In the course of a derivation I encountered the following integral and was unable to find it in standard integral tables (though I found one that looked very similar).
Here, $\beta$ is a positive real number. $$ \int_{0}^{a} \frac{1-e^{-\beta x}}{x}J_0(x) dx $$ Is there a known formula for this integral, or if not, for when the upper integral bound approaches infinity?
Thank you in advance!
To substantiate the partial answer provided by Mathematica in comments above, the following integral is known to be valid when $\beta'\geq 0$: $$\int_0^\infty e^{-\beta' x} J_0(x)\,dx=\frac{1}{\sqrt{1+\beta'^2}}$$ (This is usually found in a table of Laplace transforms.) Integrating from $\beta'=0$ to $\beta$ then yields
$$\int_0^\infty \frac{1-e^{-\beta x}}{x}J_0(x)\,dx=\int_0^\beta \frac{d\beta'}{\sqrt{1+\beta'^2}}=\sinh^{-1}\beta=\operatorname{arcsinh} \beta$$ using a standard integral representation of arcsinh. (A more scrupulous calculation would check that this integration is legitimate, but since the integrand vanishes exponentially fast as $x\to\infty$ I'm not especially concerned.)