## Inverse Laplace Transform IV

$$\hat{f}(s)=\frac{e^{-a\sqrt {s(s+r_0)}}}{\sqrt {s(s+r_0)}}$$

This one is interesting because of the presence of two branch points: one at $s=0$ and the other at $s=-r_0$. We will consider the complex integral

$$\oint_C ds \frac{e^{-a \sqrt{s (s+r_0)}}}{\sqrt{s (s+r_0)}} e^{s t}$$

where $C$ is the following contour pictured below:

This one’s a bit odd because we are removing the branch point singularities by subtracting the inner contour $C_i$ from the outer contour $C_o$, i.e. $C=C_o+C_i$.

The integral about $C$ consists of the integral about vertical path (that ultimately becomes the ILT) and the integral about the outer arc. We expect that this latter integral vanishes as the radius of the arc gets larger; this only happens when $a \lt t$ as shown below, by letting $s=R e^{i \phi}$; this integral over this arc is, for large $R$:

$$\left| \int_{\pi/2}^{3 \pi/2} d\phi \,e^{R (t-a) \cos{\phi}} \right|$$

Note that $\cos{\phi} \lt 0$ over the integration region, so that the integral only vanishes for $a \lt t$ when the contour opens to the left.

This leaves the integral over $C_i$ which consists of integrals above and below the real axis, between the branch points, and integrals about small circular arcs about the branch points. These latter integrals vanish in the limit of the radii of the arcs going to zero. Thus, we are left with

$$\int_{c-i \infty}^{c+i \infty} ds \frac{e^{-a \sqrt{s (s+r_0)}}}{\sqrt{s (s+r_0)}} e^{s t} – \underbrace{i \int_0^{r_0} dx \frac{e^{i a \sqrt{x (r_0-x)}}}{\sqrt{x (r_0-x)}} e^{-x t}}_{s=e^{-i \pi} x} – \underbrace{i \int_0^{r_0} dx \frac{e^{-i a \sqrt{x (r_0-x)}}}{\sqrt{x (r_0-x)}} e^{-x t}}_{s=e^{i \pi} x} = 0$$

We set the sum of the integrals to zero because of Cauchy’s theorem, i.e., there are no poles inside the contour $C$. The second integral is the integral over the line below the real axis; the third is the integral over the line above the real axis. Note that the phase transition was continuous from $-\pi$ to $\pi$ because we did not cross the branch cut; we did not see this explicitly because that information was included in the integrals about the branch points which vanished.

We do a little algebra and we have a real-valued integral for the ILT:

$$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{-a \sqrt{s (s+r_0)}}}{\sqrt{s (s+r_0)}} e^{s t} = \frac{1}{\pi} \int_0^{r_0} dx \frac{\cos{\left[ a \sqrt{x (r_0-x)}\right]}}{\sqrt{x (r_0-x)}} e^{-x t}$$

Using a substitution of $x=u^2$ and performing a series of trig substitutions (which I leave to the reader), I get for the integral

$$\frac{1}{\pi} e^{-r_0 t/2} \int_0^{\pi} d\theta \: \cos{\left(\frac{1}{2} a r_0 \sin{\theta}\right)} e^{(r_0 t/2) \cos{\theta}}$$

To evaluate, use the fact that

$$\cos{(b \sin{\theta})} = J_0(b) + 2 \sum_{k=1}^{\infty} J_{2 k}(b) \cos{2 k \theta}$$

and

$$\int_0^{\pi} d\theta \: \cos{n \theta} \, e^{z \cos{\theta}} = \pi I_n(z)$$

where $J_n$ is the $n$th Bessel of the first kind, and $I_n$ is the $n$th modified Bessel of the first kind.

We thus have a result for $t \gt a$. When $a \ge t$, however, we choose the integration contour to open to the right rather than the left. Because there are no poles or branch points there, the ILT for $a \ge t$ is zero.

The result is

$$f(t) = \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{-a \sqrt{s (s+r_0)}}}{\sqrt{s (s+r_0)}} e^{s t} = e^{-r_0 t/2} H(t-a) I_0\left(r_0 \sqrt{t^2-a^2}\right)$$

where $H$ is the Heaviside step function:

$$H(x) = \begin{cases} \\ 1 & x \gt 0\\ 0 & x \lt 0\end{cases}$$

• Hi Doug, thanks for writing in! I apologize for being unclear. Best approach is direct: sub $x=r_0 \cos^2{(\theta/2)}$ and the rest should be straightforward.
Hey there Ron, are you sure you aren’t missing a negative sign as a result of that substitution? we would have $dx = – r_0 \cos{(\theta/2)}\sin{(\theta/2)} d \theta$, no? I can’t seem to see where it is going to disappear further along if it does cancel out with something.