# Leading terms in asymptotic expansion of modified bessel function of the first kind

Show that leading and next-to-leading terms in an asymptotic expansion for large $x>0$ of the modified Bessel functions of the first kind $I_0(x)$ and $I_1(x)$ are:

$$I_0(x) \sim \frac{e^x}{\sqrt{2\pi x}}(1+\frac{1}{8x}) \ \ \ \text{and} \ \ \ I_1(x) \sim \frac{e^x}{\sqrt{2\pi x}}(1-\frac{3}{8x}).\tag{1}$$

I am fairly sure I should use the method of steepest descent to evaluate the integral, but not sure how to proceed. Any help would be greatly appreciated.

Modified Bessel function of the first kind ($n$ is an integer):

$$I_n(x) = \frac{1}{\pi}\int_0^\pi e^{x \ cos(\theta)} cos(n\theta) d\theta.\tag{2}$$

From $$I_n(x) ~=~\frac{1}{\pi}\int_0^{\pi} \! \mathrm{d}\theta ~\exp\left(x\cos\theta\right)\cos n\theta, \qquad n\in~\mathbb{N}_0,\tag{A}$$ we calculate \begin{align} \sqrt{x}\pi e^{-x}I_n(x) &~~=~\sqrt{x} \int_0^{\pi} \! \mathrm{d}\theta ~\exp\left(- x(1-\cos\theta)\right)\cos n\theta \cr &\stackrel{t=\sqrt{x}\theta}{=}~ \int_0^{\pi\sqrt{x}} \! \mathrm{d}t ~\exp\left(- x(1-\cos\frac{t}{\sqrt{x}})\right)\cos \frac{nt}{\sqrt{x}} \cr &~~=~ \int_0^{\infty} \! \mathrm{d}t ~\exp\left(- \frac{t^2}{2} + \frac{t^4}{24 x} + O(x^{-2})\right) \left(1- \frac{(nt)^2}{2x} + O(x^{-2})\right) \cr &~~=~ \int_0^{\infty} \! \mathrm{d}t ~\exp\left(- \frac{t^2}{2}\right) \left(1- \frac{(nt)^2}{2x}+ \frac{t^4}{24 x} + O(x^{-2})\right) \cr &\stackrel{u=t^2/2}{=}~ \int_0^{\infty} \! \frac{\mathrm{d}u}{\sqrt{2u}} ~\exp\left(- u\right) \left(1- \frac{n^2u}{x}+ \frac{u^2}{6x} + O(x^{-2})\right) \cr &~~=~\frac{1}{\sqrt{2}}\left( \Gamma(\frac{1}{2}) - \frac{n^2}{x}\Gamma(\frac{3}{2}) +\frac{1}{6x}\Gamma(\frac{5}{2}) + O(x^{-2})\right)\cr &~~=~\sqrt{\frac{\pi}{2}}\left( 1+ \frac{1-4n^2}{8x} + O(x^{-2})\right), \end{align}\tag{B} which agrees with OP's sought-for formulas (1).