Fourier Transform of exponential

While solving Example 4.1 of Signals and Systems by Alan Oppenheim. Example 4.1 is: $$x(t)=e^{-at}u(t), a>0$$ and the transform I get is: $$X(j\omega)\frac{1}{a+j\omega}, a>0$$

The problem is understanding the sketch: The sketch is drawn for both the values $+a$ and $-a$> i dont understand why it $-a$ when $a>0$ is defined.

• $a>0$, no question about it. It is the frequency variable $\omega$ which can take on negative values, e.g. $\omega=-a$ (which is of course negative if $a$ is positive). – Matt L. Nov 19 '15 at 11:47
• "The sketch is drawn for both the values $+a$ and $−a$. i dont understand why it $-a$ when $a>0$ is defined." The variable on the horizontal axis is $\omega$, the radian frequency which can take on all real number values. The $a$ and $-a$ are labels on the axes. Surely we can agree that if $a =1$, say, then there should be a point on the horizontal axis that bears the label $1$ and another with label $-1$? Ditto for labels $2a=2$ and $-2a=-2$ etc? The reason for not putting in these additional labels is that $a$ and $-a$ are of particular importance as the 3dB down points. – Dilip Sarwate Nov 19 '15 at 19:41

The plot is of $$\mid X\left(i\omega\right) \mid = \sqrt{\left(\frac{1}{a+j\omega}\right)\left(\frac{1}{a-j\omega}\right)} = \frac{1}{\sqrt{a^2 + \omega^2}}$$
against $\omega$
In particular $\omega$ can be equal to $-a$.