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

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

  • $\begingroup$ Welcome to MSE! Please include some of the things you have tried so that we can see where you are stuck. $\endgroup$
    – mrp
    Feb 12 '15 at 11:56
  • $\begingroup$ $\int_2^x {dt\over log(t)} = \text{li}(x)-\text{li}(2)$, I presume $\endgroup$ Feb 12 '15 at 11:58
  • $\begingroup$ Hint: Use the asymptotic expansion of the exponential integral $\endgroup$
    – tired
    Feb 12 '15 at 15:39
  • $\begingroup$ [Disclaimer: I am a student, not a professional mathematician.] You can get it directly by repeated partial integration. $\endgroup$
    – student
    Jan 4 '18 at 22:30

This is a very sketchy answer!

Observe, that


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.


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