Limit of integral with a Bessel function

Solution 1:

Let

$$ f(\alpha)=\int\limits_{-1}^1dx \ e^{-\alpha x}I_0(\sqrt{1-x^2}) $$

So that $A=-f'/f$. Integrate by parts

$$ f(\alpha)=-\frac{1}{\alpha}e^{-\alpha x}I_0(\sqrt{1-x^2})\bigg|_{-1}^1 -\frac{1}{\alpha}\int\limits_{-1}^1dx \ e^{-\alpha x} \frac{x I_1(\sqrt{1-x^2})}{\sqrt{1-x^2}} $$

Using $I_0(0)=1$ and multiplying by $\alpha$ and using the definition of $f$

$$ \alpha f(\alpha)=e^\alpha-e^{-\alpha}-\int\limits_{-1}^1dx \ e^{-\alpha x} \left[\frac{x I_1(\sqrt{1-x^2})}{\sqrt{1-x^2}}\right]=\int\limits_{-1}^1dx \ e^{-\alpha x}\left[ \alpha I_0(\sqrt{1-x^2}) \right] $$

When $\alpha \to \infty$ we have $[\dots]_\text{middle term}<<[\dots]_\text{RHS}$, because $I_1/\sqrt{...} < I_0$ in the integration region, and $x<<a$

$$ f(\alpha)\sim \frac{e^\alpha}{\alpha} \qquad, \qquad \alpha \to \infty $$

So we have for $A$ (if it is valid to differentiate the asymptotic relation for $f$)

$$ A(\alpha)\sim -1 + \frac{1}{\alpha} \qquad, \qquad \alpha \to \infty $$

Solution 2:

By a simple change of variables, we obtain $$ \int_{ - 1}^1 {I_0 (\sqrt {1 - x^2 } )e^{ - \alpha x} dx} = e^\alpha \int_0^2 {I_0 (\sqrt {t(2 - t)} )e^{ - \alpha t} dt} $$ and $$ \int_{ - 1}^1 {I_0 (\sqrt {1 - x^2 } )xe^{ - \alpha x} dx} = e^\alpha \int_0^2 {I_0 (\sqrt {t(2 - t)} )(t - 1)e^{ - \alpha t} dt} . $$ We have the following Maclaurin expansions about $t=0$: \begin{align*} I_0 (\sqrt {t(2 - t)} ) & = 1 + \frac{t}{2} - \frac{{3t^2 }}{{16}} - \frac{{17}}{{288}}t^3 + \ldots , \\ I_0 (\sqrt {t(2 - t)} )(t - 1) & = - 1 + \frac{t}{2} + \frac{{11t^2 }}{{16}} - \frac{{37}}{{288}}t^3 + \ldots \; . \end{align*} Therefore, by Watson's lemma, \begin{align*} \int_{ - 1}^1 {I_0 (\sqrt {1 - x^2 } )e^{ - \alpha x} dx} & \sim \frac{{e^\alpha }}{\alpha }\left( {1 + \frac{1}{{2\alpha }} - \frac{3}{{8\alpha ^2 }} - \frac{{17}}{{48\alpha ^3 }} + \ldots } \right), \\ \int_{ - 1}^1 {I_0 (\sqrt {1 - x^2 } )xe^{ - \alpha x} dx} & \sim - \frac{{e^\alpha }}{\alpha }\left( {1 - \frac{1}{{2\alpha }} - \frac{{11}}{{8\alpha ^2 }} + \frac{{37}}{{48\alpha ^3 }} + \ldots } \right) \end{align*} as $\alpha \to +\infty$. Taking the ratios and re-expanding in inverse powers of $\alpha$ yields $$ \frac{{\int_{ - 1}^1 {I_0 (\sqrt {1 - x^2 } )xe^{ - \alpha x} dx} }}{{\int_{ - 1}^1 {I_0 (\sqrt {1 - x^2 } )e^{ - \alpha x} dx} }} \sim - 1 + \frac{1}{\alpha } + \frac{1}{{2\alpha ^2 }} - \frac{1}{{\alpha ^3 }} + \ldots , $$ as $\alpha \to +\infty$.