How to arrive at this solution to an integral over the Bessel function of the first kind? I am trying to solve a complex integral over $y$, and am really struggling so would appreciate some help. The function is given by
$$
f(x) = \frac{2a}{i}e^{iax^2}\int_0^\infty p(y)\; e^{iay^2} J_0(2axy)\;y\;\; \textrm{d}y, \tag{1}
$$
where $p(y) = \textrm{circ}(y/R)$ is the circ function, $J_0$ is the zero-order Bessel function of the first kind, and $a$ and $R$ are positive, real constants.
In Eq. (2) of this paper, it is simply stated that the solution is written as
$$
f(x) = 1-e^{iax^2}e^{iaR^2} \sum_{n=0}^\infty \bigg( -i\frac{x}{R} \bigg)^n J_n(2aRx), \tag{2}
$$
and that this was arrived at using partial integration together with the differential formula for Bessel functions
$$
\frac{\textrm{d}}{\textrm{d}z}z^{n+1}J_{n+1}(z)=z^{n+1}J_n(z).\tag{3}
$$
I cannot figure out how to attack this problem, and how to obtain Eq. (2) from Eq. (1). If someone is able to see it I would appreciate being walked through the steps. Thank you!
 A: By changing $z=2axy$, an expression for the function is
\begin{align}
f(x)& = \frac{2a}{i}e^{iax^2}\int_0^R e^{iay^2} J_0(2axy)y\,{d}y\\
&= \frac{e^{iax^2}}{2iax^2}\int_0^{2axR} e^{i\frac{z^2}{4ax^2}} J_0(z)z\,{d}z
\end{align}
With $X=2axR,\lambda=i/(4ax^2)$ and
\begin{equation}
K=\int_0^Xe^{\lambda z^2}z J_0(z)\,dz
\end{equation}
we have to evaluate
\begin{equation}
f(x)=e^{iax^2} (-2\lambda) K
\end{equation}
From the quoted property (3), $zJ_0(z)=d/dz\left( zJ_1(z) \right)$, integrating by parts gives
\begin{align}
K&= \left.zJ_1(z)e^{\lambda z^2}\right|_0^X-2\lambda \int_0^Xe^{\lambda z^2}z^2 J_1(z)\,dz\\
&=XJ_1(X)e^{\lambda X^2}-2\lambda \int_0^Xe^{\lambda z^2}z^2 J_1(z)\,dz
\end{align}
Now, using again the differentiation property, integration by parts of this new integral gives
\begin{equation}
\int_0^Xe^{\lambda z^2}z^2 J_1(z)\,dz=X^2J_2(X)e^{\lambda X^2}-2\lambda \int_0^Xe^{\lambda z^2}z^3 J_2(z)\,dz
\end{equation}
By induction, admitting that the series converges,
\begin{equation}
K=e^{\lambda X^2}\sum_{k=1}^\infty(-2\lambda )^{k-1}X^kJ_k(X)
\end{equation}
Then,
\begin{align}
f(x)&=e^{iax^2+iaR^2} \sum_{k=1}^\infty(-2\lambda X )^{k}J_k(X)\\
&=e^{iax^2+iaR^2} \sum_{k=1}^\infty(-\frac{iR}{x})^{k}J_k(2axR)
\end{align}
The generating function for the Bessel functions
$$e^{\frac{1}{2}z(t-t^{-1})}=\sum_{m=-\infty}^{\infty}t^{m}J_{m}\left(z\right)$$
gives the expressions
\begin{align}
 \sum_{k=-\infty}^\infty(-\frac{iR}{x})^{k}J_k(2axR)&=J_0(2axR)+\left( \sum_{k=-\infty}^{-1}+\sum_{k=1}^\infty \right)(-\frac{iR}{x})^{k}J_k(2axR)\\
 &=e^{-ia\left( x^2+R^2 \right)}
\end{align}
from which, we deduce
\begin{equation}
\sum_{k=1}^\infty(-\frac{iR}{x})^{k}J_k(2axR)=e^{-ia\left( x^2+R^2 \right)}-J_0(2axR)-\sum_{k=-\infty}^{-1}(-\frac{iR}{x})^{k}J_k(2axR)
\end{equation}
As $J_{-n}(z)=(-1)^nJ_n(x)$ and including the term $J_0(2axR)$ in the series, we have
\begin{equation}
\sum_{k=1}^\infty(-\frac{iR}{x})^{k}J_k(2axR)=e^{-ia\left( x^2+R^2 \right)}-\sum_{k=0}^{\infty}(-\frac{ix}{R})^{k}J_k(2axR)
\end{equation}
Finally,
\begin{equation}
f(x)=1-e^{ia\left( x^2+R^2 \right)}\sum_{k=0}^{\infty}(-\frac{ix}{R})^{k}J_k(2axR)
\end{equation}
as expected.
