Inverse Laplace Transform of $e^{-\sqrt{s^2+s}}$.

I've come across this problem trying to find an integral representation to a PDE (damped wave equation with initial conditions).

What I would like to do is compute $$f(t) = \mathcal{L}^{-1}\left[e^{-\sqrt{s^2+s}}\right](t)$$

via the bromwich integral,

$$f(t) = \frac1{2 \pi i} \int_{c- i \infty}^{c + i \infty} e^{-\sqrt{s^2+s}}e^{st}ds $$

My progress so far has been stunted by the fact that we have branch points at $s = 0$, and $s = -1$. My idea so far has been to make the branch cut the interval $[-1, 0]$ on the real axis, however this has been quite annoying as the typical shape of the contour takes the shape of a reversed $D$. My idea might be to do something of a dog bone contour to avoid the branch, but I've been having trouble keeping track of the evaluation.

Doing it this way, the sum of the integrals would have to be zero, there are no poles involved. Thankfully the outer arcing contours would go to zero, it's really the contour going around the sneaky branch cut that would be cumbersome. Perhaps a dog bone would work?

My goal would be to even just come up with some sort of series/integral representation of the laplace transform. Any help is appreciated.

I will illustrate a slightly different, more rigorous derivation of the ILT result here, which is

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{-\sqrt{s (1+s)}} \, e^{s t} = \frac12 e^{-t/2} \frac{I_1 \left ( \frac12 \sqrt{t^2-1} \right )}{\sqrt{t^2-1}} H(t-1) $$

where

$$H(t-1) = \begin{cases} 1 \quad t \gt 1 \\ 0 \quad t \le 1 \end{cases} $$

To demonstrate where this result comes from, I am going to define the following contour integral:

$$ \oint_C dz \, e^{-\sqrt{z (1+z)}} \, e^{t z} $$

where we will figure out the allowed values of $t$ and the contour $C$ is as follows:

where the radius of the large arc is $R$ and the radii of the small arcs are each $\epsilon$. By Cauchy's theorem, this contour integral is zero. On the other hand, we can write out the contour integral in the limit as $\epsilon \to 0$ as follows:

$$ \int_{c-i \sqrt{R^2-c^2}}^{c+i \sqrt{R^2-c^2}} ds \, e^{-\sqrt{s (1+s)}} \, e^{s t} + i R \int_{\pi/2 - \arcsin{(c/R)}}^{\pi} d\theta \, e^{i \theta} \, e^{-R e^{i \theta} \left (1+\frac1{R e^{i \theta}} \right )^{1/2}} e^{R t e^{i \theta}} \\+ e^{i \pi} \int_R^{1} dx \, e^{\sqrt{x (x-1)}} e^{-x t} + e^{i \pi} \int_1^0 dx \, e^{-i \sqrt{x (1-x)}} e^{-x t} \\ + e^{-i \pi} \int_0^1 dx \, e^{i \sqrt{x (1-x)}} e^{-x t} + e^{-i \pi} \int_1^R dx \, e^{\sqrt{x (x-1)}} e^{-x t} \\ + i R \int_{\pi}^{3 \pi/2 + \arcsin{(c/R)}} d\theta \, e^{i \theta} \, e^{-R e^{i \theta} \left (1+\frac1{R e^{i \theta}} \right )^{1/2}} e^{R t e^{i \theta}}$$

It should be clear that the third and sixth integrals cancel. Further, we can deduce the values of $t$ for which the ILT is defined by combining the second and seventh integrals:

$$\begin{align} i R \int_{\pi/2-\arcsin{(c/R)}}^{3 \pi/2+\arcsin{(c/R)}} d\theta \, e^{i \theta} \, e^{-R e^{i \theta} \left (1+\frac1{R e^{i \theta}} \right )^{1/2}} e^{R t e^{i \theta}} &= R \int_{\frac{c}{R} + i \sqrt{1-\frac{c^2}{R^2}}}^{\frac{c}{R} - i \sqrt{1-\frac{c^2}{R^2}}} d\zeta \, e^{-R \zeta \left ( 1 + \frac1{R \zeta} \right)^{1/2}} e^{R t \zeta} \\ &=_{R \to \infty} R e^{-1/2} \int_{\frac{c}{R}+i}^{\frac{c}{R}-i} d\zeta \, e^{R (t-1) \zeta} \\ &= -i 2 e^{c (t-1)} e^{-1/2} \frac{\sin{R (t-1)}}{t-1} \\ &=_{R \to \infty} -i 2 \pi e^{c (t-1)} e^{-1/2} \delta(t-1) \end{align} $$

It should be noted that the above integrals converge as $R \to \infty$ only when $t \gt 1$. Accordingly, there will be a $H(t-1)$ term in the final result. Because $\lim_{x \to 0} H(x) \delta(x) = 0$, we may ignore the delta function contribution in this case. (When $t \lt 1$, the ILT is zero.)

