Asymptotic expansion of Bessel Function

Hi I am interested in calculating an asymptotic expansion of the following function. Or, I would at least like to know how the function behaves for large values of x. I am having trouble simplifying the expression, it's not the nicest looking.

The function is

$$f(x)=\sqrt{ \Re\left[\frac{J_1\left(\alpha \sqrt{x^2+ix}\right)}{J_1\left(\beta \sqrt{x^2+ix}\right)}\right]^2+\Im\left[\frac{J_1\left(\alpha \sqrt{x^2+ix}\right)}{J_1\left(\beta \sqrt{x^2+ix}\right)}\right]^2}$$ where $\alpha,\beta>0$ and $\alpha\neq \beta$. Note, $J_1(z)$ is the Bessel function and has the follow series expansion $$J_1(z)=\frac{z}{2}\sum_{k\geq 0} \frac{(-1)^k}{k!\Gamma(k+2)} \left(\frac{z}{2}\right)^{2k}$$ Any hints or help would be greatly appreciated. (I have plotted the function, however I am interested in some analytic results).

Thank you!

• for $x\rightarrow \infty$ you may replace your square root by $x(1+\frac{i}{2x})$ which turns out to be an excellent approximation. Now i think the usual asymptotic expansions of the bessel functions should be enough to get to the desired result. – tired Oct 30 '15 at 13:11
• $$\sqrt{\frac{\beta}{\alpha}}\sqrt{ \Im\left(\frac{\cos \left(-\frac{3 \pi }{4}+\alpha x \left(1+\frac{i}{2 x}\right)\right)}{\cos \left(-\frac{3 \pi }{4}+\beta x \left(1+\frac{i}{2 x}\right)\right)}\right)^2+ \Re\left(\frac{\cos \left(-\frac{3 \pi }{4}+ \alpha x \left(1+\frac{i}{2 x}\right)\right)}{\cos \left(-\frac{3 \pi }{4}+\beta x \left(1+\frac{i}{2 x}\right)\right)}\right)^2}$$ is the result you are looking for, i leave further simplification up to you. – tired Oct 30 '15 at 13:24

I want to give some justification to the result given above.

Let's look at the quotient of two Bessel functions

$$C(\alpha,\beta,x)=\frac{J_1(\alpha f(x) )}{J_1(\beta f(x) )}$$

now assume that $f(x)\sim x^a+iO(x^{a-\delta})$ as $x\rightarrow \infty$ with $a>\delta>0$it is now tempting to just throw away the imaginary part and proceeding with $f\sim x^a$.

The problem is that the Besselfunction with purely real argument has zeros on the real axis so $C(\alpha,\beta,x_{n,\beta})=\infty$ where $x_{n,\beta}$ is defined as $J_1(\beta x_{n,\beta})=0$ .

We therefore induce some singularities which are absent in the original problem by taking only leading order contributions.

This problem can be cured by taking into account the first non vanishing imaginary term, because it will shift away the $x_{n,\beta}$ away from the real axis. We can now safely proceed with

$$C(\alpha,\beta,x)\sim\frac{J_1(\alpha x^a(1+icx^{a-\delta-1}))}{J_1(\beta x^a (1+icx^{a-\delta-1}) )}$$

where $c$ is the first coefficient in the asymptotic expansion of the imaginary part of $f(x)$. Using the asymptotic expansions for the Besselfunction found here , we obtain

$$C(\alpha,\beta,x)\sim\frac{\sqrt{\beta}}{\sqrt{\alpha}}\times\frac{\cos\left(\alpha x^a(1+icx^{a-\delta-1})-\frac{3 \pi}{4}\right)}{\cos\left(\beta x^a (1+icx^{a-\delta-1}) -\frac{3 \pi}{4}\right)}$$

Setting $c=\frac{1}{2},a=2,\delta=1$ and calculating

$$|C(\alpha,\beta,x)|=\sqrt{\Re[C(\alpha,\beta,x)]^2+\Im[C(\alpha,\beta,x)]^2}=\\ \sqrt{\frac{\beta}{\alpha}} \times \sqrt{ \Im\left(\frac{\cos \left(-\frac{3 \pi }{4}+\alpha x \left(1+\frac{i}{2 x}\right)\right)}{\cos \left(-\frac{3 \pi }{4}+\beta x \left(1+\frac{i}{2 x}\right)\right)}\right)^2+ \Re\left(\frac{\cos \left(-\frac{3 \pi }{4}+ \alpha x \left(1+\frac{i}{2 x}\right)\right)}{\cos \left(-\frac{3 \pi }{4}+\beta x \left(1+\frac{i}{2 x}\right)\right)}\right)^2}$$

As stated in the comments follows.

Note that this result doesn't decay to zero or grows to infinity as $x\rightarrow \infty$ as the leading term is of oscillatory nature.

Edit:

I am quite confident that the error is of order $\mathcal{O}(\frac{1}{x})$

Edit2:

To give an impression how good this approximation is, i plotted

$\color{red}{|C(2,1,x)|} \quad \text{and} \quad \color{blue}{|C(1,2,x)|}$ Dotted lines are asymptotics, full lines the exact expressions

Edit3:

One may also notice that this method works for all $J_N(x)$ but the convergence gets worse because sub-leading terms are enhanced by $4 N^2$