I am looking for a compact result of this integral:
$$\int_R^\infty k_n(x) \, dx,$$
where $k_n$ is the modified spherical Bessel function of the second kind (explanation to this function).
Actually, wolframalpha has an antiderivate, but it is quite messy and I guess that the upper limit should always be zero. Can we simplify something there for $R>0$? ${{{{}}}}$
It turns out that $k_n(x)$ has a pretty nifty integral representation:
$$k_n(x) = \frac{\pi}{2} \int_1^{\infty} dt \, P_n(t) \, e^{-x t}$$
where $P_n$ is a Legendre polynomial of the first kind, so that
$$\int_R^{\infty} dx \, k_n(x) = \frac{\pi}{2} R \int_1^{\infty} dt \, P_n(t) \, \int_1^{\infty} du \, e^{-R t u}$$
(Note that I reversed the order of integration and made the substitution $x=R u$.) We may simplify this to a single integral:
$$\int_R^{\infty} dx \, k_n(x) = \frac{\pi}{2} \int_1^{\infty} dt \, P_n(t) \frac{e^{-R t}}{t}$$
Clearly, we may use the recurrence relation for the Legendre polynomials:
$$(n+1) P_{n+1}(x) = (2 n+1) P_n(x) - n P_{n-1}(x)$$
in conjunction with the initial values of the integral at $n=0$ and $n=1$:
$$\int_R^{\infty} dx \, k_0(x) = -\frac{\pi}{2}\text{Ei}(-R)$$ $$\int_R^{\infty} dx \, k_1(x) = \frac{\pi}{2} \frac{e^{-R}}{R}$$
uniquely specifies the value of the integral for arbitrary $n$.