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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 $k\geq1$

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  • $\begingroup$ It it possible that you are having a moment of confusion and thinking that $e^{0}$ is $0$ instead of $1$? $\endgroup$
    – Mike F
    Mar 21 at 18:50
  • $\begingroup$ 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 $\endgroup$
    – SPARSE
    Mar 21 at 18:55
  • $\begingroup$ What is the domain of $f(x)$? $\endgroup$ Mar 21 at 18:59
  • $\begingroup$ The domain as it appears is strictly dependent on the unit step function. $\endgroup$
    – SPARSE
    Mar 21 at 19:00
  • $\begingroup$ Yes, so the domain is all of $\Bbb R$, not just $\{x\ge 0\}$. $\endgroup$ Mar 21 at 19:00
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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. $$\lim_{x\rightarrow 0^-} e^{-ax}u(x)=0\neq 1 = f(0).$$

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  • $\begingroup$ Ahh now I see thank you $\endgroup$
    – SPARSE
    Mar 21 at 19:00

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