<s>[INST] 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 $x$.
 [/INST] 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,
$$_1F_2\left(-\frac{1}{3};-\frac{2}{3},\frac{2}{3};u\right)=\sum_{n=0}^\infty \frac{\Gamma \left(-\frac{2}{3}\right)}{(1-3 n) n! \Gamma \left(n-\frac{2}{3}\right)}u^n$$
$$_1F_2\left(\frac{4}{3};\frac{7}{3},\frac{8}{3};u\right)=\sum_{n=0}^\infty \frac{4 \Gamma \left(\frac{8}{3}\right)}{(3 n+4) n! \Gamma \left(n+\frac{8}{3}\right)} u^n$$
</s>