# Finding $\mathcal{L}^{-1}_t\left(\frac{e^{|x|\sqrt{\frac{s}{k}}}}{\sqrt{ks}}\right)$ Using the General Inversion Formula

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

• 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. – Maxim Apr 15 '19 at 16:24

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 $$0$$.
• Are there any singular points of $f$? – user654924 Apr 11 '19 at 0:25
• $f$ is a two-valued function. You have to fix a branch of the square root first before you can analyze this function. – Kavi Rama Murthy Apr 11 '19 at 0:29
• 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). – user654924 Apr 11 '19 at 0:32