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: $\operatorname{Re}(s+a)>0$

  • $\begingroup$ 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 $\endgroup$ – Maximilian Janisch Mar 25 at 17:36
  • $\begingroup$ One second I’m correcting. I downloaded the app and I misused it... $\endgroup$ – Vitali Pom Mar 25 at 17:37
  • $\begingroup$ 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 $\endgroup$ – Maximilian Janisch Mar 25 at 17:38
  • $\begingroup$ Indeed. Corrected. Thanks! $\endgroup$ – Vitali Pom Mar 25 at 17:42
  • 1
    $\begingroup$ 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 $<0$ $\endgroup$ – Maximilian Janisch Mar 25 at 17:48

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