Inverse Laplace transform of $\frac{s}{\sqrt{(s+a)^3}}$

Question

Trying to find the inverse Laplace transform of $\frac{s}{\sqrt{(s+a)^3}}$. So solving $\oint_B dz \: \frac{z}{\sqrt{(z+a)^3}} e^{z t}$ (Bromwich contour). I tried doing a u-substitution with $u=z+a$ but that gave me a non converging integral. Is there a better way to approach this problem or was something wrong with my substitution?

Answer 1

For this one, I would express the ILT as

$$s (s+a)^{-3/2} = (s+a)^{-1/2} - a (s+a)^{-3/2}$$

Now recognize that

$$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds\, (s+a)^{-1/2} e^{s t} = \frac{e^{-a t}}{i 2 \pi} \int_{a+c-i \infty}^{a+c+i \infty} ds\, s^{-1/2} e^{s t}$$

The latter integral has been evaluated using a modified Bromwich contour as follows. Consider

$$\oint_C dz \frac{e^{z t}}{\sqrt{z}}$$

where $C$ is a keyhole contour that goes up and back around the negative real axis and encircles the origin from $\arg{s}=\pi$ to $\arg{s}=-\pi$. By the residue theorem (or Cauchy's integral theorem), this integral is zero because there are no poles within $C$. $C$, however, has $5$ pieces: the original integral along $\Re{s}=a$, a circular arc of large radius $R$, a section that goes in a positive direction just above the negative real axis, a circular arc of small radius $r$ around the origin, and another section just below the negative real axis in a negative direction. In the limit as $R \rightarrow \infty$ and $ r \rightarrow 0$, the integrals along the circular arcs vanish. This leaves

$$ \int_{a+c-i\infty}^{a+c+i\infty} ds \frac{e^{s t}}{\sqrt{s}}+e^{i \pi} \int_{\infty}^0 dx \frac{e^{-x t}}{i \sqrt{x}} + \int_0^{\infty} dx \frac{e^{-x t}}{-i \sqrt{x}}=0$$

A little rearranging produces

$$ \frac{1}{i 2 \pi} \int_{a+c-i\infty}^{a+c+i\infty} ds \frac{e^{s t}}{\sqrt{s}} = \frac{1}{ \pi} \int_0^{\infty} dx \frac{e^{-x t}}{\sqrt{x}}$$

Substitute $y=\sqrt{x}$ into the integral on the RHS and finally get

$$ \frac{1}{i 2 \pi} \int_{a+c-i\infty}^{a+c+i\infty} ds \frac{e^{s t}}{\sqrt{s}} = \frac{2}{ \pi} \int_0^{\infty} dy \; e^{-t y^2}=\frac{1}{\sqrt{\pi t}}$$

To get the second piece, note that

$$\frac{a}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds\, (s+a)^{-3/2} e^{s t} = \frac{a e^{-a t}}{i 2 \pi} \int_{a+c-i \infty}^{a+c+i \infty} ds\, s^{-3/2} e^{s t}$$

Note that

$$\frac{d}{dt}\int_{a+c-i \infty}^{a+c+i \infty} ds\, s^{-3/2} e^{s t} = \int_{a+c-i \infty}^{a+c+i \infty} ds\, s^{-1/2} e^{s t}$$

so that

$$\frac{1}{i 2 \pi} \int_{a+c-i \infty}^{a+c+i \infty} ds\, s^{-3/2} e^{s t} = \int \frac{dt}{\sqrt{\pi t}} = 2 \sqrt{\frac{t}{\pi}}$$

Putting this all together, and noting that this has been done only for $t \gt 0$, we have

$$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s (s+a)^{-3/2} = \left (\pi t\right)^{-1/2} (1-2 a t) e^{-a t} \theta(t)$$

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{\int_{\gamma -\ic\infty}^{\gamma + \ic\infty} {s\expo{st} \over \root{\pars{s + a}^3}}\,{\dd s \over 2\pi\ic} =\int_{\gamma -\ic\infty}^{\gamma + \ic\infty} s\pars{s + a}^{-3/2}\expo{st}\,{\dd s \over 2\pi\ic}\,,\quad\gamma > \verts{a}\ \mbox{and}\ t > 0.\quad}$

