TITLE: Missing term in series expansion
QUESTION [8 upvotes]: I asked a similar question before, but now I can formulate it more concretely. I am trying to perform an expansion of the function $$f(x) = \sum_{n=1}^{\infty} \frac{K_2(nx)}{n^2 x^2},$$ for $x \ll 1$. Here, $K_2(x)$ is the modified Bessel function of the second kind. This series is a result of solving the integral $$f(x) = \frac{1}{3}\int_1^\infty \frac{(t^2-1)^{3/2}}{\mathrm{e}^{xt}-1}\mathrm{d}t.$$ The stated result is $$f(x) \approx \frac{\pi^4}{45 x^4} - \frac{\pi^2}{12 x^2}+\frac{\pi}{6x}-\frac{1}{32}\left(  \frac{3}{2}-2\gamma+2\ln4\pi-\ln x^2\right)+\mathcal{O}(x^2),$$ where $\gamma$ is the Euler-Mascheroni constant. It agrees numerically with $f(x)$ for small $x$. However, by using the series expansion of the Bessel function $$K_2(nx) = \frac{2}{n^2x^2}-\frac{1}{2}+\frac{1}{2}\sum_{k=0}^\infty \left[\psi(k+1)+\psi(k+3)-\ln\frac{n^2x^2}{4}\right]\frac{\left(\frac{n^2 x^2}{4}\right)^{k+1}}{k!(k+2)!},$$ with $\psi(x)$ being the digamma function and using the zeta regularization for summation over $n$, I am able to reproduce all the terms except $\frac{\pi}{6x}$. I.e., my result is $$f(x) = \frac{2\zeta(4)}{x^4} - \frac{\zeta(2)}{2x^2} + \frac{1}{8}\sum_{k=0}^\infty \left[\left(\psi(k+1)+\psi(k+3)-\ln\frac{x^2}{4}\right)\zeta(-2k) + 2 \zeta'(-2k)\right]\frac{\left(\frac{x^2}{4}\right)^{k}}{k!(k+2)!}.$$ It seems very strange that the $\frac{\pi}{6x}$ term should appear in the expansion since only even powers of $x$ appear in $K_2(nx)$. But, numerically, it is certainly there. How did I miss it?
Edit #1: I just got an idea where the $\frac{\pi}{6x}$ term might come from! Approximating the integral representation of $f(x)$ for $xt \ll 1$ and using the UV regulator $\Lambda$, we have $$f(x) \approx \frac{1}{3} \int_1^\Lambda \frac{(t^2-1)^{3/2}}{xt}\mathrm{d}t \approx \frac{\Lambda^3}{9x} - \frac{\Lambda}{2x} + \frac{\pi}{6x}.$$ OK, so now the question is why doesn't the Bessel series see this term and what to do about it?
Edit #2: The missing term might indicate that the zeta regularization isn't used properly. The term $\frac{\pi}{6x}$ appears right in the middle, separating the convergent sums $\zeta(4)$ and $\zeta(2)$ from the (regularized) divergent sums $\zeta(-2k)$ and $\zeta'(-2k)$. So, the missing term may be the price to pay for using the zeta regularization. Unfortunately, I don't know enough math to come to any decisive conclusion.
Edit #3: In edit #1 I argued that the missing term $\frac{\pi}{6x}$ comes from expansion of the exponential in the denominator. When I expand the numerator in the binomial series $$f(x) = \frac{1}{3} \int_1^\infty \frac{t^3}{\mathrm{e}^{xt}-1}\sum_{k=0}^\infty \binom{3/2}{k}(-t^{-2})^k\mathrm{d}t$$ and perform the integration, I again obtain my original result without $\frac{\pi}{6x}$. So, depending on what I choose to expand, I obtain different (incomplete) results??

