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I have been having trouble with the definitions of the inverse Laplace Transform (LT). LT being: $$F\left(s\right)=\intop_{0}^{\infty}f\left(t\right)e^{-st}dt$$ And the Inverse transform: $$f\left(t\right)=\frac{1}{2\pi i}\lim_{T\rightarrow\infty}\intop_{\gamma-iT}^{\gamma+iT}e^{st}F\left(s\right)ds$$ Now if I want to apply this for instance to $f\left(t\right)=e^{at}$. The LT transform exists only for $\Re\left(s\right)>a$ (Region of Convergence, ROC) and is $\frac{1}{s-a}$ (if we try to apply for $\Re\left(s\right)\le a$ we would get $\infty$). But now when I want to apply the inverse LT, we use $\gamma>a$ and close contour to the left half-plane to get the pole and recover the original expression. My question is why are we allowed to do it over the area where the LT is not defined over? and if I was to sort of continue the LT it would be infinity over that region as the transform suggests wouldn't it?

I know if we just insert the definitions and exchange the integration order I'd always recover the same function, but in this process I don't modify it to be defined outside of the ROC and here we have to do it manually.

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  • $\begingroup$ We analytically continue the function represented by LT outside the ROC. Outside the ROC, it no longer reflects the LT itself. But the inversion works on the anlytically continued version of the function represented by a convergent LT $\endgroup$ Commented Aug 15 at 22:19
  • $\begingroup$ I understand that the inversion works if we analytically continue. My question was why, as we don't need to analytically continue when e.g. we show that the inversion works for a general function. $\endgroup$ Commented Aug 16 at 7:14

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Note that if $f(s)$ is an analytic function in some vertical strip $a \le \sigma \le b$ (or has finitely many isolated singularities inside the strip) and $f(\sigma+it) \to 0, |t| \to \infty$ uniformly in $\sigma \in [a,b]$, then the convergence of the integral $\int_{c-i\infty}^{c+i\infty}f(s)ds$ for one single $a \le c \le b$ implies the convergence and the fact that the integral is the same for all $a \le \sigma \le b$; if we have isolated singularities at some $\sigma_1, ..\sigma_k$, the integral above still exists for all $\sigma \ne \sigma_k$ but the value changes each time we pass through a singularity (by $2\pi i$ times) appropriate residue.

Proof: Fix $a \le \sigma \le b$ and wlog assume $\sigma \le c$ Take large $T_2 >T_1$. By Cauchy $\int_{R_{\pm}} f(s)ds=0$ where $R_{\pm}$ are the oriented rectangles with vertices at $\sigma \pm iT_{1,2}, c \pm iT_{1,2}$ where we choose the same sign in each.

By hypothesis (uniform convergence to $0$ as $|T| \to \infty$ of $f(x+iT)$) we have $$\int_{\sigma}^c|f(x\pm iT_{1,2})|dx \to 0, T_2>T_1 \to \infty$$ and since the convergence of $\int_{c-i\infty}^{c+i\infty}f(s)ds$ is equivalent to the fact that $\int_{c -iT_2}^{c-iT_1}f(s)ds+\int_{c +iT_1}^{c+iT_2}f(s)ds \to 0$ we get that the same limit holds for $\sigma$ integrals so $\int_{\sigma-i\infty}^{\sigma+i\infty}f(s)ds$ converges too.

Now using the rectangle with vertices at $\sigma \pm iT_1, c \pm iT_1$ we see that up to the horizontal integrals that go to $0$ the integrals on line $\sigma$ and $c$ are either equal if there is no singularity in between or differ by the appropriate residues if there are finitely many singularities, hence part $2$ (where we choose the height larger than the singularities heights which are finite in number by hypothesis)

In particular for $F(s)=\frac{e^{sx}}{s-a}$ we notice that $|F(s)| \sim \frac{e^{\sigma x}}{|t|}$ for large $t$ where $s=\sigma+it$ as usual (note that we use $x$ instead of $t$ vs OP) and when $\sigma$ stays in a bounded interval (and $x$ is fixed or stays in a bounded interval too), this converges uniformly to $0$ as $|t| \to \infty$ hence the result above applies and we can move integrals to the left of $a$ but their values changes by the approriate residue.

This procedure is very general and allows for the explicit computation of many integrals as above by moving the line of integration and picking up residues, though of course we require the uniform convergence to $0$ in the hypothesis.

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  • $\begingroup$ Though I didn't ask if the line integral converges or how the singularities change sign when we pass them, it did help me understand why the analytic continuation here is valid. $\endgroup$ Commented Aug 16 at 7:20
  • $\begingroup$ the inverse Laplace or Mellin transforms are valid when the integral converges - it is very easy to make mistakes (even in published papers btw) and assume that if the direct transform converges for some $\Re s \in [a,b]$ the inverse integral converges for $c \in [a,b]$ and that is simply not true in general - you need to prove the convergence in a way or another (usually simple estimates but sometimes more is required since many times the integrals in question are only conditionally convergent) - the OP case is typical since the integral on the line is not absolutely convergent btw) $\endgroup$ Commented Aug 16 at 14:25
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Since what we are interested in is essentially the integration of:$$f\left(t\right)=\frac{1}{2\pi i}\lim_{T\rightarrow\infty}\intop_{\gamma-iT}^{\gamma+iT}e^{st}F\left(s\right)ds$$ The poles become a tool for computation (I checked that it could also be done directly over the line for a simple pole). So even though the LT might not exist outside the ROC, we can analytically continue the function to that which we have over the path integral and apply the residue theorem there (In a sense the path integral doesn't know of the divergences of the LT).

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