Sum of modified Bessel functions of the second kind

Question

Does anyone have any suggestions for solving the following sum: $$\sum_{m=1}^\infty \frac{1}{m^{2a-\frac{1}2}} K_{a}(m\, b)$$ where $a$ and $b$ are both real and positive.

Answer 1

Well, we know that:

$$\text{K}_\sigma\left(x\right):=\int_0^\infty\frac{\cosh\left(\sigma t\right)}{\exp\left(x\cosh\left(t\right)\right)}\space\text{d}t\tag1$$

So, we get:

$$\mathscr{I}:=\sum_{\text{m}=1}^\infty\frac{\text{K}_\text{a}\left(\text{b}\cdot\text{m}\right)}{\text{m}^{2\text{a}-\frac{1}{2}}}=\sum_{\text{m}=1}^\infty\left\{\frac{1}{\text{m}^{2\text{a}-\frac{1}{2}}}\cdot\int_0^\infty\frac{\cosh\left(\text{a}\cdot t\right)}{\exp\left(\text{b}\cdot\text{m}\cdot\cosh\left(t\right)\right)}\space\text{d}t\right\}=$$ $$\int_0^\infty\cosh\left(\text{a}\cdot t\right)\cdot\left\{\sum_{\text{m}=1}^\infty\frac{1}{\text{m}^{2\text{a}-\frac{1}{2}}}\cdot\frac{1}{\exp\left(\text{b}\cdot\text{m}\cdot\cosh\left(t\right)\right)}\right\}\space\text{d}t\tag2$$

Let $\text{u}=\text{b}\cdot\cosh\left(t\right)$so we get:

$$\sum_{\text{m}=1}^\infty\frac{1}{\text{m}^{2\text{a}-\frac{1}{2}}}\cdot\frac{1}{\exp\left(\text{u}\cdot\text{m}\right)}\tag3$$