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I'm trying to compute the inverse Laplace transform of a function:

$$ g(s)=e^{-\tau s\sqrt{\frac{s+q}{s+p}}} $$

where $\tau$, $p$ and $q$ are all positive real numbers, and $q>p$. The ILT is given by:

$$ \mathcal{L}^{-1}\left[g(s)\right] = \frac{1}{2\pi i}\int_{\epsilon-i\infty}^{{\epsilon-i\infty}}dse^{ts}e^{-\tau s\sqrt{\frac{s+q}{s+p}}} $$

I chose a deformed integral contour avoiding the branch cut of $g(s)$ to evaluate the Bromwich integration. integral contour

The integrals along $C_2$, $C_{10}$, $C_{4}$, $C_{8}$ and $C_{6}$ all vanish as the radius of the large circle goes to infinity and that of the small circle goes to 0. The integrals over $C_3$ and $C_9$ cancel each other. Letting $s=-p+(x-p)e^{i\pi}$ over $C_5$ and $s=-p+(x-p)e^{-i\pi}$ over $C_7$, I get this result:

$$ \mathcal{L}^{-1}\left[g(s)\right]=-\frac{1}{2\pi i}\int_q^pdx e^{-tx}e^{i\tau x\sqrt{1+\frac{p-q}{x-p}}}+\frac{1}{2\pi i}\int_q^pdx e^{-tx}e^{-i\tau x\sqrt{1+\frac{p-q}{x-p}}} =-\frac{1}{\pi i}Im\left(\int_q^pdx e^{-tx}e^{i\tau x\sqrt{1+\frac{p-q}{x-p}}}\right) $$

Is my derivation correct? And does anyone know how to evaluate this integration?

Yang Xiao
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    The integrals over $C_2$ and $C_{10}$ do not vanish. The inverse transform of $g$ does not exist as an ordinary function since $g(s)$ behaves like $e^{-\tau s}$ for large $s$, which, on a vertical line, is just a periodic oscillation. – Maxim Jul 20 '19 at 18:04
  • @Maxim thanks for commenting. Could plz explain why the integrals over $C_2$ and $C_{10}$ do not vanish? Because I think they do if we place the vertical line on the left plane and assume $t>\tau$. And I don't understand why it's like a periodic oscillation for large $s$ because to my knowledge $e^{-\tau s}$ is a time-shift factor. Plz let me know if I misunderstand you. – Yang Xiao Jul 21 '19 at 01:40
  • Evaluate the integral of $e^{-s}$ along a path from $\gamma - i A$ to $\gamma + i A$ (the line segment, the left semicircle, or any other path connecting those points). Consider what happens when you increase $A$. – Maxim Jul 21 '19 at 02:07
  • @Maxim I tried to compute the ILT of $e^{-\tau s}$ by evaluating the integral over the vertical line, and I got this result $2i\frac{e^{(t-\tau)\gamma}}{t-\tau}\sin\left((t-\tau)A\right)$, which is indeed a periodic oscillation. But now I am very confused because shouldn't the integral be $\delta(t-\tau)$ as $A\rightarrow \infty$? – Yang Xiao Jul 21 '19 at 04:18
  • In a certain sense, the limit of that expression is indeed $2 \pi i \delta(t - \tau)$, but this is a distributional limit. The delta function is not an $\mathbb R \to \mathbb R$ function, you can't expect to obtain it as a pointwise limit. – Maxim Jul 21 '19 at 22:48

1 Answers1

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When working with distributions, the integral transforms are, in fact, not defined as integrals. They can be formally written as integrals, but that's just a symbolic notation for the transforms. To separate the regular and singular parts, notice that $$ \lim_{A \to \pm \infty} \left( g(\gamma + i A) - e^{-\tau (\gamma + i A) + \tau (p - q)/2} \right) = 0,$$ therefore $$\mathcal L^{-1}[g](t) = e^{\tau (p - q)/2} \delta(t - \tau) + \mathcal L^{-1}[s \mapsto e^{\tau s} g(s) - e^{\tau (p - q)/2}](t - \tau).$$ The inverse transform on the rhs exists as an ordinary function (the Bromwich integral converges by Dirichlet's test) but is unlikely to be evaluable in closed form.

Maxim
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  • Thanks a lot! But I cannot understand the limit expression. As $A\rightarrow \infty$, isn't $g(\gamma +iA)$ supposed to be $e^{-\tau (\gamma +iA)}$ ? – Yang Xiao Jul 23 '19 at 03:04
  • We want to compute the limit, not just understand the qualitative behavior of $g$ for large $s$. Write the expression as $$g(s) - e^{-\tau s + \tau (p - q)/2} = e^{-\tau s} (e^{-\tau s \left( \sqrt {\smash[b] {(s + q)/(s + p)}} - 1 \right)} - e^{\tau (p - q)/2}).$$ The $e^{-\tau s}$ factor is bounded. To find the limit of the second factor, compute $$\lim_{\operatorname{Im} s \to \pm \infty} -\tau s \left( \sqrt {\frac {s + q} {s + p}} - 1 \right).$$ This will prove that the integral you started with is divergent. – Maxim Jul 23 '19 at 11:52
  • Thanks! Your answer is of great help! – Yang Xiao Jul 25 '19 at 06:27