# Inverse Laplace of $\frac{1}{\sqrt{s} - 1}$?

I found this in textbook.
Not derived from any differential equation.
Also found the answer $$\frac{1}{\sqrt{\pi}\sqrt{t}} + e^t * erf(\sqrt{t})$$
(but don't know how)

Start from Babak's observation that

$$\frac{1}{\sqrt{s}-1} = \frac{\sqrt{s}}{s-1} + \frac{1}{s-1}$$

The ILT of the second term is simply $e^t$. For the first term, consider the following integral:

$$\oint_C dz \,e^{z t} \frac{\sqrt{z}}{z-1}$$

where $C$ is as follows:

We will define $\text{Arg}{z} \in (-\pi,\pi]$, so the branch is the negative real axis. There are $6$ pieces to this contour, $C_k$, $k \in \{1,2,3,4,5,6\}$, as follows.

$C_1$ is the contour along the line $z \in [c-i R,c+i R]$ for some large value of $R$.

$C_2$ is the contour along a circular arc of radius $R$ from the top of $C_1$ to just above the negative real axis.

$C_3$ is the contour along a line just above the negative real axis between $[-R, -\epsilon]$ for some small $\epsilon$.

$C_4$ is the contour along a circular arc of radius $\epsilon$ about the origin.

$C_5$ is the contour along a line just below the negative real axis between $[-\epsilon,-R]$.

$C_6$ is the contour along the circular arc of radius $R$ from just below the negative real axis to the bottom of $C_1$.

We will show that the integral along $C_2$,$C_4$, and $C_6$ vanish in the limits of $R \rightarrow \infty$ and $\epsilon \rightarrow 0$.

On $C_2$, the real part of the argument of the exponential is

$$R t \cos{\theta}$$

where $\theta \in [\pi/2,\pi)$. Clearly, $\cos{\theta} < 0$, so that the integrand exponentially decays as $R \rightarrow \infty$ and therefore the integral vanishes along $C_2$.

On $C_6$, we have the same thing, but now $\theta \in (-\pi,-\pi/2]$. This means that, due to the evenness of cosine, the integrand exponentially decays again as $R \rightarrow \infty$ and therefore the integral also vanishes along $C_6$.

On $C_4$, the integral vanishes as $\epsilon$ in the limit $\epsilon \rightarrow 0$. Thus, we are left with the following by the residue theorem:

$$\left [ \int_{C_1} + \int_{C_3} + \int_{C_5}\right] dz \: \frac{\sqrt{z}}{z-1} e^{z t} = i 2 \pi \, e^t$$

because of the pole at $z=1$.

On $C_3$, we parametrize by $z=e^{i \pi} x$ and the integral along $C_3$ becomes

$$\int_{C_3} dz \: \frac{\sqrt{z}}{z-1} e^{z t} = \int_{\infty}^0 dx \: \frac{i \sqrt{x}}{x+1} e^{-x t}$$

On $C_5$, however, we parametrize by $z=e^{-i \pi} x$ and the integral along $C_5$ becomes

$$\int_{C_5} dz \: \frac{\sqrt{z}}{z-1} e^{z t} = \int_0^{\infty} dx \: \frac{-i \sqrt{x}}{x+1} e^{-x t}$$

We may now write

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

Now consider the integral on the RHS:

\begin{align}\int_0^{\infty} dx \frac{\sqrt{x}}{x+1} e^{-x t} &= \int_{-\infty}^{\infty} du \frac{u^2}{1+u^2} e^{-t u^2}\\ &=\int_{-\infty}^{\infty} du e^{-t u^2} -\int_{-\infty}^{\infty} du \frac{1}{1+u^2} e^{-t u^2}\\ &= \sqrt{\frac{\pi}{t}} - \pi \,e^t \, \text{erfc}{\sqrt{t}}\end{align}

Putting this all together, we get

$$\frac{1}{\sqrt{\pi t}} + e^t \text{erf}{\sqrt{t}} + e^t = \frac{1}{\sqrt{\pi t}} + e^t (1+\text{erf}{\sqrt{t}})$$

Your answer is missing an $e^t$.

• I really love the way you play with integrals. Honestly, wherever I face a question including integrals, complex integrals, I expect to see your complete post, Ron. +1 's given 1 hour ago. :-) Commented Sep 25, 2013 at 14:39
• @BabakS.: many thanks. You're not bad yourself. Anyway, my post is shortened; although I did do that latter integral out by hand, I only put the end result there. Maybe later I will put in all the steps. Commented Sep 25, 2013 at 14:43
• I have been looking at this solution intermittently and I still cannot figure out "On C2, the real part of the argument of the exponential" part. I upper bounded the integrand by $$\frac{e^{R \left(1-\frac{2 }{\pi }\theta\right)}}{\sqrt{R}}$$ which is $0$ when integrated $\theta$ in $[\pi/2,\pi)$ but I don't understand how $\sqrt R$ ended up in the exponent. Commented Jun 2, 2015 at 15:36
• @grdgfgr: Thank you! I had a leftover piece from another calculation that got mixed up in this. I apologize for the confusion and have corrected the mistake. Commented Jun 2, 2015 at 15:42
• @RonGordon the answer is great. Just one point: for the integrals over $C_3$ and $C_5$ isn't there a tiny $y$ component on the parameterization? I mean isn't it $z = e^{\pm i\pi}x+ i\delta$? Of course as $\epsilon\to 0$ one has $\delta\to 0$ but can we simply commute the limit with the integral and set $\delta = 0$ altogether?
– Gold
Commented Oct 3, 2019 at 2:11

$$\frac{1}{\sqrt{s}-1}=\frac{\sqrt{s}+1}{s-1}=\frac{\sqrt{s}}{s-1}+\frac{1}{s-1}=\frac{s}{\sqrt{s}(s-1)}+\frac{1}{s-1}=\frac{1}{\sqrt{s}(s-1)}+\frac{1}{\sqrt{s}}+\frac{1}{s-1}$$ And we know that $$\mathcal{L}(\text{erf}(\sqrt{t}))=\frac{1}{s\sqrt{s+1}},~\mathcal{L}(e^{at}f(t))=\mathcal{L}(f(t))|_{s\to s-1},~\mathcal{L}\left(\frac{1}{\sqrt{t}}\right)=\sqrt{\frac{\pi}{s}}$$

• your answer is right ! means answer in my book is wrong... wasted several hours to get the wrong answer :( Commented Sep 25, 2013 at 13:09
• @palatok: Sometimes, happens. :-) Commented Sep 25, 2013 at 13:11