I need to find asymptotic expansion of $e^{-x}[I_{0}-I_{1}](x)$ when $x\rightarrow \infty$. Where $I_{0}(x)$ and $I_{1}(x)$ are modified Bessel functions of first kind of order zero and one respectively. I know that $I_{0}$ and $I_{1}$ both are asymptotically equal to $\dfrac{e^{x}}{\sqrt (2\pi x)}$ when $x\rightarrow \infty $. Also that if $f_{1}\sim g_{1} $ and $f_{2}\sim g_{2}$ then it won't imply $f_{1}-f_{2}\sim g_{1}-g_{2}$. I am not sure how to solve this.
Any help is appreciated.
Approximation of subtraction of modified Bessel functions of first kind
approximationasymptoticsbessel functionsspecial functions
Best Answer
It is known that $$ I_\nu (x) \sim \frac{{e^x }}{{\sqrt {2\pi x} }}\left( {1 - \frac{{4\nu ^2 - 1}}{{8x}} + \frac{{(4\nu ^2 - 1)(4\nu ^2 - 9)}}{{128x^2 }} - \cdots } \right) $$ as $x\to +\infty$ with fixed $\nu$ (cf. http://dlmf.nist.gov/10.40.E1). Thus $$ I_0 (x) \sim \frac{{e^x }}{{\sqrt {2\pi x} }}\left( {1 + \frac{1}{{8x}} + \frac{9}{{128x^2 }} + \cdots } \right),\;\, I_1 (x) \sim \frac{{e^x }}{{\sqrt {2\pi x} }}\left( {1 - \frac{3}{{8x}} - \frac{{15}}{{128x^2 }} - \cdots } \right), $$ and therefore, $$ e^{ - x} (I_0 (x) - I_1 (x)) \sim \frac{1}{{\sqrt {2\pi x} }}\left( {\frac{1}{{2x}} + \frac{3}{{16x^2 }} + \cdots } \right) = \frac{1}{{2\sqrt {2\pi } x^{3/2} }}\left( {1 + \frac{3}{{8x}} + \cdots } \right) $$ as $x\to +\infty$.