I am trying to understand if the integral $$f(R)=\int_a^R\frac{K_1(r)dr}{r}$$ has a finite limit for $R\rightarrow\infty$.
With Wolfram Mathematica I got the following primitive:
$$\frac{1}{4} G_{1,3}^{2,1}\left(\frac{r}{2},\frac{1}{2}| \begin{array}{c} 1 \\ -\frac{1}{2},\frac{1}{2},0 \\ \end{array} \right)$$
but I can't get the limit for $r\rightarrow\infty$. If I plot it, it seems to go to me value around $-1.57$ but, again, Mathematica does not seem to handle these Meijer G-functions very well and after $r\sim 30$ it blows up to infinity.
Of course, the first derivative of this function is:
$$f^\prime(R)=\frac{K_1(R)}{R}$$
which goes to zero for $R\rightarrow\infty$. But this does not give me the constant to which it tends.
Any ideas or reference?
I believe the value of the constant (finite limit) should be $$ \int_1^\infty dt \frac{e^{-a t}\sqrt{t^2-1}}{t}\ . $$ [checked with Mathematica for a few values] You can just use the integral representation $$ K_1(r)=r \int_1^\infty e^{-rt}\sqrt{t^2-1}dt $$ and swap the integrals. The integral in $r$ is easy $$ \int_a^R dr e^{-r t}=\frac{e^{-a t}-e^{-R t}}{t} $$ and the surviving integral in $t$ has a dependence in $R$ that can be easily evaluated.