Consider the two following functions defined by double integrals \begin{align} \varphi_a (R) &= \int_0^\frac{\pi}{2} \int_0^R e^{-at\cos\theta} \sin^2\theta \, J_0(at\sin\theta) \sin\theta \, \mathrm{d}t \, \mathrm{d}\theta \, , \\ \psi_a (R) &= \int_0^\frac{\pi}{2} \int_0^R e^{-at\cos\theta} \sin^2\theta \, J_1(at\sin\theta) \cos\theta \, \mathrm{d}t \, \mathrm{d}\theta \, , \end{align} wherein $a, R \in \mathbb{R}_+^*$. It can be shown that $\lim_{R\to\infty} \varphi_a = 2/ (3a)$ and that $\lim_{R\to\infty} \psi_a = -1/ (3a)$

But how do $\varphi_a(R)$ and $\psi_a(R)$ behave asymptotically to leading order as $R \to \infty$?

Any help or suggestions are highly appreciated.

Thank you,

  • 2
    $\begingroup$ Minor point: $$ \varphi _a (R) = \frac{1}{a}\varphi _1 (aR),\quad \psi _a (R) = \frac{1}{a}\psi _1 (aR), $$i.e., we can assume that $a=1$. $\endgroup$
    – Gary
    Sep 6, 2021 at 9:00
  • $\begingroup$ @Gary i agree. Thanks for the comment $\endgroup$
    – preuss
    Sep 6, 2021 at 9:14

1 Answer 1


By using the exponential generating function for the Legendre polynomials: \begin{equation} e^{xz}J_{0}\left(z\sqrt{1-x^{2}}\right)=\sum_{n=0}^{\infty}\frac{P_{n}\left(x\right)}{n!}z^{n} \end{equation} with $z=-t,x=\cos\theta$ the first integral (taken with $a=1$, as remarked by @Gary in a comment) reads \begin{align} \varphi (R) &= \int_0^\frac{\pi}{2} \int_0^R e^{-t\cos\theta} \sin^3\theta\, J_0(t\sin\theta) \, dt \, d\theta \\ &=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_0^\frac{\pi}{2} \int_0^Rt^nP_n(\cos\theta)\sin^3\theta \, dt \, d\theta\\ &=\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)!}R^{n+1}\int_0^1(1-x^2)P_n(x)\,dx \end{align} we supposed valid the interchange of integration and summation and change $x=\cos\theta$ to obtain the latter expression.

When $n$ is even ($n=2p$), we have, by symmetry, \begin{align} w_{2p}&=\int_0^1(1-x^2)P_{2p}(x)\,dx\\ &=\frac{1}{2}\int_{-1}^1(1-x^2)P_{2p}(x)\,dx \end{align} The orthogonality properties of the Legendre polynomials indicate that the integrals with $p\ge2$ vanish and direct calculation shows that $w_0=2/3,w_2=-2/15$.

When $n$ is odd ($n=2p+1$), we use the tabulated integral (DLMF) \begin{equation} \int_{0}^{1}P_{2p+1}\left(x\right)x^{z-1}\,dx=\frac{(-1)^{p}{\left(1-\frac{1}{2}z\right)_{p}}}{2{\left(\frac{1}{2}+\frac{1}{2}z\right)_{p+1}}} \end{equation} where $(.)_n$ is Pochhammer's symbol, to express \begin{align} w_{2p+1}&=\int_0^1(1-x^2)P_{2p+1}(x)\,dx\\ &=\frac{(-1)^{p}{\left(\frac{1}{2}\right)_{p}}}{2{\left(1\right)_{p+1}}}-\frac{(-1)^{p}{\left(-\frac{1}{2}\right)_{p}}}{2{\left(2\right)_{p+1}}} \end{align} Then, converting the series into generalized hypergeometric functions (GHF), \begin{align} \varphi(R)&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)!}R^{n+1}w_n\\ &=w_0R+\frac16w_2R^3-\sum_{p=0}^\infty\frac{1}{(2p+2)!}w_{2p+1}R^{2p+2}\\ &=\frac23R-\frac1{45}R^3-\frac12\sum_{p=0}^\infty\frac{(-1)^p}{(2p+2)!}R^{2p+2}\left[\frac{\left(\frac{1}{2}\right)_{p}}{{\left(1\right)_{p+1}}}-\frac{\left(-\frac{1}{2}\right)_{p}}{{\left(2\right)_{p+1}}}\right]\\ &=\frac23R-\frac1{45}R^3-\frac{R^2}4\left[ \phantom{}_{2}F_{3}\! \left(\frac{1}{2},1;\frac{3}{2},2,2;-\frac{R^{2}}{4}\right) -\frac12\phantom{}_{2}F_{3}\! \left(-\frac{1}{2},1;\frac{3}{2},2,3;-\frac{R^{2}}{4}\right) \right] \end{align} Then the asymptotic behavior follows: \begin{equation} \varphi(R)=\frac23+\sqrt{\frac{2}{\pi}}\frac{\sin(R-\pi/4)}{R^{3/2}}-\frac{21}{8}\sqrt{\frac{2}{\pi}}\frac{\sin(R+\pi/4)}{R^{5/2}}+O(R^{-7/2}) \end{equation} This result was obtained with a CAS, however it should not be too difficult to derive it from the asymptotic expansion of the GHF.

This method should also apply to evaluate the second integral, by differentiating the starting identity \begin{equation} J_0(t\sin\theta)=e^{t\cos(\theta)}\sum_{n\ge0}\frac{(-1)^n}{n!}P_n(\cos\theta)t^n \end{equation} w.r.t. $t$ or $\theta$ to obtain an expansion of $J_1$, but I didn't try it.

  • $\begingroup$ Merci Paul! You will be rewarded PTS 50 in about 23 hours for your great answer. $\endgroup$
    – preuss
    Sep 13, 2021 at 16:06
  • 1
    $\begingroup$ You're welcome. Merci for the generous bounty. Tell me if you are in trouble with $\psi(R)$, I can try. $\endgroup$
    – Paul Enta
    Sep 13, 2021 at 18:27
  • 1
    $\begingroup$ Il n'y a pas de quoi ;) No need for doing that for $\psi$. The picture is clear now. Best greetings, A $\endgroup$
    – preuss
    Sep 14, 2021 at 16:10

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