$\sum _{m=0}^{\infty} \frac{b^m }{(m!)^2}K_{m+\frac{1}{2}}(a)$ with Bessel K How can this expression reformulated without sum or integral? $K$ is a modified Bessel function of the $2^{\text{nd}}$ kind with positive half-integer order. Due to the squared factorial the sum converges very quickly.
\begin{align*}\sum _{m=0}^{\infty} \frac{b^m}{(m!)^2}K_{m+\frac{1}{2}}(a)& =\frac{1}{\sqrt{2 \pi  a}}\int_{-\infty }^{\infty } \frac{\exp  \left(\frac{2b}{a(t^2+1)}+i a t\right)}{t^2+1} \, {\rm d}t\\ & =\frac{1}{2}\int_0^\infty e^{-a \cosh(t)} e^{-\frac t2}\bigg(\text I_0\big(2\sqrt be^{-\frac t2}\big)+e^t\text I_0\big(2\sqrt be^\frac t2\big)\bigg){\rm d}t,
\end{align*}
with $a$, $b>0$ and $i$ being the imaginary unit.
The last expression contains Bessel ${\rm I}_0$ functions and was derived in the post below. An integral-free and sum-free representation is still missing.
The Bessel $K$ function can be expressed as a finite sum however then one has to deal with a double sum.
$$K_{m+\frac{1}{2}}(a)=\sqrt{\frac{\pi }{2 a}} \frac{1}{e^a}\sum _{k=0}^m \frac{(m+k)!}{k!(m-k)!}\frac{1}{(2a)^k}$$
 A: Let’s try to find a solution using an integral representation of Bessel K:
$$\sum _{m=0}^{\infty} \frac{b^m }{(m!)^2}\text K_{m+\frac{1}{2}}(a)= \sum _{m=0}^{\infty} \frac{b^m }{(m!)^2}\int_0^\infty e^{-a \cosh(t)}\cosh \bigg(\bigg( m+\frac12\bigg)t\bigg) dt$$
The inside is a Bessel K sum with $\text{Re}(a)>0$ with term by term integration:
$$\sum _{m=0}^{\infty} \frac{b^m }{(m!)^2}\int_0^\infty e^{-a \cosh(t)}\cosh \bigg(\bigg( m+\frac12\bigg)t\bigg) dt =\int_0^\infty e^{-a \cosh(t)}\sum _{m=0}^{\infty} \frac{b^m }{(m!)^2}\cosh \bigg(\bigg( m+\frac12\bigg)t\bigg) dt =\boxed{\frac 12\int_0^\infty e^{-a \cosh(t)} e^{-\frac t2}\bigg(\text I_0\big(2\sqrt be^{-\frac t2}\big)+e^t\text I_0\big(2\sqrt be^\frac t2\big)\bigg)dt}$$
Which works as seen with this $a=4,b=2$ case with the Bessel I function Sorry if this is not a closed form, but now you can try to evaluate the above integral. The integral seems to have no closed form for special cases of $b,a$ and there is no way I see to put it into a general hypergeometric form. Please correct me and give me feedback!
Also see:

Evaluating the integral $\int_0^{2\pi}e^{-\sqrt{a-b\cos t}}\mathrm dt$

Where a very similar sum is the result for the integral, but no closed form was provided in the answer by @Yuriy.
