# Integral of absolute e-function

I have to integrate the following function:

$$\int e^{-|x|}$$

I tried this and I don't think, that this is right. So can you tell me, where my fault is? $$\int e^{-|x|} = \int_{-\infty}^{\infty}e^{-|x|} = \int_{-\infty}^0 e^{-(-x)} + \int_{0}^{\infty}e^{-x} = e^x - e^{-x}$$

The problem is, that $e^{-|x|}$ has no root. $e^{-|x|} \in (0,1]$

-
The calculations are correct, but you have to plug in the boundary values, i.e. $e^x\big|_{-\infty}^0-e^{-x}\big|_{0}^\infty=1+1=2$ –  user8 Jan 15 '13 at 12:23
Thx for advice.. i got it –  Giesi Jan 15 '13 at 12:31
Are you trying to compute a definite integral, such as $\int_{-\infty}^\infty \exp(-|x|)\,\mathrm dx$, or a primitive of the function $x\mapsto\exp(-|x|)$? –  jathd Jan 15 '13 at 12:32
It's the indefinite integral –  Giesi Jan 15 '13 at 12:33
@Giesi If you want to calculate the indefinite integral, i.e. $\int\exp{(-|x|)}dx$ the result would be $\exp{(x)}-\exp{(-x)}$ as you calculated. But then you have to cancelt the boundaries of the integral! –  user8 Jan 15 '13 at 12:35

I'm not sure why you turn your indefinite integral into a definite one. What you should do (as I think you attempted) is consider when $x$ is in certain intervals.
For $-\infty<x\le 0$, $e^{-|x|}=e^{x}$, so $\int e^{-|x|}dx=\int e^{x}dx=e^{x}+C=e^{-|x|}+C$
For $0\le x < \infty$, $e^{-|x|}=e^{-x}$, so $\int e^{-|x|}dx=\int e^{-x}dx=-e^{-x}+C=-e^{-|x|}+C$
To express this in a simple formula, we can make use of the signum function, which is defined to be $-1$ when $x<0$, $0$ at $x=0$ and $1$ when $x>0$. Clearly, your antiderivative can be expressed as $$\int e^{-|x|}dx=-sgn(x)e^{-|x|}$$
This is also a consequence of the fact that $sgn(x)=\frac{x}{|x|}=\frac{d}{dx}|x|$; can you see why?