Is there any analytical approximation to this integral? I have an integral:$$ I(x) = x\int_{x}^{\infty}K_{5/3}(\xi)\,\rm{d}\xi~,$$
where $K$ is modified Bessel function.
I need to evaluate this integral for tens of thousand different $x$ ultimately to use it for some numerical calculation. In the extreme limits of $x\gg 1$ or $x\sim 0$, the approximations are:
$$\lim_{x\rightarrow 0}I(x) = \frac{4\pi}{\sqrt{3}\Gamma(1/3)}\left(\frac{x}{2}\right)^{1/3}~,$$
and
$$\lim_{x\rightarrow \infty}I(x) = \sqrt{\frac{\pi x}{2}}e^{-x}~.$$
However, I would like to know if there are any approximate expressions $I_\text{approx}(x)$ valid for all $x$ such that $$\frac{|I_\text{approx}(x)-I(x)|}{I(x)}\lesssim 0.1~,$$
for all $x$.
 A: By http://dlmf.nist.gov/10.29.E1 (or http://functions.wolfram.com/03.04.20.0006.01), we obtain
$$
\int_x^\infty  {K_{5/3} (t)\mathrm{d}t}  =  - \int_x^\infty  {(K_{1/3} (t) + 2K'_{2/3} (t))\mathrm{d}t}  = 2K_{2/3} (x) - \int_x^\infty  {K_{1/3} (t)\mathrm{d}t} .
$$
By the change of integration variables $t = \frac{2}{3}s^{3/2}$ and http://dlmf.nist.gov/9.6.E15, this becomes
$$
2K_{2/3} (x) - \pi \sqrt 3 \int_{(3x/2)^{2/3} }^\infty  {\operatorname{Ai}(s)\mathrm{d}s} ,
$$
where $\operatorname{Ai}$ is the Airy function. By http://dlmf.nist.gov/9.10.E1 and then by http://dlmf.nist.gov/9.6.E15 and http://dlmf.nist.gov/9.6.E16, this further equals to
\begin{align*}
& 2K_{2/3} (x) - \pi ^2 \sqrt 3 \left( {\operatorname{Ai}((3x/2)^{2/3} )\operatorname{Gi}'((3x/2)^{2/3} ) - \operatorname{Ai}'((3x/2)^{2/3} )\operatorname{Gi}((3x/2)^{2/3} )} \right)
\\ & = 2K_{2/3} (x) \!-\! \pi\! \left( {(3x/2)^{1/3} K_{1/3} (x)\!\operatorname{Gi}'((3x/2)^{2/3} ) + (3x/2)^{2/3} K_{2/3} (x)\!\operatorname{Gi}((3x/2)^{2/3} )} \right)\!.
\end{align*}
Here $\operatorname{Gi}$ denotes the Scorer function (see http://dlmf.nist.gov/9.12).
A: Using hypergeometric functions, the antiderivative exists. So,
$$I(x)=-\frac{\pi  x}{\sqrt{3}}-\frac{2^{2/3} \sqrt{3} \pi  x^{1/3} \,
   _1F_2\left(-\frac{1}{3};-\frac{2}{3},\frac{2}{3};\frac{x^2}{4}\right)}{\Gamma
   \left(-\frac{2}{3}\right)}-$$ $$\frac{81 x^{11/3} \Gamma \left(\frac{4}{3}\right) \,
   _1F_2\left(\frac{4}{3};\frac{7}{3},\frac{8}{3};\frac{x^2}{4}\right)}{320\
   2^{2/3}}$$
As series,
$$_1F_2\left(-\frac{1}{3};-\frac{2}{3},\frac{2}{3};u\right)=\sum_{n=0}^\infty \frac{\Gamma \left(-\frac{2}{3}\right)}{(1-3 n) n! \Gamma \left(n-\frac{2}{3}\right)}u^n$$
$$_1F_2\left(\frac{4}{3};\frac{7}{3},\frac{8}{3};u\right)=\sum_{n=0}^\infty \frac{4 \Gamma \left(\frac{8}{3}\right)}{(3 n+4) n! \Gamma \left(n+\frac{8}{3}\right)} u^n$$
