# moving the branch cut in an inverse Laplace transform

I would like to calculate the inverse Laplace transform of the function $$e^{-\sqrt s}$$. I understand that this is given by $${\cal L}^{-1} [e^{-\sqrt s}] = \frac{1}{2\pi i} \int_{-i\infty}^{i\infty} e^{-\sqrt s + s t}ds = \frac{1}{2\sqrt \pi}\exp\left(-\frac{1}{4t}\right)t^{-3/2}.$$ The way I know how to do this is to take the branch cut of the the square root function on the line $$(-\infty,0]$$ and then close the contour of integration while avoiding this branch cut (and therefore coming along the negative real axis from $$-\infty$$ to $$0$$ and then encircling the origin before going back out to $$-\infty$$.) This then allows us to figure out the value of the vertical portion of the contour, since the total value is $$0$$.

However, my question is whether one can do the same problem, not by deforming the contour to avoid the branch cut but by moving the branch cut. Suppose we take the branch cut of the square root to be on the positive real axis. Then it seems to me that we can close the contour in the left half plane without coming back to the origin. Since the total value of the integral should still vanish, one obtains $${\cal L}^{-1}[e^{-\sqrt s}] = -\frac{1}{2\pi i}\lim_{r\to\infty}\int_{\pi/2}^{3\pi/2}\exp\left[-\sqrt r e^{i\theta/2} + re^{i\theta} t\right] i r e^{i\theta} d\theta,$$ where $$r$$ and $$\theta$$ are real, and this is meant to represent the closure of the contour in the left half plane. However it seems to me this should vanish, and so I am confused. Is this correct? If so, how does one proceed?

Integrate along the imaginary axis in Mellin's inversion formula.
Because the inversion is real valued, we only integrate the real part of the integrand.

With the integrand $$F(s,t)=e^{-\sqrt{s}}\cdot e^{s\ t}$$ we obtain

$$\Re(F(i \omega ,t))=\Re\left(e^{i t \omega -\sqrt{i \omega }}\right)= \begin{cases} e^{-\frac{\sqrt{-\omega }}{\sqrt{2}}} \cos \left(t \omega +\frac{\sqrt{-\omega }}{\sqrt{2}}\right) & \omega <0 \\ e^{-\frac{\sqrt{\omega }}{\sqrt{2}}} \cos \left(\frac{\sqrt{\omega }}{\sqrt{2}}-t \omega \right) & \omega \geq 0 \\ \end{cases}$$

We choose the integration path with $$s= i \cdot \omega$$ and $$ds= i\cdot d\omega$$ and integrate, the branch cut is only touched at $$s=0$$:

$$\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(i \omega, t)\ d\omega=\frac{1}{2\pi}\left[\int_{-\infty }^0 e^{-\frac{\sqrt{-\omega }}{\sqrt{2}}} \cos \left(t \omega +\frac{\sqrt{-\omega }}{\sqrt{2}}\right) \, d\omega+\int_0^{\infty } e^{-\frac{\sqrt{\omega }}{\sqrt{2}}} \cos \left(\frac{\sqrt{\omega }}{\sqrt{2}}-t \omega \right) \, d\omega \right]=\frac{1}{2\pi}\left[\frac{\sqrt{\pi } e^{-\frac{1}{4 t}}}{2 t^{3/2}}+\frac{\sqrt{\pi } e^{-\frac{1}{4 t}}}{2 t^{3/2}} \right]=\frac{e^{-\frac{1}{4 t}}}{2 \sqrt{\pi } t^{3/2}}$$

Hence we get $$\mathcal{L}_s^{-1}\left[e^{-\sqrt{s}}\right](t)= \frac{e^{-\frac{1}{4 t}}}{2 \sqrt{\pi } t^{3/2}}$$, as expected.

• If one closes the loop in the left half plane, then the contribution from closing the contour should cancel with this to produce a net value of $0$. But it seems to me like this contribution would vanish. How does one resolve this? May 13, 2023 at 20:59

Any closed loop integral around a domain of analycity is zero.

In order to have a formula, the integral has to enclose the singularities to the left. That is, what you find in the tables.

If you have a first order pole to the left, you get its residue; if there is a branch cut, you get an difference integral of the jump value in the argument along the negative real line:

Map the complex plane to the Riemann sphere by

$$r \to -\tan\theta/2: \ \ z=0 \to \theta=\pi,\ \ z=\infty \to \theta =0$$.

There are two types of branch cuts $$z\to z^\alpha, \alpha \notin \mathbb Z, -1<\alpha<1$$ and $$z\to \log z$$ and combinations and function chains with entire analytic functions, that are analytic along the negative axis.

If one is not referring to multiple spheres representing the complete Riemann manifold with overcrossings on the meridian $$\phi =\pm\pi$$ one simply makes a cut along this meridian (date meridian on the earth: time is by no means a trivial concept in rotating systems).

Since the absolute values of analytic functions with such a branch cut along the $$\pm\pi$$ meridian are equal an both sides of the cut, along the cut the difference integral simply sums the jump of the arg angle along the path.

By the analyticity argument, the value does not change by a path deformation, as long as the cut is inside.

The open question after this general considerations: whats the role of the point at $$\infty$$.

Answer: even if the abolute value goes to zero there, the $$\arg$$ jump over the line is typically not depending on the $$\text{abs}$$. Just look at $$\arg \log(1/z) =-\phi$$ and $$\arg \sqrt 1/z=-\phi/2$$.

By this argument, the path of integration is fixed to pass the north pole. If not, it be can deformed to any closed loop somwhere in the domain of analycity yielding zero.

A plot reveals why the path has to pass the north pole on a real path: In most cases there is sitting the essential singurarity of the exponential.

By the infinitely often changing signs on a circle at infinity, while the the absolute value $$exp$$ lodes, only following the negative real line exactly, $$\pm \varepsilon$$, yields a definite limiting value in for $$z \to \infty$$.

The second reason for scrutinity is the measure at infinity, reflected to zero, yielding an additional $$z^{-2}$$ factor at z=0.