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.
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Why did someone voted down this question? Looks perfectly legit to me. – bartgol Aug 29 '14 at 18:51
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$\operatorname{Ei}(x)$ is a special function and is generally agreed to be considered useful enough to have it's own place amongst the special functions. – Ali Caglayan Aug 29 '14 at 19:05
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"Solve" is the wrong word here. "Evaluate" is appropriate. – Michael Hardy Aug 29 '14 at 20:28
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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:
$$\begin{align} \int\frac{e^x-e^{-x}}{x}\mathrm{d}x &=2\int\frac{\sinh{x}}{x}\mathrm{d}x\\ &=2\operatorname{Shi}{(x)}+\color{grey}{\text{constant}}. \end{align}$$
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@user3601507 The function Shi(x) is the hyperbolic sine integral. It is defined as the integral $\operatorname{Shi}{(x)}=\int_{0}^{x}\frac{\sinh{t}}{t}\mathrm{d}t$. – David H Aug 29 '14 at 18:16
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1@DavidH If you want to apply bold or italic within a word, Markdown won't do it (at least in its SE version), but HTML can be used as fallback:
<b> </b>, as in hyperbolic sine integral... Big whoops, HTML would do this in a post, but not a comment. :-( – Aug 29 '14 at 18:34 -
Conveniently, $x$ has been used instead of $t$ here. (immature, I know) – PepperSausage Aug 29 '14 at 20:49
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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").
The_Sympathizer
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The integral you gave as a definition for $\mathrm{Ei}(x)$ diverges for all $x$, and so must be incorrect. It should read: $\mathrm{Ei}(x) = \int_{-\infty}^{x} \frac{e^{t}}{t} dt$. – David H Aug 29 '14 at 20:01
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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
$$ \int \frac{e^{x}-e^{-x}}{x}dx = \sum_{k=0}^{\infty}\frac{1}{(2k+1)!}\int x^{2k}dx + C = 2\,\rm Shi(x)+C . $$
Mhenni Benghorbal
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