Integral of a modified Bessel function of the second kind 1st order

Question

in Gradshteyn-Ryzhik see Bessel functions and exponentials 6.611 (9), there is the formula for $$\int_0^\infty e^{-\alpha x}K_0(\beta x)\,dx$$ My question is: what about $$\int_0^\infty e^{-\alpha x}K_1(\beta x)\,dx$$

Answer 1

$$I=\int_0^\infty e^{-\alpha x}K_1(\beta x)\,dx$$ does not converge.

Let $ x=\frac t \beta$ and $k=\frac \alpha \beta$ $$I=\frac 1 \beta\int_0^\infty e^{-k t}K_1(t)\,dt$$ Around $t=0$ $$K_1(t)=\frac{1}{t}+\frac{1}{4} t (2 \log (t)+2 \gamma -1-2 \log (2))+O\left(t^3\right)$$ $$J=\int e^{-kt} \Big[\frac{1}{t}+\frac 12 t\log(t)+c\Big]\,dt$$ $$J=\frac{\left(2 k^2+1\right) \text{Ei}(-k t)-e^{-k t} (2 c k+k t \log (t)+\log (t)+1)}{2 k^2}$$ which tends to $-\infty$ when $t\to 0$.