Solving $\int e^{|x|} dx$

Question

Solving $\displaystyle\int e^{|x|} dx$.

I just solved a quite similar definite integral, and it wasn't hard, but it was because I could divide the integral in two integrals and get rid of modulus. Problem is I can't do that here, since it's indefinite now. I tried already substitution, parts, series and got nothing. How can I aproach such integral?

Thanks.

Edit: I'd like to obtain Wolfram's answer, not a function defined by parts.

Answer 1

Let us express the antiderivative as a definite integral plus a constant.

$$\int e^{|x|}dx=\int_0^x e^{|x|}dx+C.$$

Then for positive $x$,

$$e^x-1+C$$

and for negative $x$,

$$1-e^{-x}+C.$$

Hence

$$\text{sgn}(x)\left(e^{|x|}-1\right)+C.$$

As we can check, the derivative is $\text{sgn}^2(x)e^{|x|}$ (and $1$ for $x=0$).

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Note that WA seemed to strive to avoid the absolute value in the exponent. We can achieve the same effect with

$$\text{sgn}(x)\frac{(1+\text{sgn}(x))(e^x-1)+(1-\text{sgn}(x))(e^{-x}-1)}{2} =\frac{(\text{sgn}(x)+1)(e^x-1)-(\text{sgn}(x)-1)(e^x-1)e^{-x}}2 =e^{-x}\frac{\text{sgn}(x)(e^x-1)^2+e^{2x}-e^x-e^x-1}2.$$

This is ugly.

Answer 2

You should simply integrate separately for $x\ge 0$ and $x<0$. The result is then the function defined in different ways for $x\ge 0$ and $x<0$. (But it will be actually possible to write it as an unique expression).

Answer 3

Write $e^{|x|}=e^{\operatorname{sgn}(x) x}$. Then

$$\int e^{|x|} dx = \int e^{\operatorname{sgn}(x)x} dx = \frac{e^{\operatorname{sgn}(x)x}}{\operatorname{sgn}(x)} + C = \operatorname{sgn}(x)e^{|x|} + C$$

edit: And well, this is broken.

Answer 4

The result given by Wolframalpha might seem attractive, but it isn't. Consider $$ F(x)=\frac{1}{2}e^{-x}((e^x-1)^2\operatorname{sgn}(x)-2e^x+e^{2x}-1) $$ (the integration constant is irrelevant); then, for $x>0$, $$ F(x)=\frac{1}{2}e^{-x}(e^{2x}-2e^x+1-2e^x+e^{2x}-1)= \frac{1}{2}e^{-x}(2e^{2x}-4e^x)=e^x-2 $$ For $x<0$, $$ F(x)=\frac{1}{2}e^{-x}(-e^{2x}+2e^x-1-2e^x+e^{2x}-1)=-e^{-x} $$ Also $F(0)=-1$ and it can be easily checked that $F'(x)=e^{|x|}$ for all $x$.

On the other hand, doing algebraic manipulations we obtain $$ F(x)=\frac{1}{2}((e^x-2+e^{-x})\operatorname{sgn}(x)-2+e^x-e^{-x}) $$ Note that $e^x-e^{-x}=(e^{|x|}-e^{-|x|})\operatorname{sgn}(x)$ and that $e^x+e^{-x}=e^{|x|}+e^{-|x|}$, so we obtain \begin{align} F(x) &=\frac{1}{2}((e^{|x|}+e^{-|x|}-2+e^{|x|}-e^{-|x|})\operatorname{sgn}(x)+2) \\[6px] &=(e^{|x|}-1)\operatorname{sgn}(x)+1 \end{align} which is much simpler, isn't it?