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I am trying to classify the singularities in the function $$\mathcal{L}^{-1}_t\left(\frac{e^{|x|\sqrt{\frac{s}{k}}}}{\sqrt{ks}}\right), \ k>0$$ as the final part of solving a PDE. To use the General Inversion Formula, $$f(t)=\frac{1}{2\pi I}\int_{\gamma-i\infty}^{\gamma+i\infty} e^{st}F(s) \ ds,$$ for $t\geq 0$, I need to calculate the residues of $$F(s)=\mathcal{L}^{-1}_t\left(\frac{e^{|x|\sqrt{\frac{s}{k}}}}{\sqrt{ks}}\right).$$ I think the only singular point is when $s=0$. Is this a simple pole?

I considered the equation $g(s)=\sqrt{ks}$. I used the definition below:

$z_0$ is a zero of order $k$ for $f(z)$ if and only if $f(z_0)=f'(z_0)=...=f^{k-1}(z_0)=0$ and $f^k(z_0)\neq 0$.

However, $g'(0)\rightarrow\infty$.

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    $\begingroup$ If you take the principal value of the square root, the argument of $\sqrt s$ tends to $\pm \pi/4$ on the line $s = \gamma + i \sigma$. The absolute value of $\exp(\sqrt s)$ grows as $\exp(\sqrt{\sigma/2})$, the Bromwich integral diverges. $\endgroup$ – Maxim Apr 15 '19 at 16:24
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There is no deleted neighborhood of $s=0$ in which the function is defined and analytic. so we cannot talk about the type of singularity at $0$.

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  • $\begingroup$ Are there any singular points of $f$? $\endgroup$ – user654924 Apr 11 '19 at 0:25
  • $\begingroup$ $f$ is a two-valued function. You have to fix a branch of the square root first before you can analyze this function. $\endgroup$ – Kavi Rama Murthy Apr 11 '19 at 0:29
  • $\begingroup$ I am trying to find $\mathcal{L}^{-1}_t(f(s))$ as part of solving a PDE. I was hoping to use the general inversion formula, which relies on the Residue theorem. In $f(s)$, $k>0$ (constant). $\endgroup$ – user654924 Apr 11 '19 at 0:32

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