Inverse Laplace Transform of $1/(s+a)$

Question

Is the $a$ always a real value? So, if $a = 6$, then $\frac{1}{s+6}$ transforms to $e^{-6t}$. But can $a$ also take on complex values, such as $a = 4 + 5i$?

Answer 1

That's an interesting question. If you go to the Wikipedia article for the Inverse Laplace Transform, you will find that the "Mellin's inverse formula" uses a limit involving complex numbers. Moreover, let us attempt to find the Laplace transform of $e^{-(a+bi)t}$ from the definition: $$\mathscr{L}\left\{e^{-(a+bi)t}\right\}=\int_0^{\infty}e^{-(a+bi)t}e^{-st}dt=\int_0^{\infty}e^{-(s+a+bi)t}dt=-\bigg[\frac{1}{s+a+bi}e^{-(s+a+bi)t}\bigg]_0^{\infty}$$ Now the question becomes: What happens to this result as $t\to\infty$? $$\lim_{t\to\infty}e^{-(s+a+bi)t}=\lim_{t\to\infty}e^{-st}\cdot e^{-at}\cdot e^{-bti}=^{?}\lim_{t\to\infty}e^{-st}\cdot \lim_{t\to\infty}e^{-at}\cdot \lim_{t\to\infty}e^{-bti}.$$ For the integral to be convergent, this upper limit should also converge. Hence, we would need the equality marked with a question mark to hold, and for each limit to be non-infinite. Notice that the first limit is zero, because $s>0$. As for the second limit, this is the one that could give us problems, as $e^{-at}$ will explode at $t\to\infty$ if $a<0$. The last limit requires us to notice that $e^{tbi}=\cos(tb)+i\sin(tb)$, which doesn't converge when $t\to\infty$, but it remains bounded by the disk $|z|\leq1$ (so this one doesn't give any restrictions). Hence, the Laplace transform of $e^{-(a+bi)t}$ will only exist if $a\geq 0$. In the end, we got the upper limit $t\to\infty$ to vanish, so we are only left with the lower limit $t=0$, and then: $$\mathscr{L}\left\{e^{-(a+bi)t}\right\}=\frac{1}{s+(a+bi)}.$$ Thus, the inverse Laplace transform $\mathscr{L}^{-1}\left\{\frac{1}{s+(a+bi)}\right\}$ only exists if $a\geq 0$, and it is $e^{-(a+bi)t}$. So yes, we can extend the inverse Laplace transform of $\frac{1}{s+z}$ to the complex numbers, but only those with non-negative real part, and the result will also be the corresponding exponential.

Edit: I changed some negative/positive signs to be consistent with the original question.