How to arrive at this solution to an integral over the Bessel function of the first kind?

Question

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!

Answer 1

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.

Answer 2

As it is mentioned, it is just partial integration. So start with $$\int_0^\infty p(y)\; e^{iay^2} J_0(2axy) \,y \, {\rm d}y = \int_0^R e^{iay^2} J_0(2axy) \,y \, {\rm d}y \\ \stackrel{z=2axy}{=} \frac{1}{(2ax)^2} \int_0^{2axR} z \, e^{\frac{iaz^2}{(2ax)^2}} \, J_0(z) \, {\rm d}z=\frac{1}{2ia}\int_0^{2axR} \left( \frac{{\rm d}}{{\rm d}z} \, e^{\frac{iaz^2}{(2ax)^2}} \right) J_0(z) \, {\rm d}z \\ = \frac{e^{\frac{iaz^2}{(2ax)^2}}}{2ia} \, J_0(z)\bigg|_0^{2axR} - \frac{1}{2ia} \int_0^{2axR} z\, e^{\frac{iaz^2}{(2ax)^2}} z^{-1}J_{-1}(z) \, {\rm d}z \\ = \frac{e^{iaR^2}}{2ia} \, J_0(2axR) - \frac{1}{2ia} - e^{\frac{iaz^2}{(2ax)^2}} \frac{(2ax)^2}{(2ia)^2} \, z^{-1} J_{-1}(z) \bigg|_0^{2axR} \\+ \frac{(2ax)^2}{(2ia)^2} \int_0^{2axR} z \, e^{\frac{iaz^2}{(2ax)^2}} z^{-2}J_{-2}(z) \, {\rm d}z \\ = -\frac{1}{2ia}\sum_{k=0}^{n-1} \frac{(2ax)^{2k}}{(2ia)^k} \, \frac{2^{-k}}{k!} + \frac{e^{iaR^2}}{2ia} \sum_{k=0}^{n-1} \left(i\,\frac{x}{R}\right)^k \, J_{-k}(2axR) \\ + (-1)^n \frac{(2ax)^{2n-2}}{(2ia)^n} \int_0^{2axR} z\, e^{\frac{iaz^2}{(2ax)^2}} z^{-n}J_{-n}(z) \, {\rm d}z$$ which you can prove by induction and the formulas $$\lim_{z\rightarrow 0} \frac{J_n(z)}{z^n} = \frac{2^{-n}}{n!} \\ J_{-n}(z)=(-1)^n J_n(z) \\ \frac{\textrm{d}}{\textrm{d}z}z^{-n}J_{-n}(z)=z^{-n}J_{-n-1}(z) \, .$$

Estimating the remainder integral using $$|J_n(z)|\leq \frac{2^{-n}z^n}{n!}$$ shows that it vanishes in the limit $n\rightarrow \infty$. Hence $${-2ia} \, e^{iax^2} \int_0^\infty p(y)\; e^{iay^2} J_0(2axy) \,y \, {\rm d}y \\ =-2ia \, e^{iax^2} \left( -\frac{1}{2ia}\sum_{k=0}^{\infty} \frac{(-iax^2)^k}{k!} + \frac{e^{iaR^2}}{2ia} \sum_{k=0}^{\infty} \left(-i\,\frac{x}{R}\right)^k \, J_{k}(2axR) \right) \\ =1-e^{ia(R^2+x^2)} \sum_{k=0}^{\infty} \left(-i\,\frac{x}{R}\right)^k \, J_{k}(2axR) \tag{1}$$

and we are done.