Now, as $R \to \infty$ we are down to three integrals which we can relate by Cauchy's theorem:

$$ \int_{c-i \infty}^{c+i \infty} ds \, e^{-\sqrt{s (1+s)}} \, e^{s t} = \int_0^1 dx \, e^{i \sqrt{x (1-x)}} e^{-x t} - \int_0^1 dx \, e^{-i \sqrt{x (1-x)}} e^{-x t}$$

so that the ILT for $t \gt 1$ is now in terms of a single, real integral:

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{-\sqrt{s (1+s)}} \, e^{s t} = \frac1{\pi} \int_0^1 dx \, \sin{\left ( \sqrt{x (1-x)} \right )} e^{-t x} H(t-1)$$

We may now focus on evaluating this real integral. We can put the integral into a more familiar form with the substitution $x=\sin^2{(\theta/2)}$:

$$\begin{align}\frac1{\pi} \int_0^1 dx \, \sin{\left ( \sqrt{x (1-x)} \right )} e^{-t x} &= \frac1{2 \pi} e^{-t/2} \int_0^{\pi} d\theta \, \sin{\theta} \, \sin{\left ( \frac12 \sin{\theta} \right )} e^{(t/2) \cos{\theta}} \\ &= \frac1{4 \pi} e^{-t/2} \operatorname{Im} \int_{-\pi}^{\pi} d\theta \, \sin{\theta} \, e^{i \frac12 \sin{\theta}} e^{(t/2) \cos{\theta}} \end{align} $$

We may evaluate the integral on the right by expressing it as a complex integral and taking it from there...

$$\begin{align} \int_{-\pi}^{\pi} d\theta \, \sin{\theta} \, e^{i \frac12 \sin{\theta}} e^{(t/2) \cos{\theta}} &= -i \oint_{|z|=1} \frac{dz}{z} \left ( \frac{z-z^{-1}}{i 2} \right ) e^{(t/2) \left (\frac{z+z^{-1}}{2} \right )} e^{(i/2) \left (\frac{z-z^{-1}}{i 2} \right )} \\ &= -\frac12 \oint_{|z|=1} \frac{dz}{z} \left ( z-z^{-1} \right ) e^{\frac14 (t+1) z + \frac14 (t-1) z^{-1}} \end{align} $$

The contour integral is equal to $i 2 \pi$ times the sum of the residues of the poles inside of the unit circle. In this case, the integrand has an essential singularity at $z=0$. To compute the residue at $z=0$, we expand the exponential in a Laurent series and we need only determine the coefficient of $z^{-1}$ in that expansion. In short, we wish to compute the coefficient of $z^0$ in the Laurent expansion of

$$\left ( z-z^{-1} \right ) e^{a z + b z^{-1}} $$

where $a=\frac14 (t+1)$ and $b=\frac14 (t-1)$. This expansion looks like

$$\begin{align} \left ( z-z^{-1} \right ) e^{a z + b z^{-1}} &= \left ( z-z^{-1} \right ) \sum_{k=0}^{\infty} \frac1{k!} \left ( a z + b z^{-1} \right )^k \\ &= \left ( z-z^{-1} \right ) \sum_{k=0}^{\infty} \frac1{k!} \sum_{m=0}^k \binom{k}{m} (a z)^m \left ( b z^{-1} \right )^{k-m} \\ &= \left ( z-z^{-1} \right ) \sum_{k=0}^{\infty} b^k \sum_{m=0}^k \frac1{m! (k-m)!} \left ( \frac{a}{b} \right )^m z^{2 m-k} \end{align} $$

To simplify things a bit, we note that only odd values of $k$ will result in a nonzero coefficient of $z^0$. We can then rewrite that last sum as

$$ \left ( z-z^{-1} \right ) e^{a z + b z^{-1}} = \left ( z-z^{-1} \right ) \sum_{k=0}^{\infty} b^{2 k+1} \sum_{m=0}^k \frac1{m! (2 k+1-m)!} \left ( \frac{a}{b} \right )^m z^{2 m-2 k - 1} $$

We have nonzero coefficients of $z^0$ when $m=k$ or $m=k+1$. Therefore, by the residue theorem, the integral above is equal to

$$\int_{-\pi}^{\pi} d\theta \, \sin{\theta} \, e^{i \frac12 \sin{\theta}} e^{(t/2) \cos{\theta}} = -\frac12 i 2 \pi (b-a) \sum_{k=0}^{\infty} \frac{(a b)^k}{k! (k+1)!} = -i \pi \frac{b-a}{\sqrt{a b}} I_1 \left ( 2 \sqrt{a b} \right )$$

Plug in $a=\frac14 (t+1)$ and $b=\frac14 (t-1)$ and we find that

$$\frac1{\pi} \int_0^1 dx \, \sin{\left ( \sqrt{x (1-x)} \right )} e^{-t x} = \frac12 e^{-t/2} \frac{I_1 \left ( \frac12 \sqrt{t^2-1} \right )}{\sqrt{t^2-1}} $$

and the ILT is as asserted above.