# How to get asymptotic expansion of the sum of modified Bessel function $\sum_{n=1}^\infty K_0(s\, n)$ as $s\to 0^+$?

I guess for the modified Bessel funcion $$K_0(z)$$, $$\sum_{n=1}^\infty K_0(s\, n) \sim \frac{-2\log 2 - \log \pi + \gamma}{2} + \frac{\log s}{2} + \frac{\pi}{2\, s}, \quad s\to 0^+,$$ if taking $$\sum_{n=1}^\infty 1 =\zeta(0),$$ and $$\sum_{n=1}^\infty \log n = \log \prod_{n=1}^\infty n=\log \sqrt{2\pi},$$ I can get all the rest terms, but where does the term $$\frac{\pi}{2\,s}$$ come from?

The term $$\frac{\pi}{2s}$$ comes from the pole of $$\zeta(s)$$. Let's use the fact that Mellin transform of $$K_0(s)$$ equals $$\int_0^{+\infty} K_0(s)s^{t-1}ds=2^{t-2}\Gamma^2(t/2).$$

From this we get that if $$f(s)=\sum_{n} K_0(sn)$$ then for $$\mathrm{Re}\,s>1$$ we have

$$\int_0^{+\infty} f(s)s^{t-1}ds=2^{t-2}\Gamma^2(t/2)\zeta(t)$$

Therefore, by Mellin inversion we get

$$f(s)=\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty} 2^{t-2}\Gamma^2(t/2)\zeta(t)s^{-t}dt.$$

Now, if you move your contour far to the left, you will encounter poles of the integrand at the points $$t=1,0,-2,-4,-6\ldots$$. Using the fact that $$\Gamma$$ decays exponentially on the vertical lines and $$\zeta$$ grows at most polynomially, we obtain for any $$N$$

$$f(s)=\mathrm{Res}_{t=1} 2^{t-2}\zeta(t)\Gamma^2(t/2)s^{-t}+\sum_{n=0}^{N} \mathrm{Res}_{t=-2n} 2^{t-2}\zeta(t)\Gamma^2(t/2)s^{-t}+O(s^{-1-2N})$$

Computation of the first term reveals the desired $$\frac{\pi}{2s}$$ term, as $$\Gamma(1/2)^2=\pi$$.