Inverse Laplace transform of $\frac{s}{\sqrt{(s+a)^3}}$ 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?
 A: 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
$$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \,  s (s+a)^{-3/2} = \left (\pi t\right)^{-1/2} (1-2 a t) e^{-a t} \theta(t)$$
A: $\newcommand{\+}{^{\dagger}}%
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$\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 $\pars{1}$:
$$
\int_{\gamma -\ic\infty}^{\gamma + \ic\infty}
{s\expo{st} \over \root{\pars{s + a}^{3}}}\,{\dd s \over 2\pi\ic}
=
\pars{1 + 2a\,\partiald{}{a}}{\expo{-at} \over \root{\pi t}}
=
{1 \over \root{\pi}}\,{\pars{1 - 2at}\expo{-at} \over \root{t}}
$$

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

