# logarithmic integral function and asymptotic expansion

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?

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

## 1 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.