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Addon:

Similarly we can integrate by parts the other way around $$\int_0^\infty p(y)\; e^{iay^2} J_0(2axy) \,y \, {\rm d}y = \int_0^R e^{iay^2} J_0(2axy) \,y \, {\rm d}y \stackrel{z=2axy}{=} \frac{1}{(2ax)^2} \int_0^{2axR} e^{\frac{iaz^2}{(2ax)^2}} zJ_0(z) \, {\rm d}z \\ =\frac{1}{(2ax)^2} \, e^{\frac{iaz^2}{(2ax)^2}} \, zJ_1(z) \bigg|_0^{2ax R} - \frac{2ia}{(2ax)^4} \int_0^{2axR} e^{\frac{iaz^2}{(2ax)^2}} z^2 J_1(z) \, {\rm d}z\\ =\frac{R}{2ax} \, e^{iaR^2} \, J_1(2axR) - \frac{2ia \, R^2}{(2ax)^2} \, e^{iaR^2} \, J_2(2axR) + \frac{(2ia)^2}{(2ax)^6} \int_0^{2axR} e^{\frac{iaz^2}{(2ax)^2}} z^3 J_2(z) \, {\rm d}z = \dots$$ and so on. Therefore we can guess $$=e^{iaR^2} \sum_{k=0}^{n-1} (-2ia)^{k}\left( \frac{R}{2ax} \right)^{k+1} J_{k+1}(2axR) + \frac{(-2ia)^n}{(2ax)^{2n+2}} \int_0^{2axR} e^{\frac{iaz^2}{(2ax)^2}} z^{n+1} J_{n}(z) \, {\rm d}z$$ which can be proved again by induction. The case $n=1$ is evident. Suppose it is true for $n$, then for $n\rightarrow n+1$ we have $$e^{iaR^2} \sum_{k=0}^{n-1} (-2ia)^{k}\left( \frac{R}{2ax} \right)^{k+1} J_{k+1}(2axR) \\ + e^{iaR^2} (-2ia)^n \, \left(\frac{R}{2ax} \right)^{n+1} \, J_{n+1}(2axR) + \frac{(-2ia)^{n+1}}{(2ax)^{2n+4}} \int_0^{2axR} e^{\frac{iaz^2}{(2ax)^2}} z^{n+2} J_{n+1}(z) \, {\rm d}z \\ =e^{iaR^2} \sum_{k=0}^{n} (-2ia)^{k}\left( \frac{R}{2ax} \right)^{k+1} J_{k+1}(2axR) + \frac{(-2ia)^{n+1}}{(2ax)^{2n+4}} \int_0^{2axR} e^{\frac{iaz^2}{(2ax)^2}} z^{n+2} J_{n+1}(z) \, {\rm d}z$$ which is what we needed to show.

Now we estimate the remainder term and substitute $z=2axRu$ s.t. $$\left|\frac{(-2ia)^{n}}{(2ax)^{2n+2}} \int_0^{2axR} e^{\frac{iaz^2}{(2ax)^2}} z^{n+1} J_{n}(z) \, {\rm d}z\right|\leq R^2 \left(\frac{R}{x}\right)^n \int_0^1 u^{n+1} |J_n(2axRu)| \, {\rm d}u \leq R^2 \left(\frac{R}{x}\right)^n $$ since $|J_n|\leq 1$. Now if $x>R$ this vanishes exponentially as $n\rightarrow\infty$ giving $${-2ia} \, e^{iax^2} \int_0^\infty p(y)\; e^{iay^2} J_0(2axy) \,y \, {\rm d}y \\= e^{ia(x^2+R^2)} \sum_{k=1}^{\infty} \left( -i\,\frac{R}{x} \right)^{k} J_{k}(2axR) = e^{ia(x^2+R^2)} \sum_{k=-\infty}^{-1} \left( -i\,\frac{x}{R} \right)^{k} J_{k}(2axR) \tag{2}$$ since $J_{-k}(x)=(-1)^kJ_k(x)$.

By combining the previous two results (1)=(2), you'll get the Laurent expansion $$\sum_{k=-\infty}^{\infty} \left(-i\,\frac{x}{R}\right)^k \, J_{k}(2axR)=e^{-ia(R^2+x^2)} $$ i.e. the generating function for the Bessel-functions.