What is $\operatorname{Ei}(x)$?

Question

I was trying to solve

$$\int\frac{e^x - e^{-x}}{x}\,dx$$

But I have no idea how to do it and the calculator said to use a common integral that I don't know what it means.

Answer 1

There is no elementary anti-derivative for that function. Probably the most compact way to represent the integral in terms of special functions is in terms of the hyper-sine integral:

$$\begin{align} \int\frac{e^x-e^{-x}}{x}\mathrm{d}x &=2\int\frac{\sinh{x}}{x}\mathrm{d}x\\ &=2\operatorname{Shi}{(x)}+\color{grey}{\text{constant}}. \end{align}$$

Answer 2

The exponential integral function $\mathrm{Ei}(x)$ is defined by

$$\mathrm{Ei}(x) = \int_{-\infty}^{x} \frac{e^{t}}{t} dt$$.

These functions are not elementary, so they cannot be reduced to a finite combination of the arithmetic operations including exp and log (and the trig functions, but these are related to exp/log via the complex numbers).

The best you can do in that regard is an infinite sum of some form, such as by integrating the Laurent series for $\frac{e^{-x}}{x}$ and taking the Cauchy principal value:

$$\mathrm{Ei}(x) = \gamma + \ln(|x|) + \sum_{k=1}^{\infty} \frac{x^k}{k\ k!}$$

where $\gamma$ is the Euler gamma constant.

There is also a proof that these integrals cannot be solved in elementary terms, but it requires abstract algebra ("Differential Galois Theory").

Answer 3

Related technique. You can use the power series approach. First note that

$$ e^{x}-e^{-x}=\sum_{k=1}^{\infty}( 1-(-1)^k )\frac{x^k}{k!} = 2\sum_{k=0}^{\infty}\frac{x^{2k+1}}{(2k+1)!}.$$

Back to our integral we have

$$ \int \frac{e^{x}-e^{-x}}{x}dx = \sum_{k=0}^{\infty}\frac{1}{(2k+1)!}\int x^{2k}dx + C = 2\,\rm Shi(x)+C . $$