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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.

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$z$ is a complex number, right? –  Babak S. Mar 15 '13 at 11:34
    
yes, $z$ is a complex number. –  pbs Mar 15 '13 at 12:58

1 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

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

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thanks for the nice derivation/explanation as I thought the same thing - very odd (+1). –  Amzoti Mar 15 '13 at 13:45
    
@Amzoti: thanks. This one would be a good one to feed to students to test to see if they really remember what an inverse LT is. –  Ron Gordon Mar 15 '13 at 13:46
    
@pbs: what you are saying makes no sense. The ILT takes a function of $s$ to a function of $t$, unless I am misunderstanding what you are saying. –  Ron Gordon Mar 16 '13 at 20:25
    
@RonGordon How is the Dirac delta function defined here for complex arguments? Since $z$ is complex, we have, for example, $\delta(t-\log z)$, but $t$ is purely real, whereas $\log z$ can be complex. I'm not really sure what $\delta(t - \log z)$ means for complex $z$. –  pbs Mar 17 '13 at 14:54
    
The point is that we are exploiting the sampling property of the delta function: $$\int dx \: \delta (x-a) f(x) = f(a)$$ Really nothing more to it than that. –  Ron Gordon Mar 17 '13 at 16:17

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