# Improper Integral

1. Mar 20, 2006

### G01

$$\int_{-\infty}^{\infty} e^{-|x|} dx$$

Could someone tell me why this integral, when you split it comes out to be:

$$\int_{-\infty}^0 e^x dx + \int_0^{\infty} e^{-x} dx$$

I keep thinking it should be e^(-x) in the first integral. I don't know why its positive. I can solve this integral otherwise. Thanks again.

2. Mar 20, 2006

### nocturnal

The absolute value is defined for a real number $x$ by
$$|x|= \left \lbrace \begin{array}{l l} x & \mbox{if} \ x \geq 0 \\ -x & \mbox{if} \ x\leq 0 \end{array} \right.$$

when $x$ is in the interval $(-\infty,0)$, then $|x| = -x$. Likewise, when $x$ is in the interval $(0,\infty)$, then $|x| = x$. Therefore,

$$\int_{-\infty}^0 e^{-|x|}dx = \int_{-\infty}^0 e^{-(-x)}dx = \int_{-\infty}^0 e^x dx$$

Last edited: Mar 20, 2006
3. Mar 20, 2006

### G01

Im sorry that definition confuses me...... I always thought the absolute value was the positive distance of that number from the origin, leading me to believe that |-x|= x. This seems contrary.

4. Mar 20, 2006

### nocturnal

This is true.
This is false. For example take $x=-3$. Then your saying that $|-(-3)| = -3$, which is the same as saying that $|3| = -3$, which is clearly false. I used paranthenses to delimit the value which was substituted for x, namely -3.

However, if we use the definition of absolute value, we arive at the correct result. For example, if $x=-3$, we have that $x<0$ so we use the second case of the definition to arrive that $|x| = -x$, which gives $|-3| = -(-3) = 3$.

Try plugging in different values for x into the definition to convince yourself that this definition works.

The definition for when $x>0$ should be clear. The absolute value of a positive number is a positive quantity and is equal to that number. However if $x<0$, then $-x>0$, so $-x$ is a positive number (don't let the negative sign confuse you, remember that the negative of a negative number is positive).

Last edited: Mar 20, 2006