The pdf of the K-distribution is given by
$$ f_X(x) = \frac{2}{x} \frac{1}{\Gamma(L)\Gamma(v)}\left(\frac{Lvx}{\mu} \right)^{\frac{v+L}{2}} K_{v-L}\left(2\sqrt{\frac{Lvx}{\mu}} \right) $$
for which I attempted to calculate the cdf both for the general case but most importantly for $L=1$. I managed to express the cdf in terms of generalized hypergeometric functions but I also found a paper which stated that (for the case $L=1$)
$$ F_X(z) = \int_0^z f_X(x) dx = 1-\frac{2}{\Gamma(v)}\left(\frac{vz}{\mu}\right)^{v/2} K_{v}\left(2\sqrt{\frac{vz}{\mu}}\right) $$
but without proof. Both the found expression and the one I managed to obtain give the same answer when I check numerically for various values. I'd rather use the one above, but I don't quite seem to obtain it.
My attempt was the following:
Using the identities $K_{\alpha}(x) = \frac{\pi}{2} \frac{I_{-\alpha}-I_{\alpha}}{\sin(\alpha\pi)}$ and then having $I_{\alpha} = \frac{(x/2)^{\alpha}}{\Gamma(\alpha+1)}{}_0F_1(-;\alpha+1;\frac{1}{4}x^2)$ I obtained that the indefinite integral
$$ \int x^{s-1}I_v(2\sqrt{x})dx = \frac{1}{\Gamma(v+1)}\int x^{s+v/2-1} \sum_{n=0}^{\infty} \frac{1}{(v+1)_n} \frac{x^n}{n\!} dx $$
which after changing the order of integration and summation gives
$$ I = x^{s+v/2}\Gamma(s+v/2){}_1\tilde{F}_2(s+v/2;v+1,s+v/2+1;x) $$
where ${}_1\tilde{F}_2(s+v/2;v+1,s+v/2+1;x)$ is the regularized version of a generalized hypergeometric function. For the specific case and using the identities above, I obtained the cdf as (using an indefinite integral where I thought of inserting the limits later on)
\begin{equation} \displaylines{ F_X(x) = \frac{2}{x} \frac{1}{\Gamma(L)\Gamma(v)}\left(\frac{Lv}{\mu} \right)^{\frac{v+L}{2}} \int \left(x\right)^{\frac{v+L}{2}}K_{v-L}\left(2\sqrt{\frac{Lvx}{\mu}} \right)dx \\ = \pi\csc(\pi(v-L)) \left(\left(\frac{Lv}{\mu}\right)^L \frac{x^L}{\Gamma(v)}{}_1\tilde{F}_2\left(L;L-v+1,L+1;\frac{Lvx}{\mu}\right) -\left(\frac{Lv}{\mu}\right)^v \frac{x^v}{\Gamma(L)}{}_1\tilde{F}_2\left(v;v-L+1,v+1;\frac{Lvx}{\mu}\right)\right) + C } \end{equation}
Then, with $L=1$ things simplify a bit further to give
\begin{equation} \pi\csc(\pi(v-1)) \left(\left(\frac{vx}{\mu}\right) \frac{1}{\Gamma(v)}{}_1\tilde{F}_2\left(1;2-v,2;\frac{vx}{\mu}\right) -\left(\frac{vx}{\mu}\right)^v {}_1\tilde{F}_2\left(v;v,v+1;\frac{vx}{\mu}\right)\right) \end{equation}
For the second hypergeometric function, it is possible to simplify and one obtains
$$ \left(\frac{vx}{\mu} \right)^v {}_1\tilde{F}_2\left(v;v+1,v;\frac{vx}{\mu} \right) = \frac{z^{v/2}}{\Gamma(v)}I_v\left(2\sqrt{\frac{vx}{\mu}}\right) $$
which is quite hopeful, and I would like to think that it is possible to obtain $I_{-v}(\cdot )$ from the other hypergeometric function. Alas, I haven't made much progress beyond this point...
Suggestions on how to proceed or alternative ways to find the cdf are much appreciated!