What's the inverse Laplace transform of $\sqrt{s^2-s}$

Question

I know the inverse Laplace transform of $s$ is the delta function $\delta^\prime (t),$ i.e. $\mathcal{L}^{-1}(s)=\delta^\prime(t).$

And I know actually $\mathcal{L}^{-1}(s^n)=\delta^{(n)}(t),$ where $\delta^{(n)}$ means the nth-derivative of $\delta.$

My question is that, what is the inverse Laplace transform of $\sqrt{s^2-s},$ i.e. what's $\mathcal{L}^{-1}(\sqrt{s^2-s})\,?$

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My attempts:

$$\sqrt{s^2-s}=s\sqrt{1-\frac{1}{s}}=s[1-\frac{1}{2s}+o(1)\frac{1}{s}]=s-\frac{1}{2}+o(1)\,\ \ \ \ \ \ \ \ \ \ \cdot\cdot\cdot\cdot\cdot\cdot\,\ \ \ (A)\,,$$ where the second equality is by the Taylor theorem. (I don't know if the first equality is vailed in the complex space.)

Then, $$\mathcal{L}^{-1}(\sqrt{s^2-s})=\mathcal{L}^{-1}(s)-\mathcal{L}^{-1}(\frac{1}{2})+\mathcal{L}^{-1}[o(1)]=\delta^\prime(t)-\frac{1}{2}\delta(t)+\mathcal{L}^{-1}[o(1)]$$

On one hand, we know $$\forall n\ge 0, \mathcal{L}^{-1}[\frac{1}{s^{n+1}}]=\frac{t^n}{n!}.$$

On the other hand, we know $o(1)$ is nothing than a power series in $\frac{1}{s}.$

So it seems like we have solved the question. But the result looks a mess, and I am not really confident of the first equality in $(A)$. And the application of Taylor theorem in $(A)$ is not vailed for those $s$ with $\lvert s \rvert$ not large enough.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

• $\ds{\root{z}}$ and $\ds{\root{z - 1}}$ are their Principal Branchs.
• The big arc integration vanishes out as its radius $\ds{\to \infty}$ mainly due to the factor $\ds{\expo{ts}}$ with $\ds{t > 0}$.
• There's not any pole inside the contour. The integration is reduced to the subtraction of integrals above and belows the above mentioned branch-cuts.

Namely, \begin{align} &\bbox[5px,#ffd]{\left.\int_{1^{+}\ -\ \infty\ic}^{1^{+}\ +\ \infty\ic} \root{s}\root{s - 1}\expo{ts}{\dd s \over 2\pi\ic} \,\right\vert_{\,t\ >\ 0}} \\[5mm] = &\ -\int_{-\infty}^{0} \root{-s}\expo{\ic\pi/2}\root{1 - s}\expo{\ic\pi/2}\,\expo{ts}{\dd s \over 2\pi\ic} \\[2mm] &\ -\int_{0}^{1} \root{s}\root{1 - s}\expo{\ic\pi/2}\,\expo{ts} {\dd s \over 2\pi\ic} \\[2mm] &\ -\int_{1}^{0} \root{s}\root{1 - s}\expo{-\ic\pi/2}\,\expo{ts} {\dd s \over 2\pi\ic} \\[2mm] &\ -\int_{0}^{-\infty} \root{-s}\expo{-\ic\pi/2}\root{1 - s}\expo{-\ic\pi/2}\,\expo{ts}{\dd s \over 2\pi\ic} \\[5mm] = &\ -\,{1 \over \pi}\int_{0}^{1}\root{s - s^{2}}\expo{ts}\dd s \\[5mm] = &\ -\,{1 \over 2\pi}\expo{t/2} \int_{-1/2}^{1/2}\root{1 - \pars{2s}^{2}}\expo{ts}\dd s \\[5mm] \stackrel{2s\ \mapsto\ s}{=}\,\,\, &\ -\,{1 \over 4\pi}\expo{t/2}\ \underbrace{\int_{-1}^{1}\root{1 - s^{2}}\expo{ts/2}\dd s} _{\ds{{\pi \over t/2}\on{I}_{1}\pars{t/2}}} \end{align} $\ds{\on{I}_{\nu}}$ is a Modified Bessel Function. The last result is A & S Table Identity $\ds{{\bf\color{black}{9.6.18}}}$. Finally, \begin{align} &\bbox[5px,#ffd]{\left.\int_{1^{+}\ -\ \infty\ic}^{1^{+}\ +\ \infty\ic} \root{s}\root{s - 1}\expo{ts}{\dd s \over 2\pi\ic} \,\right\vert_{\,t\ >\ 0}} = \bbx{-\,{1 \over 2}\,{\expo{t/2} \over t} \on{I}_{1}\pars{t \over 2}} \\ & \end{align} The integral $\ds{\underline{vanishes\ out}}$ whenever $\ds{t < 0}$.

Answer 2

From MacLaurin expansion we have $$s\sqrt{1-\frac{1}{s}}=\sum _{n=0}^{\infty } -\frac{s^{2-n} \Gamma \left(n-\frac{3}{2}\right)}{2 \sqrt{\pi } \;\Gamma (n)}$$ $$\mathcal{L}^{-1}\left(s\sqrt{1-\frac{1}{s}}\right)=-\sum _{n=0}^{\infty }\frac{\mathcal{L}^{-1}\left(s^{2-n}\right) \Gamma \left(n-\frac{3}{2}\right)}{2 \sqrt{\pi } \;\Gamma (n)}=\ldots$$ Knowing that $$\mathcal{L}_s^{-1}\left(s^{2-n}\right)(t)=\frac{t^{n-3}}{\Gamma (n-2)};\;n>2$$ we have $$\ldots=-\sum _{n=0}^{\infty }\frac{\frac{t^{n-3}}{\Gamma (n-2)} \Gamma \left(n-\frac{3}{2}\right)}{2 \sqrt{\pi } \;\Gamma (n)}=-\frac{e^{t/2} I_1\left(\frac{t}{2}\right)}{2 t}$$ where $I_1(x)$ is the modified Bessel function of the first kind.