Inverse Laplace Transform of $(s+1)/z^s$

Question

I'm trying to compute this ILT $$\mathcal{L}^{-1}\left\{\frac{s+1}{z^s}\right\},$$ where $|z|>1$. However, I'm not sure this is possile? Any help would be appreciated.

Answer 1

This is an odd one. I worked the actual Bromwich integral directly because the residue theorem is no help here. You get something in terms of $c$, the offset from the imaginary axis of the integration path. So I appealed to something more basic. Consider

$$\hat{f}(s) = \int_0^{\infty} dt \: f(t) e^{-s t}$$

Just the plain Laplace transform of some function $f$. Let's throw away any conditions on continuity, etc. on $f$, and consider

$$f(t) = \delta(t-\log{z})$$

for some $z$. Then

$$\hat{f}(s) = z^{-s}$$

Interesting. Now consider $f(t) = \delta'(t-\log{z})$; then

$$\hat{f}(s) = s z^{-s} + \delta(-\log{z})$$

It follows that

$$\mathcal{L}\left\{\delta(t-\log{z})+ \delta'(t-\log{z})\right\} = \frac{s+1}{z^s} + \delta(-\log{z})$$

Therefore

$$\mathcal{L}^{-1}\left\{\frac{s+1}{z^s}\right\} = \delta(t-\log{z})+ \delta'(t-\log{z}) - \delta(-\log{z}) \delta(t)$$