# Why is $e^{-ax}u(x)$ not continuous for $a\in\mathbb{C}$?

Let $$a\in\mathbb{C}$$ with $$\Re(a)>0$$

Define the unit step function : $$u(x):=\begin{cases} 1&\text{if x\geq0}\\ 0&\text{if otherwise} \end{cases}$$ We wish to study continuity of : $$f(x)=\frac{x^{k}}{k!}e^{-ax}u(x)$$ But I am only concerned regarding the case $$k=0$$ where we get $$e^{-ax}u(x)$$, with $$a\in\mathbb{C}$$ this would "virtually" mean that we have a decaying exponential multiplied with a complex sinusoidal but why would this imply that $$f$$ is not continuous for $$k=0$$? My textbook says $$f$$ must be continuous for all $$k\geq1$$

• It it possible that you are having a moment of confusion and thinking that $e^{0}$ is $0$ instead of $1$? Mar 21 at 18:50
• I maybe having one large mental hiccup but let's hope not, All I see is $\frac{x^{0}}{0!}e^{-ax}u(x)\implies e^{-ax}$ if $x\geq0$ @MikeF Mar 21 at 18:55
• What is the domain of $f(x)$? Mar 21 at 18:59
• The domain as it appears is strictly dependent on the unit step function. Mar 21 at 19:00
• Yes, so the domain is all of $\Bbb R$, not just $\{x\ge 0\}$. Mar 21 at 19:00

When $$k=0$$ the term $$x^k$$ vanishes and thats the one that guarantees continuity at zero, so the discontinuity of $$u$$ at $$0$$ comes into play. $$\lim_{x\rightarrow 0^-} e^{-ax}u(x)=0\neq 1 = f(0).$$