Integral over product of 1/x, exp(-x), J0 and J1 Bessel function Is there an analytic expression for the integral
$$\int_0^\infty dx \frac{1}{x} e^{-\lambda x} J_1(xy) J_0(xz)$$
with $\lambda\geq0$ and $y,z>0$?
I know that there is a (complicated) closed form for the integral without the $\frac{1}{x}$, but that expression can't be integrated with respect to $\lambda$, so expressing $\frac{1}{x}$ as an integral of $e^{-\lambda x}$ over $\lambda$ is not an option.
Numerical integration is slow and unreliable due to the oscillatory behavior of the integrand.
Edit: Possibly, the original integral is to general. In the end, I am trying to solve it for a specific value of $\lambda$, namely
$$\int_0^\infty dx \frac{1}{x} (1-e^{-x}) J_1(xy) J_0(xz)$$
 A: Your integral is in Gradshteyn-Ryzhik formula 6.626.1. Note that there is a mistake in the book. The correct result is (valid for $z\geq y$)
$$ \int_0^\infty \frac{dx}{x} e^{-\lambda x} J_1(x y) J_0(x z) = \frac{y}{2\lambda} \sum_{m=0}^\infty C_m \;{}_2 F_1(m+\tfrac12,m+1;1;-z^2/\lambda^2) \left( - \frac{y^2}{4 \lambda^2}\right)^m \tag{1}$$
with $C_m$ the Catalan numbers. The series is alternating an decaying quickly; it is thus well-suited for numerical calculation.
In particular, the op is interested in the result for $\lambda=0$. We can take the limit of (1) and obtain (also valid for $z \geq y$)
$$ \int_0^\infty \frac{dx}{x}  J_1(x y) J_0(x z) = \sum_{m=0}^\infty
 2 (m+1) (C_m)^2\,\left(\frac{y}{4z}\right)^{2m+1}.$$
For $z\leq y$, we need the alternative expansion
$$ \int_0^\infty \frac{dx}{x} e^{-\lambda x} J_1(x y) J_0(x z) = \frac{y}{2\lambda} \sum_{m=0}^\infty (m+1)C_m \;{}_2 F_1(m+\tfrac12,m+1;2;-y^2/\lambda^2) \left( - \frac{z^2}{4 \lambda^2}\right)^m ;\tag{2}$$
and evaluating it for $\lambda=0$, we obtain the result (valid for $z\leq y$)
$$ \int_0^\infty \frac{dx}{x}  J_1(x y) J_0(x z) = \sum_{m=0}^\infty \frac{1}{1-2m}
 (m+1)^2 (C_m)^2\,\left(\frac{z}{4y}\right)^{2m}.$$
