Expansion for Modified Bessel Function Around Infinity

Question

I'm somewhat new to Mathematica, and I don't understand why I'm getting inconsistent series expansions for the modified Bessel Function of first kind near $x=\infty$.

First problem: I get different expansions if I multiply the modified Bessel function by any constant:

Series[BesselI[0,x],{x, ∞, 0}]

$$ e^{-x} \left(O\left(\left(\frac{1}{x}\right)^0\right)+e^{2 x} \left(\frac{\sqrt{\frac{1}{x}}}{\sqrt{2 \pi }}+O\left(\left(\frac{1}{x}\right)^1\right)\right)\right) $$

Series[2 BesselI[0,x],{x, ∞, 0}]

$$ e^{-x} \left(e^{2 x} \left(\sqrt{\frac{2}{\pi }} \sqrt{\frac{1}{x}}+O\left(\left(\frac{1}{x}\right)^{3/2}\right)\right)+\left(i \sqrt{\frac{2}{\pi }} \sqrt{\frac{1}{x}}+O\left(\left(\frac{1}{x}\right)^{3/2}\right)\right)\right) $$

Second problem: Even for $x \in \Re$, the expansion always gives me an imaginary component, for any order of my expansion. This is clearly wrong. For example:

Series[BesselI[0, x], {x, ∞, 3}]

$$ e^{-x} \left(e^{2 x} \left(\frac{\sqrt{\frac{1}{x}}}{\sqrt{2 \pi }}+\frac{\left(\frac{1}{x}\right)^{3/2}}{8 \sqrt{2 \pi }}+\frac{9 \left(\frac{1}{x}\right)^{5/2}}{128 \sqrt{2 \pi }}+O\left(\left(\frac{1}{x}\right)^{7/2}\right)\right)+\left(\frac{i \sqrt{\frac{1}{x}}}{\sqrt{2 \pi }}-\frac{i \left(\frac{1}{x}\right)^{3/2}}{8 \sqrt{2 \pi }}+\frac{9 i \left(\frac{1}{x}\right)^{5/2}}{128 \sqrt{2 \pi }}+O\left(\left(\frac{1}{x}\right)^{7/2}\right)\right)\right) $$

Is there a problem with Mathematica, or I am misunderstanding how Series works? For the reference, I'm using Mathematica 10.3.0.0.

Thanks!

"for $x\in\mathbb R$" - you are assuming that, but Mathematica assumes all variables are complex unless told otherwise (e.g. via Assuming[]). — J. M.'s missing motivation, Aug 05 '16 at 19:11
what I mean is that, if you evaluate that expression with $x \in \Re$, you get a complex number, which is obviously wrong. And if I explicitly include "Assumptions->x>0", I still get the same expansion. — jornada, Aug 05 '16 at 19:16
I see what you mean now after doing my own tests. That is bizarre indeed. — J. M.'s missing motivation, Aug 06 '16 at 04:45

bmf · Answer 1 · 2023-01-15T04:01:09.340

I am using 13.2.0

First problem: I get different expansions if I multiply the modified Bessel function by any constant:

a1 = Series[BesselI[0, x], {x, ∞, 0}] // Normal;
a2 = Series[2 BesselI[0, x], {x, ∞, 1}] // Normal;
a3 = 2 Series[BesselI[0, x], {x, ∞, 0}] // Normal;
FullSimplify@{a2 - a3, a2 - 2 a1}

Second problem: Even for $x \in \Re$, the expansion always gives me an imaginary component, for any order of my expansion. This is clearly wrong.

A potential workaround -albeit not automatic and pretty

Consider

$$ \begin{equation} I_n(x) = \frac{1}{\pi} \int^{\pi}_0 d\theta e^{x \cos\theta}\cos(n\theta) \end{equation} $$

After the changes of variables, $t=\sqrt{x}\theta$ and $u=\tfrac{t^2}{2}$ the above can be re-written as and some massaging we end up having:

$$ \begin{equation} \sqrt{x} \pi e^{-x} I_{n}(x) = \int^{\infty}_0 du \frac{1}{\sqrt{2u}} e^{-u} \left(1 + \frac{u^2}{6x} - n^2 \frac{u}{x} + \texttt{stuff} \right) \end{equation} $$

Mathematica is now happy to do the integral

Integrate[(1/
    Sqrt[2 u] Exp[-u] (1 - (n^2 u)/x + u^2/(
      6 x)(*+more stuff relevant for ≠ 0*))) /. n -> 0, {u, 0, 
  Infinity}]// Expand

I am uncertain how you can impose these changes of variables automatically in Mathematica. Not sure if there's even a way to do it.

Michael E2 has written a nice code that does substitutions in the integrand, and you might want to experiment a bit with that.

Since, v13.1.0 we also have at our disposal IntegrateChangeVariables which can be used in this case, however, some manual manipulations are still needed.

Expansion for Modified Bessel Function Around Infinity

1 Answers1