# Why is $\left.\frac {-1} {s+a} e^{-(s+a)t}\right\vert_{t=0}^{t=\infty}= \frac 1 {s+a}$ and that the ROC is: $\operatorname{Re}(s+a)>0$?

I know this is a self study question, but help would be much of appreciated. I don’t know why this is true.

For complex $$s, a$$, show that: $$\left.\frac {-1} {s+a} e^{-(s+a)t}\right\vert_{t=0}^{t=\infty}=\frac 1 {s+a}$$ and that the region of convergance is: $$\operatorname{Re}(s+a)>0$$

• Are you sure it is not $e^{\color{red}-(s+a)t}$ ? Also, use \vert instead of \vline, _ instead of \from and ^ instead of \to – Maximilian Janisch Mar 25 at 17:36
• One second I’m correcting. I downloaded the app and I misused it... – Vitali Pom Mar 25 at 17:37
• I am also still not sure if this is true. I mean it could go off to $\infty$ if $s$ and $a$ are chosen apropriately. The correct command for $\infty$ is \infty – Maximilian Janisch Mar 25 at 17:38
• Indeed. Corrected. Thanks! – Vitali Pom Mar 25 at 17:42
• Hint: Use the fact that $\exp(0)=1$ and $\exp(c t)\xrightarrow{t\to\infty}0$ for all complex numbers $c$ with real part $<0$ – Maximilian Janisch Mar 25 at 17:48