We'll use the identity: \begin{align} &\int_{\gamma -\ic\infty}^{\gamma + \ic\infty} {s\expo{st} \over \root{\pars{s + a}^3}}\,{\dd s \over 2\pi\ic} = \int_{\gamma -\ic\infty}^{\gamma + \ic\infty}\expo{st} \bracks{{1 \over \pars{s + a}^{1/2}} - {a \over \pars{s + a}^{3/2}}} \,{\dd s \over 2\pi\ic} \\[3mm]&= \int_{\gamma -\ic\infty}^{\gamma + \ic\infty}\expo{st} \braces{{1 \over \pars{s + a}^{1/2}} + 2a\,\partiald{}{a}\bracks{{1 \over \pars{s + a}^{1/2}}}} \,{\dd s \over 2\pi\ic} \\[3mm]&= \pars{1 + 2a\,\partiald{}{a}}\int_{\gamma -\ic\infty}^{\gamma + \ic\infty} {\expo{st} \over \pars{s + a}^{1/2}}\,{\dd s \over 2\pi\ic}\tag{1} \end{align}

In the complex plane $\pars{~s \in {\mathbb C}~}$, the $\ds{\pars{s + a}^{-1/2}}$-branch-cut is defined by $$ \pars{s + a}^{-1/2} = \verts{s + a}^{-1/2}\exp\pars{-\,{\phi \over 2}\,\ic}\quad \mbox{where}\quad s \not= -a\quad\mbox{and}\quad\verts{\phi} < \pi $$ such that \begin{align} &\int_{\gamma -\ic\infty}^{\gamma + \ic\infty} {\expo{st} \over \pars{s + a}^{1/2}}\,{\dd s \over 2\pi\ic} \\[3mm]&= -\int_{-\infty}^{-a}\pars{-x - a}^{-1/2}\ \overbrace{\exp\pars{-\,{\pi \over 2}\,\ic}}^{\ds{=\ -\ic}}\ \expo{xt} \,{\dd x \over 2\pi\ic} -\int_{-a}^{-\infty}\pars{-x - a}^{-1/2}\ \overbrace{\exp\pars{-\,{\bracks{-\pi} \over 2}\,\ic}}^{\ds{=\ \ic}}\ \expo{xt}\,{\dd x \over 2\pi\ic} \\[3mm]&= -\,{1 \over \pi}\int_{-a}^{-\infty}\pars{-x - a}^{-1/2}\expo{xt}\,\dd x = {1 \over \pi}\int_{a}^{\infty}\pars{x - a}^{-1/2}\expo{-xt}\,\dd x \\[3mm]&= {\expo{-at} \over \pi}\ \overbrace{\qquad\int_{0}^{\infty}x^{-1/2}\expo{-xt}\,\dd x\qquad}^{\ds{=\ t^{-1/2}\Gamma\pars{1/2} = \root{\pi/t}}} \end{align}

\begin{align} \int_{\gamma -\ic\infty}^{\gamma + \ic\infty} {\expo{st} \over \pars{s + a}^{1/2}}\,{\dd s \over 2\pi\ic} = {\expo{-at} \over \root{\pi t}}\,,\qquad t > 0 \end{align}

With the identity $\pars{1}$: $$ \int_{\gamma -\ic\infty}^{\gamma + \ic\infty} {s\expo{st} \over \root{\pars{s + a}^{3}}}\,{\dd s \over 2\pi\ic} = \pars{1 + 2a\,\partiald{}{a}}{\expo{-at} \over \root{\pi t}} = {1 \over \root{\pi}}\,{\pars{1 - 2at}\expo{-at} \over \root{t}} $$

$$\color{#0000ff}{\large% \int_{\gamma -\ic\infty}^{\gamma + \ic\infty} {s\expo{st} \over \root{\pars{s + a}^{3}}}\,{\dd s \over 2\pi\ic} \color{#000000}{\ =\ } {1 \over \root{\pi}}\,{\pars{1 - 2at}\expo{-at} \over \root{t}} \,,\qquad\color{#000000}{t > 0}}$$