I am following the book Introduction to Infrared and Electro-Optical Systems by Driggers. Below is a derivation of Fresnel diffraction using Huygen's principle. In (4.24), the $t$ should be a $z$
Previously, he derived $r_{12} \approx z (1 + \frac{1}{2} \big( \big( \frac{x-x_a}{z} \big)^2 + \big( \frac{y-y_a}{z} \big)^2 \big) )$. I understand $\frac{1}{r_{12}} \approx \frac{1}{z}$.
What I don't understand is the simplification of the exponent. It probably has to do with the $z^3$ inequality that he derived, but I don't see how to apply it. My derivation of the exponent term is:
$$ jk[r_{12}] \approx jk[ z \big(1 + \frac{1}{2} \big( \big( \frac{x-x_a}{z} \big)^2 + \big( \frac{y-y_a}{z} \big)^2 \big) \big)]$$
$$ jk \big(z + \frac{z}{2} \big( \big( \frac{x-x_a}{z} \big)^2 + \big( \frac{y-y_a}{z} \big)^2 \big) \big)$$
$$ jk \big(z + \frac{z}{2z^2} \big( ( x-x_a )^2 + ( y-y_a)^2 \big) \big)$$
$$ jkz + \frac{jk}{2z} \big( ( x-x_a )^2 + ( y-y_a)^2 \big) $$
Therefore my answer is
$$ A \frac{e^{2jkz}}{z} \iint e^{\frac{jk}{2z} \big( ( x-x_a )^2 + ( y-y_a)^2 \big) } dx_a dy_a, $$
which doesn't match (4.26a).
Your derivation is conceptually correct, but the factor of $2$ in the exponent outside the integral of your last expression is unfounded.
The author has supressed the time dependence $e^{-j\omega t}$ in Equation $4.24$ and begins, therefore with the expession
$$u(x,y)=A\iint_A \frac{e^{jkr_{12}}}{r_{12}}\,dx_a\,dy_a \tag1$$
Now, using your approximations yields
$$\begin{align} u(x,y)&\approx A \iint_A \frac{e^{jk\left(z+\frac{(x-x_a)^2+(y-y_a)^2}{2z}\right)}}{z}\,dx_a\,dy_a\\\\ &=A\frac{e^{jkz}}{z}\iint_A e^{\frac{jk}{2z}[(x-x_a)^2+(y-y_a)^2]}\,dx_a\,dy_a \end{align}$$
which agrees with Equation $4.26(a)$.