logarithmic integral function and asymptotic expansion

Question

Show that Logarithmic integral function $$int_2^x {1over log(t)} , dt = Li(x)$$  has asymptotic expansion of the form $${xover log(x)}cdotsum_{j=0}^infty a_jcdot (log(x))^{-j}.$$

I tried different stuff but I did not conclude to solve it. Any help for solving this?

tired · Accepted Answer

This is a very sketchy answer!

Observe, that

$$text{Li(x)}=-int_{log(2)}^{log(x)}frac{e^{-y}}{y}=text{Ei}(log(x))-text{Ei}(log(2))$$

where $text{Ei(x)}=int_{x}^{infty}frac{e^{-q}}{q}$   the Expontential integral .

Doing intgration by parts N times we obtain 
$$
text{Ei(x)}=frac{e^{-x}}{x}-frac{e^{-x}}{x^2}+frac{2!e^{-x}}{x^3}+....+(-1)^{N-1}frac{e^{-x}(N-1)!}{x^n}+R(N)
$$
where $R(N)$ is the remaining integral.

We conclude that
$$
text{Ei(x)}=frac{e^{-x}}{x}sum_{k=1}^{infty}(-1)^{k+1}frac{k!}{x^k}
$$

in an asymptotic sense. 
Now plug in $log(x)$ and $log(2)$ and you are done.

logarithmic integral function and asymptotic expansion

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