# Logarithmic and exponential integrals

1. Jun 13, 2012

### Millennial

In this text, I will ask a question about the power series expansion of exponential and logarithmic integrals.

Now, to avoid confusion, I will first give the definitions of the two:
$$\mathrm{Ei}(x)=\int_{-\infty}^{x}\frac{e^t}{t}dt$$
$$\mathrm{Li}(x)=\int_{0}^{x}\frac{dt}{\log(t)}$$
where Ei denotes the exponential integral function and Li denotes the logarithmic integral function.

I think it is clear from now that these integrals are not expressable in elementary terms as indefinite ones. My question is about expressing these as power series.

Now, one can substitute $t=e^u$ in the logarithmic integral to get
$$\mathrm{Li}(x)=\int_{0}^{x}\frac{dt}{\log(t)}=\int_{-\infty}^{\log(x)}\frac{e^u}{u}du=\mathrm{Ei}(\log(x))$$

so my question can be downgraded to only the power series expansion of the exponential integral rather than both. My question should be obvious by now: How do we obtain a power series expansion for the exponential integral, since term-by-term integration does not work after plugging in the Taylor series for the exponential function?

2. Jun 13, 2012

### haruspex

For x > 0, I believe Ei is only meaningful in the sense of a Cauchy principal value. So maybe you can break it into
$$\mathrm{Ei}(x)=\int_{-\infty}^{-x}\frac{e^t}{t}dt + \int_{-x}^{x}\frac{e^t}{t}dt$$
$$\int_{-x}^{x}\frac{e^t}{t}dt = \int_{0}^{x}\frac{e^t-e^{-t}}{t}dt$$
Then the Taylor series should work for (et-e-t)/t.
The remaining portion is the same as the x < 0 case, which by swapping sign becomes:
$$\int_{x}^{\infty}\frac{e^{-t}}{t}dt = \int_{1}^{\infty}\frac{e^{-t}}{t}dt - \int_{1}^{x}\frac{e^{-t}}{t}dt$$
Now the Taylor expansion works in the second integral. The first integral is some constant.