I have this integral $$ \int_0^\infty e^{-\alpha x}K_1(\beta \sqrt{x}) \, dx. $$
for $\Re[\alpha] >0$, and $\Re[\beta]>0$
According to the (Table of Integrals, Series, and Products, Seventh Edition), equation 6.614.5 the solution of the above integral is: $$ \frac{\sqrt{\pi}\beta}{8\alpha^{\frac{3}{2}}} \exp \left( \frac{\beta^2}{8\alpha}\right)\left[ K_1 \left(\frac{\beta^2}{8\alpha} \right)- K_0 \left( \frac{\beta^2}{8\alpha}\right)\right] \,, $$
However, when using Mathematica to solve the above equation we get:
$$ \frac{\pi}{2\beta\sqrt{\alpha}} U\left(0.5,0,\frac{\beta^2}{4\alpha}\right).$$
Why this difference?
There is a special relation between Tricomi's U and the Bessel K functions, see e.g. http://functions.wolfram.com/07.33.03.0006.01: $$U(a, 2 a - 1, z) = \frac{e^{\frac{z}{2}} z^{\frac{3}{2} - a}}{2(a - 1) \sqrt{\pi}}\left(K_{a - \frac{1}{2}} \left(\frac{z}{2}\right) - K_{a - \frac{3}{2}}\left(\frac{z}{2}\right)\right)$$ With your $a=1/2$ this simplifies to $$U\left(\frac{1}{2}, 0, z\right) = -\frac{e^{\frac{z}{2}} z}{ \sqrt{\pi}}\left(K_0 \left(\frac{z}{2}\right) - K_1\left(\frac{z}{2}\right)\right)\cdot$$ Now substitute $z=\frac{\beta^2}{4\alpha}$ and multiply by $\frac{\pi}{2\beta\sqrt{\alpha}}$ and you get $$\frac{\pi}{2\beta\sqrt{\alpha}} U\left(\frac{1}{2}, 0, \frac{\beta^2}{4\alpha}\right) = \frac{\sqrt{\pi}\beta}{8\alpha^{\frac{3}{2}}} \exp \left( \frac{\beta^2}{8\alpha}\right)\left( K_1 \left(\frac{\beta^2}{8\alpha} \right)- K_0\left( \frac{\beta^2}{8\alpha}\right)\right)\cdot$$ Therefore there is actually no difference in the results.