# Asymptotics of $\int xdx ~ f(x) J_\nu(x) J_\nu(\alpha x)$ type integral for $\nu \to \infty$

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 $$\nu \to \infty$$?

Edit

Perhaps using the product expression foun here

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

$$\int_0^\infty e^{-x} J_\nu(x) \, J_\nu(a\,x) dx = \frac{1}{\pi \sqrt{a}} \, Q_{\nu-1/2}(1/a+a/2)$$ 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.
• 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. – z.v. May 28 at 7:27
• @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. – skbmoore May 28 at 16:08