REPLY [3 votes]: Using Mellin transforms we can find the asymptotics in an intuitive and straightforward manner. We have that
$$ \mathfrak{M}(K_2(x)/x^2; s) = 2^{s-4} \Gamma(s/2) \Gamma(s/2-2),$$
so that viewing $f(x)$ as a harmonic sum, we get
$$ \mathfrak{M}(f(x); s) = f^*(s) = 2^{s-4} \Gamma(s/2) \Gamma(s/2-2) \zeta(s).$$
Now invert to get the expansion of $f(x).$
I will give a table of the contributions from the poles down to the pole at $s=-6.$
$$ \begin{array}
\operatorname{Res}(f^*(s) \, x^{-s}; s=4) & = &
1/45\,{\frac {{\pi }^{4}}{{x}^{4}}} \\
\operatorname{Res}(f^*(s) \, x^{-s}; s=2) & = &
-1/12\,{\frac {{\pi }^{2}}{{x}^{2}}} \\
\operatorname{Res}(f^*(s) \, x^{-s}; s=1) & = &
1/6\,{\frac {\pi }{x}} \\
\operatorname{Res}(f^*(s) \, x^{-s}; s=0) & = &
-1/16\,\ln  \left( 4\,\pi  \right) +1/16\,\gamma+1/16\,
\ln  \left( x \right) -{\frac {3}{64}} \\
\operatorname{Res}(f^*(s) \, x^{-s}; s=-2) & = &
{\frac {1}{96}}\,\zeta  \left( 1,-2 \right) {x}^{2} \\
\operatorname{Res}(f^*(s) \, x^{-s}; s=-4) & = &
{\frac {1}{3072}}\,\zeta  \left( 1,-4 \right) {x}^{4} \\
\operatorname{Res}(f^*(s) \, x^{-s}; s=-6) & = &
{\frac {1}{184320}}\,\zeta  \left( 1,-6 \right) {x}^{6}.
\end{array}$$
As to how the Mellin transform of $K_2(x)$ is calculated, I can offer some ideas.
Start with the known integral representation
$$K_\alpha(x) = \int_0^\infty e^{-x \cosh t} \cosh (\alpha t) \; dt,$$
so that 
$$\mathfrak{M}(K_2(x); s) =
\int_0^\infty  \int_0^\infty e^{-x \cosh t} \cosh (2t) \; dt \; x^{s-1} \; dx.$$
This becomes
$$ \int_0^\infty \cosh (2t)  \int_0^\infty  e^{-x \cosh t}  x^{s-1} \; dx \; dt
= \Gamma(s) \int_0^\infty \frac{\cosh (2t)}{(\cosh t)^s} dt.$$
Now using $$\cosh(2t) = 2\cosh(t)^2 - 1,$$ we obtain
$$ \Gamma(s) 
\left(2 \int_0^\infty \frac{1}{(\cosh t)^{s-2}} dt
-  \int_0^\infty \frac{1}{(\cosh t)^s} dt \right).$$
Furthermore,
$$ \int_0^\infty \frac{1}{(\cosh t)^s} dt =
2^s  \int_0^\infty \frac{1}{(e^t + e^{-t})^s} dt =
2^s  \int_0^\infty \frac{1}{e^{ts}} \frac{1}{(1 + e^{-2t})^s} dt$$
which is
$$ 2^s  \int_0^\infty \frac{1}{e^{ts}} 
\sum_{q\ge 0} (-1)^q \binom{q+s-1}{q} e^{-2qt} dt =
2^s  \int_0^\infty
\sum_{q\ge 0} (-1)^q \binom{q+s-1}{q} e^{-(2q+s)t} dt $$
or
$$ 2^s \sum_{q\ge 0} (-1)^q \binom{q+s-1}{q}  \int_0^\infty e^{-(2q+s)t} dt =
 2^s \sum_{q\ge 0} (-1)^q \binom{q+s-1}{q}  \frac{1}{2q+s}.$$
Now this last term is
$$ 2^s \frac{1}{s} \; _2F_1(s/2, s; 1+s/2; -1). $$
But we have
$$ _2F_1(a, 2a; a+1; -1) = \frac{1}{2a} \frac{\Gamma(a+1)^2}{\Gamma(2a)}$$
as can be seen e.g. here.
Hence $$\int_0^\infty \frac{1}{(\cosh t)^s} dt = 
2^s  \frac{1}{s^2}  \frac{\Gamma(s/2+1)^2}{\Gamma(s)}.$$
Concluding, we have shown that
$$\mathfrak{M}(K_2(x);s) =
2 \Gamma(s) 2^{s-2} \frac{1}{(s-2)^2}  \frac{\Gamma(s/2)^2}{\Gamma(s-2)}
- \Gamma(s) 2^s  \frac{1}{s^2}  \frac{\Gamma(s/2+1)^2}{\Gamma(s)} \\ =
2^{s-1}  \frac{s-1}{s-2}\Gamma(s/2)^2
- 2^s  \frac{1}{s^2} \Gamma(s/2+1)^2 =
2^{s-1}  \frac{s-1}{s-2}\Gamma(s/2)^2 - 2^{s-2} \Gamma(s/2)^2
= 2^{s-2} \Gamma(s/2)^2 \left( 2  \frac{s-1}{s-2} - 1\right) 
= 2^{s-2} \Gamma(s/2)^2 \frac{s}{s-2} =
2^{s-2} \Gamma(s/2)^2 \frac{s/2}{s/2-1} =
2^{s-2} \Gamma(s/2+1) \Gamma(s/2-1).$$