{equation}
I(d) \;=\;
e^{i \pi/4} \sqrt{d}\, \int_{-\infty}^{+\infty}du \;
\frac{e^{-i\, 2\,\pi\, d\, \sqrt{1+u^2}}}{({1+u^2)}^{3/4}}\, .
\end{equation}
The stationary phase approximation asserts that when $d$ is large, so that the oscillatory portion of the integral oscillates very rapidly, the value of the integral is dominated by the region where the derivative of the phase function $2\,\pi\, d\, \sqrt{1+u^2}$ is zero. This occurs at $u = 0$. Near $u=0$ we have
\begin{equation}
\sqrt{1+u^2} \thickapprox 1 + \frac{1}{2}u^2
\end{equation}
and
\begin{equation}
\frac{1}{({1+u^2)}^{3/4}} \thickapprox 1 ,
\end{equation}
and the integral becomes
\begin{align}
I(d) &\thickapprox
e^{i \pi/4} \sqrt{d}\, e^{-i\, 2\,\pi\, d}\,\int_{-\infty}^{+\infty}du \;
e^{-i\, \pi\, d\, u^2}\\
&= e^{i \pi/4} \sqrt{d}\, e^{-i\, 2\,\pi\, d}\,\frac{1}{\sqrt{i d}}\\
&= e^{-i\, 2\,\pi\, d}\, ,
\end{align}
which is the desired result. (Note that $e^{i \pi/4} = \sqrt{i}$.)
By the way, this question that I asked a while ago is related, and comes from a consideration of the Huygens-Fresnel principle in 3D.
