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I'm studying this article and I've understood most of it precisely, but I can't truly understand one thing about the formula for the Inverse Laplace transform they use.

I know the traditional definition of inverse Laplace transform, which is the Fourier-Mellin's integral $$Q(x)=\mathcal{L}^{-1}\{\tilde{Q}(z)\}=\frac{1}{2\pi i}\lim_{T\rightarrow\infty}\int_{\gamma-iT}^{\gamma+iT}e^{zx}\tilde{Q}(z)dz$$ but they claim $$Q(x) \sim \frac{1}{2\pi i} \int_C\tilde{Q}(z)dz$$ with $C$ a generic path as long as

(i) it runs in the domain of analyticity of $\tilde{Q}(z)$,

(ii) $\tilde{Q}(z)$ has the same (finite) values at the two endpoints.

I don't really understand how one can formally state that this formulas are equivalent. My guess is that they are using a residue argument and Cauchy's theorem, but I would appreciate if someone can explain this a little bit better and formally.

To give you more insight on my confusion my case is with this Laplace transform $$\tilde{Q}(z)\sim z^{v-2}exp\biggl(z+\frac{\delta^2x}{4z}\biggr)$$ so they state "Since there might be a branch cut along the negative real axis the contour cannot cross that. One can also see that the endpoints have to go either to infinity or zero in the $Re(z) < 0$ half plane because only then, due to the exponential function, will the values vanish at both ends." and then they present the following contours: Possible contours for the integral.

Donson
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  • I'm not sure I fully understand your question, but first note that $\underset{z\to -\infty}{\text{lim}} \left(z^{v-2} e^{z+\frac{\delta^2 x}{4 z}}\right)=0$, and second I believe the function $\tilde{Q}(z)=z^{v-2}, e^{z+\frac{\delta^2 x}{4 z}}$ has singularities at $z=0$ and also for $z<0$ when $v\notin \mathbb{Z}$. – Steven Clark Feb 01 '25 at 17:54
  • @StevenClark thanks for the comment, yes I understand the function has a cut in the domain (as one can see in correspondence of the dashes in the image) coming from the essential singularity in the origin. What I can't understand really is how one can formally state that integrating $\tilde{Q}(z)$ on a vertical line or on a general path $C$ (ofc respecting (i) and (ii) which are resonable requests) leads to the same solution. Maybe the answer is much simpler then I'm expecting but I don't see it. – Donson Feb 01 '25 at 18:50
  • Ok, so the path $C$ could be circle of radius $1$ around $z=2$ for example? – Steven Clark Feb 01 '25 at 19:21
  • @StevenClark I don't think it can since that contour doesn't contain any possible singularities of $\tilde{Q}(z)$, so the integral, by Cauchy's theorem, should give you $0$, which is clearly not the inverse Laplace transform of my function. – Donson Feb 02 '25 at 13:55
  • Yes, but it seems to me it satisfies (i) and (ii), and therefore I think some clarification is needed. – Steven Clark Feb 02 '25 at 14:00
  • @StevenClark Oh! Sorry, now I get what your asking, I think they just went fast over this passage in the article, probably they meant a generic open path, not a closed one, as every path in the image seems to be open (I know path $C_2$ seems closed, but I think it doesn't really reach the singularity $0$, as well as $C_1$ and $C_3$, they just tend towards $0$) – Donson Feb 02 '25 at 14:18

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