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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!

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    $\begingroup$ "for $x\in\mathbb R$" - you are assuming that, but Mathematica assumes all variables are complex unless told otherwise (e.g. via Assuming[]). $\endgroup$ Aug 5, 2016 at 19:11
  • $\begingroup$ 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. $\endgroup$
    – jornada
    Aug 5, 2016 at 19:16
  • $\begingroup$ I see what you mean now after doing my own tests. That is bizarre indeed. $\endgroup$ Aug 6, 2016 at 4:45

1 Answer 1

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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}

res1

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

res2

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.

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