Limit of integral with a Bessel function I am interested in finding the limit of the following ratio:
\begin{equation}
A=\lim_{\alpha\to+\infty}\frac{\int_{-1}^1\mathrm d x I_0(\sqrt{1-x^2})e^{-\alpha x}x}{\int_{-1}^1\mathrm d x I_0(\sqrt{1-x^2})e^{-\alpha x}}
\end{equation}
Where $I_\nu(x)$ is the modified Bessel function of the first kind.
A numerical check suggests that it should be $A=-1$, but I could not find a way to prove it. A series expansion of $I_0$ did not seem to help, neither the decomposition into positive and negative part of the integration domain. Any idea on how to tackle it?
 A: 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 $f$)
$$
A(\alpha)\sim -1 + \frac{1}{\alpha} \qquad, \qquad \alpha \to \infty
$$
A: 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 $\alpha \to +\infty$.
