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Could someone give me an example of a contour integral of a simple complex
function that requires a branch cut, and a choice of argument -- but for different parts of my contour, I make that argument choice differently.

I'm trying to understand when/if this is acceptable... and a concrete example would really help me figure out another question I have posted. I'm interested in what I need to consider, and what restrictions I have. An example of what I mean:

For example, say I'm integrating the function $z^\frac{1}{2}$ along the contour $H_1 \cup H_2 \cup H_3$ where $H_1$ is the line segment $1+i$ to $i$, $H_2$ is the semi circle with radius 1 from $i$ to $-i$, and $H_3$ is the line segment from $-i$ to $1-i$. The branch cut could be real axis $x \geq 0$. Then supposing I make the argument choice for my function different on $H_1$ and $H_3$.

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Consider the following contour integral used to compute the inverse Laplace transform of $\hat{f}(s) = s^{-1/2}$:

$$\oint_C ds \frac{e^{s t}}{\sqrt{s}}$$

where $C$ is the following contour:

enter image description here

By the residue theorem (or Cauchy's integral theorem), this integral is zero because there are no poles within $C$. $C$, however, has $6$ pieces: the original integral along $\Re{s}=a$, a circular arc of large radius $R$, a section that goes in a positive direction just above the negative real axis, a circular arc of small radius $r$ around the origin, and another section just below the negative real axis in a negative direction. In the limit as $R \rightarrow \infty$ and $ r \rightarrow 0$, the integrals along the circular arcs vanish. This leaves

$$ \int_{a-i\infty}^{a+i\infty} ds \frac{e^{s t}}{\sqrt{s}}+e^{i \pi} \int_{\infty}^0 dx \frac{e^{-x t}}{i \sqrt{x}} + \int_0^{\infty} dx \frac{e^{-x t}}{-i \sqrt{x}}=0$$

A little rearranging produces

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

Substitute $y=\sqrt{x}$ into the integral on the RHS and finally get

$$ \frac{1}{i 2 \pi} \int_{a-i\infty}^{a+i\infty} ds \frac{e^{s t}}{\sqrt{s}} = \frac{2}{ \pi} \int_0^{\infty} dy \; e^{-t y^2}=\frac{1}{\sqrt{\pi t}}$$

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  • $\begingroup$ Hi Ron, is there anywhere in here where you have made different choices for your branch, or for your argument? It doesn't appear so to me... and that is specifically what I was trying to understand better. $\endgroup$ – MathStudent Oct 14 '13 at 14:09
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    $\begingroup$ @David: absolutely, unless I am misunderstanding what you are asking for. I chose a branch cut along the negative real axis, so that the arguments of the complex numbers are in the interval $(-\pi,\pi]$. This is crucial in the evaluation of the integral, which is why I used this example to illustrate. $\endgroup$ – Ron Gordon Oct 14 '13 at 14:19
  • $\begingroup$ I'm working through a proof in a textbook where the argument choice is different (with the same branch cut) depending on which part of the contour is being integrated. I've had a question regarding this proof up on MSE for a couple weeks, but haven't gotten any answers. I posted this question hoping to get a better understanding of when/if this can be done. $\endgroup$ – MathStudent Oct 14 '13 at 15:37
  • $\begingroup$ @David: can you point to this proof? Hard to answer your question without knowing the specifics. $\endgroup$ – Ron Gordon Oct 14 '13 at 15:43
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    $\begingroup$ @David: yes, note that $-1$ is represented as $e^{i \pi}$ above the negative real axis and $e^{-i \pi}$ below. Thus, when we take the square root as demanded by the integral, we get $\pm i$ depending on which side of the negative real axis we are on. This is why I chose the contour for integration as I did. $\endgroup$ – Ron Gordon Oct 14 '13 at 18:27
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As far as I understand it branch lines are an artifact of multivalued complex functions. Riemann had a solution to this, now called a Riemann surface. On the Riemann surface a multifunction becomes single valued and continuous. The branch lines disappear but critical points remain.

I'm going to use my Presentations Mathematica Application to illustrate one method of handling multifunctions.

Here I generate a set of close-by points on the arc of a circle about the origin. The points just cross the conventional (Mathematica) branch line on the negative real axis.

enter image description here

Giving

enter image description here

The value is discontinuous with the imaginary part reversing direction on the branch line. Presentations has a Multivalues capability. It keeps track of the last set of multivalues and reorders the new set to be close to the old set. It does this by considering it as a linear programming assignment problem. It works well for cases of a small set of multivalues. In the following I define sqrtz as a multifunction initialized in the given order on first use. With each evaluation it picks the best order and keeps a record of all the values and the permutation used. Successive evaluations must also be reasonably close. The following is the result when evaluated on the same set of points.

enter image description here

Giving:

enter image description here

The function is now continuous when going across the negative real axis.

The following shows another illustration of this. A movable red point is shown on top of a modulus contour plot of Sqrt[z]. The arrow attached to the point represents the value of the function at the red point. In a sense this is a local 4-dimensional plot because the surface is 2-dimensional and the arrow is 2-dimensional. We can drag (in Mathematica) the red point around the plot and see how the function value changes. It is everywhere smooth and continuous. If we drag it around the origin once we obtain the picture on the right. The arrow now points in the opposite direction. We have smoothly transitioned to what would have been called the second value. If we drag it around again we return to the picture on the left. Within a radius of 3 we have explored all parts of the Riemann surface for the function and it was everywhere single valued and continuous - just as RIemann said.

enter image description here

Now let's evaluate the line integral around various circles in the plane. Mathematica has an NIntegrate command in which one can specify a path in the complex plane. However, their advance methods outfox Multivalues. So I retreat to a more primative method and just sum the function values times the step in the complex plane. In the following: step1 calculates function values around the circle; step 2 picks the mid value; step 3 calculates the incremental step; the last step sums the product. The following is for a circle in the right hand plane that does not enclose the origin.

enter image description here

In the following I just show the paths used and the results: They represent 1) a circle in the right hand plane that does not enclose the origin, giving 0. 2) a circle in the left hand plane that does not enclose the origin, giving 0. 3) a circle once about the origin, giving a nonzero result. 4) a circle twice about the origin giving zero.

enter image description here

Giving: {0, 0, -0.235714 + 0.00868186 I, 0}

The following results use the regular Sqrt[z] function without using Multivalues. The same paths were used. 1) a circle in the right hand plane that does not enclose the origin, giving 0. 2) a circle in the left hand plane that does not enclose the origin, incorrectly giving nonzero. 3) a circle once about the origin, giving an incorrect nonzero result. 4) a circle twice about the origin giving an incorrect nonzero result.

Giving: {0, 0.00633665 - 0.988311 I, 0.00866618 - 0.235288 I, 0.0173324 - 0.470576 I}

I don't consider myself to be a great expert on complex function theory but do believe that the Riemann surface is a way to tame branch cuts. I would welcome any corrections I might get on this.

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  • $\begingroup$ Hi David -- it may be that riemann surfaces are a better way to deal with this, but I'm trying to understand some of the details if one were going to stick with branch cuts. $\endgroup$ – MathStudent Oct 14 '13 at 14:11

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