ODE satisfied by $\,f(\xi) = \int_0^1 \frac{e^{-\xi x}}{\sqrt{1 - x^2}}dx$ An exercise in a textbook asked me to find the differential equation satisfied by $$f(\xi) = \int_0^1 dx \, \frac{e^{-\xi x}}{\sqrt{1 - x^2}}$$
This seems very difficult since I don't know how to find the integral exactly.  It looks like a Laplace transform.
If do trig substitution, I set $x = \sin \theta$.  I get $dx = \cos \theta = \sqrt{1 - x^2}$.  This is great because I get some cancellation.
$$ f(\xi) = \int_0^{\frac{\pi}{2}} d\theta \, e^{-\xi \sin \theta} $$
This is even worse than I started.  Does this integral have a name?  I need a differential equation in $\xi$.  Thanks.
 A: If we have:
$$ f(\xi)=\int_{0}^{\pi/2}e^{-\xi\sin\theta}d\theta $$
then:
$$ f(\xi)=\sum_{k=0}^{+\infty}\frac{(-1)^k \xi^k}{k!}\int_{0}^{\pi/2}\sin^k\theta\,d\theta=\frac{\pi}{2}\sum_{k=0}^{+\infty}\frac{(-\xi/2)^k}{\Gamma(1+k/2)^2},$$
so:
$$ f(\xi) = \frac{\pi}{2}-\xi+\sum_{k\geq 2}\frac{(-\xi/2)^k}{\Gamma(1+k/2)^2} = \frac{\pi}{2}-\xi+4\sum_{k\geq 0}\frac{(-\xi/2)^{k+2}}{(2+k)^2\Gamma(1+k/2)^2}$$
and:
$$ f'(\xi) = -1-2\sum_{k\geq 0}\frac{(-\xi/2)^{k+1}}{(2+k)\Gamma(1+k/2)^2},\qquad \frac{\xi}{2}f'(\xi)=-\frac{\xi}{2}-2\sum_{k\geq 0}\frac{(-\xi/2)^{k+2}}{(2+k)\Gamma(1+k/2)^2}.$$
Differentiating again we get:
$$\frac{1}{2}f'(\xi)+\frac{\xi}{2}f''(\xi)=-\frac{1}{2}+\frac{\xi}{2}f(\xi),$$
hence $f(\xi)$ satisfies the ODE:

$$\xi\, f''(\xi)+f'(\xi)-\xi\, f(\xi) = -1.$$

A: If
$\,\,\displaystyle f(\xi) = \int_0^1 \frac{\mathrm{e}^{-\xi x}\,dx}{\sqrt{1 - x^2}},\,\,$
then
\begin{align}
f'(\xi) &= -\int_0^1 x\frac{\mathrm{e}^{-\xi x}\,dx}{\sqrt{1 - x^2}}
=\int_0^1 \mathrm{e}^{-\xi x}\Big(\sqrt{1-x^2}\Big)'\,dx\\
&=\mathrm{e}^{-\xi x}\big(\sqrt{1-x^2}\big)\Big|_{\,0}^{1}+\xi\int_0^1
\mathrm{e}^{-\xi x}\sqrt{1-x^2}\,dx=\xi\int_0^1 \mathrm{e}^{-\xi x}\sqrt{1-x^2}\,dx-1.
\end{align}
Also, straight from the definition of $f$,
$$
f''(\xi)=\int_0^1\frac{x^2 \mathrm{e}^{-\xi x}}{\sqrt{1-x^2}}\,dx,
$$
and hence
$$
f(\xi)-f''(\xi)=\int_0^1 \mathrm{e}^{-\xi x}\frac{1-x^2}{\sqrt{1-x^2}}\,dx=\int_0^1 
\mathrm{e}^{-\xi x}\sqrt{1-x^2}\,dx.
$$
Thus,
$$
\xi\big(f(\xi)- f'' (\xi)\big)=\xi\int_0^1 \mathrm{e}^{-\xi x}\sqrt{1-x^2}\,dx=f'(\xi)+1.
$$
Therefore, here is the ODE:
$$
\xi\, f''(\xi)+f'(\xi)-\xi\, f(\xi)=-1.
$$
A: According with Maple the integral gives
$$f \left( \xi \right) =\frac{\pi}{2} \,{{\rm I}_{0}\left(\,\xi\right)}-\frac{\pi}{2}  \,{\it StruveL}_{0} \left(\xi \right) 
$$
and the differential equation is
$${\xi}^{2}{\frac {d^{2}}{d{\xi}^{2}}}f \left( \xi \right) +\xi\,{\frac 
{d}{d\xi}}f \left( \xi \right) -{\xi}^{2}f \left( \xi \right) =-\xi
$$
