I am interested in obtaining the asymptotic expansion of integrals of the form

$$ I_\nu(\alpha) = \int_0^\infty xdx ~ f(x) J_\nu(x) J_\nu(\alpha x),$$

for $\nu \to \infty$ and some fixed, real, $\alpha$ parameter. To have a concrete example we can specify $f(x)$ as;

$$ f(x) = \frac{x}{1+x^3},~~~{\rm or~~alternatively~~as}~~~ f(x) = x e^{-x},$$

or whatever is simpler. An approach that seemed promising to me was to use an integral representation of $J_\nu(x)$, i.e. something like $$ J_\nu(z) =\frac {z^\nu} {2^\nu \sqrt \pi \, \Gamma {\left( \nu + \frac 1 2 \right)}} \int_{-1}^1 (1 - t^2)^{\nu - 1/2} \cos z t \, dt~, $$ or perhaps an integral representation in the complex plane $$ J_\nu (z) = \frac{1}{2\pi i} \int_{\mathcal C} e^{z \sinh t - \nu t} dt~. $$

This allows doing the integral in $x$ then, but I have difficulties proceeding with the asymptotics of the two remaining integrals from the representation of $J_\nu(x)$.

How can once obtain the asymptotic of $I_\nu(\alpha)$ for $\nu \to \infty$?


Perhaps using the product expression foun here

$$ J_{\nu}\left(z\right)J_{\nu}\left(\zeta\right)=\frac{1}{2\pi i}\int_{c-i\infty% }^{c+i\infty}\exp\left(\frac{1}{2}t-\frac{z^{2}+\zeta^{2}}{2t}\right)I_{\nu}% \left(\frac{z\zeta}{t}\right)\frac{\mathrm{d}t}{t}, $$

might be helpful.


This question is far too broad, with two parameters, and an unspecified function. However, there is a simple case which is close to the exponentially decaying function you mention. I'll basically quote a result from the literature of special functions.

From Gradshteyn and Rhyzik 6.612.3

$$ \int_0^\infty e^{-x} J_\nu(x) \, J_\nu(a\,x) dx = \frac{1}{\pi \sqrt{a}} \, Q_{\nu-1/2}(1/a+a/2)$$ where $Q_\nu$ is the Legendre Q function. Incidentally, the G&R equation has a parameter in the exponential, so one of the proposed functional forms can be solved exactly.

Then used the known asymptotics $$Q_n(\cosh{x}) \sim \sqrt\frac{\pi}{2\,n\,\sinh{x}} e^{-(n+1/2)x} $$

I've coded it up and the asymptotic approximation is decent; e.g., for $a=2$ and $\nu=40,$ the difference between the asymptotic expansion and the true answer is about 7.5%. For $\nu=400,$ about 0.7% difference.

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  • $\begingroup$ tnx! Indeed, there are a couple of examples of $f(x)$ where one can find the closed solution. Then finding the asymptotics is usually easier. However, I was hoping that getting the asymptotic form directly would we far easier than first obtaining the exact solution. Precisely because, I thought, the method would not change very much if I vary a form of f(x) slightly (say e.g. f(x) is meromorphic and such that the integral exists). A canonical example would be using the saddle point, however, I could not find a nice representation of the integral where saddle point method would just work. $\endgroup$ – z.v. May 28 at 7:27
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    $\begingroup$ @z.v. I think an analytic solution is difficult, and only worse for the algebraic decay example you gave. A long time ago I had to do 'double Bessel' integrals for a code, and if numerical results are what you want, then do this: Find about 50 zeros (asymptotics are well known) for the Bessels, integrate over each region between zeros, use a series acceleration (Wynn epsilon, say) to extrapolate the sum. For a function that falls off sufficiently fast, this works O.K. If $a$ is very close to 1, then you could use the 'multiplication theorem' that might give you a fast convergent series. $\endgroup$ – skbmoore May 28 at 16:08

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