What happens to poles lying on branch cuts in contour integration?

Question

Inverse the Laplace Transform $$\frac{1}{\sqrt{s}}\cdot\frac{1}{1 + s}$$ back to time domain requires evaluation of Bromwich integration: $$\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty} \frac{1}{\sqrt{s}}\cdot\frac{1}{1 + s} e^{st} ds$$ The $\sqrt{s}$ term causes a branch cut along the negative half $x$-axis, including the origin. My question is what happens to the point $(-1, 0)$: it is both a pole, and lies on the branch cut. More specifically, when constructing the contour integration path, could I just ignore the pole because the branch cut removes it (figure A), or I should deform the contour as half circles around the pole, once on each side of the branch cut (figure B), as if the pole is duplicated? Figure A and B: contour integration path

Answer 1

It would be the situation in B: you would deform around the pole. It works as follows.

The inverse Laplace transform is given by Cauchy's theorem. I present the parametrization of each piece of the contour, assuming that the radius of the semicircular detour about the pole $z=-1$ and the branch point $z=0$ is $\epsilon$:

$$\int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} + e^{i \pi} \int_{\infty}^{1+\epsilon} dx \frac{e^{-t x}}{e^{i \pi/2} \sqrt{x} (1-x)}+i \epsilon \int_{\pi}^0 d\phi ,円 e^{i \phi} \frac{e^{t (-1+ \epsilon e^{i \phi})}}{\sqrt{e^{i \pi}+\epsilon e^{i \phi}} (\epsilon e^{i \phi})}\\+ e^{i \pi} \int_{1-\epsilon}^{\epsilon} dx \frac{e^{-t x}}{e^{i \pi/2} \sqrt{x} (1-x)}+i \epsilon \int_{\pi}^{-\pi} d\phi ,円 e^{i \phi} \frac{e^{t \epsilon e^{i \phi}}}{\sqrt{\epsilon e^{i \phi}} (1+\epsilon e^{i \phi})} +e^{-i \pi} \int_{\epsilon}^{1-\epsilon} dx \frac{e^{-t x}}{e^{-i \pi/2} \sqrt{x} (1-x)}\\+ i \epsilon \int_{2 \pi}^{\pi} d\phi ,円 e^{i \phi} \frac{e^{t (-1+ \epsilon e^{i \phi})}}{\sqrt{e^{-i \pi}+\epsilon e^{i \phi}} (\epsilon e^{i \phi})}+ e^{-i \pi} \int_{1+\epsilon}^{\infty} dx \frac{e^{-t x}}{e^{-i \pi/2} \sqrt{x} (1-x)} = 0$$

Note that the integrals about the semicircular detours above and below the axis (the 3rd and the 7th integrals) cancel. In the limit as $\epsilon \to 0,ドル the integral about the branch point (the 5th integral) also vanishes. We are then left with, as $\epsilon \to 0,ドル

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} + \frac1{2 \pi} PV \int_{\infty}^0 dx \frac{e^{-t x}}{\sqrt{x} (1-x)} - \frac1{2 \pi} PV \int_0^{\infty} dx \frac{e^{-t x}}{\sqrt{x} (1-x)} = 0$$

where $PV$ denotes the Cauchy principal value of the integral. Thus, the ILT is given by (subbing $x=u^2$)

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = \frac1{\pi} PV \int_{-\infty}^{\infty} du ,円 \frac{e^{-t u^2}}{1-u^2} $$

To evaluate the integral, we rewrite as

$$e^{-t} PV \int_{-\infty}^{\infty} du ,円 \frac{e^{t (1- u^2)}}{1-u^2} = e^{-t} I(t)$$

where

$$I'(t) = e^{t} PV \int_{-\infty}^{\infty} du ,円 e^{-t u^2} = \sqrt{\pi} t^{-1/2} e^{t} $$

and $I(0) = 0$. Thus,

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = e^{-t}\frac1{\pi} \sqrt{\pi} \int_0^t dt' ,円 t'^{-1/2} e^{t'} = e^{-t} \frac{2}{\sqrt{\pi}} \int_0^{\sqrt{t}} dv ,円 e^{v^2} $$

or, finally,

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = e^{-t} \operatorname{erfi}{\left (\sqrt{t} \right )} $$

• Taozi: Dear Ron, thank you very much for your answer, beautiful technique to simplify the final integration! I am sure the answer is correct (checked with Mathematica), what I am not sure however is that integration 2 and 4 can be readily joined together (so does 6 and 8), neither 2 or 4 's integration range include the pole x = 1, but joining them would add x = 1 to the integration range, something I am not so comfortable with.
• Ron Gordon: @Taozi: thanks for the kind words. Regarding the combination of the 2nd and the 4th integrals, that is what becomes the Cauchy principal value of the integral over combined region. This is specifically designed to deal with the pole at $x=1$. Note that $$PV \int_0^{\infty} dx \frac{e^{t x}}{\sqrt{x} (1-x)} = \lim_{\epsilon \to 0} \left [\int_0^{1-\epsilon} dx \frac{e^{t x}}{\sqrt{x} (1-x)} + \int_{1+\epsilon}^{\infty} dx \frac{e^{t x}}{\sqrt{x} (1-x)} \right ]$$ I hope this helps.
• Taozi: Exactly what I missed, thank you. In this integration the two half-circles cancel each other, and in effect do not contribute to the result, this is just a coincidence, right? If it's $\sqrt[3]{z}$ in the denominator such nice cancellation would not be there. Another question is many times a branch cut causes a circle (with an opening) at the origin, and according to my experience the integration around it is always zero, are there any counter examples or I can take it as a rule of thumb? No intention to be lazy here of course.
• Ron Gordon: @Taozi; yes, that's right. The fact that the sqrt is in the denominator makes those integrals cancel. The contribution of the arc about the origin will vanish so long as the power of $z$ in the denominator is less than 1ドル$.
• Taozi: Thank you for your help. Having been reading many of your answers to other complex analysis questions, it's a great learning experience :-)