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Given $\int\log{\frac{a^2+z^2}{z^2}}e^{ivz}$dz enter image description here

The integral along the branch cut from 0 to a and back: $$\int_0^a\log(\lvert\frac{a^2+z^2}{z^2}\rvert)+i\pi$$ $$\int_a^0\log(\lvert\frac{a^2+z^2}{z^2}\rvert)-i\pi$$

How were the angle's $\pm\pi$ determined? Was $\arg[1-(a/y)^2]$ calculated by picking an angle that was convenient to calculate or is there a unique value on the left and right for the branch cut? A second example: Given $1/2\int\frac{e^{iwx}}{\sqrt{x^2+1}}$dzenter image description here On the right side of the branch cut: $\int_R^1\frac{e^{-wy}}{(\sqrt{y^2-1})i}$ On the left side of the branch cut: $\int_1^R\frac{e^{-wy}}{-(\sqrt{y^2-1})i}$ How were the signs determined in the denominator? Here it could be $e^{i\pi/2+in\pi}$ measured from the origin in the figure but I don't think that's how it was determined because the same reasoning doesn't work in the first example.

user5389726598465
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  • Are you asking were the $\pm i\pi $ came from? Because I have a few documents that solve integrals with branch cuts in the contour. – Highvoltagemath Oct 25 '19 at 13:14
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    @Hig ok, post a link in the comments – user5389726598465 Oct 26 '19 at 02:17
  • https://highvoltagemath.wixsite.com/highvoltagemath/the-funky-logarithm https://highvoltagemath.wixsite.com/highvoltagemath/euler-s-reflection-formula-take-1 https://highvoltagemath.wixsite.com/highvoltagemath/the-funky-log-2 There are also a lot of cool pages you might like to check out too! (I gave you three links) – Highvoltagemath Oct 26 '19 at 13:32
  • Oh! I just realized something! In some of the documents that I write on my website (Highvoltagemath) I use the letter $j$ instead of $i$ for $\sqrt{-1}$. But I usually say in those documents that that's the case. – Highvoltagemath Oct 26 '19 at 13:40
  • @user5389726598465 did you figured out the reasoning behind the $\pm i \pi$ angle? I'm dealing with the same problem. And yeah, I'm a bit late to ask this lol. – ppmbb Oct 13 '21 at 03:13
  • Nevermind, I figured out. I'm gonna post it as an answer for the record. – ppmbb Oct 13 '21 at 05:05

1 Answers1

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We are looking for expressions for the $\log(\frac{a^2+z^2}{z^2})$ inside the integral along the contour segments at the sides of the vertical branch cut that goes from $-ia$ to $ia$. Say the left segment is $C_1$ and the right segment is $C_2$ (both at the upper half plane). Then we have the following cases.

Segment $C_1$:

We select an arbitrary point $z=|z|e^{i\theta}$ of the segment $C_1$. The vectors that go from $ia$ and $-ia$ to $z$ form the corresponding complex numbers $(z-ia)=|z-ia|e^{i\theta_1}$ and $(z+ia)=|z+ia|e^{i\theta_2}$. Then we can write: $$\frac{z^2+a^2}{z^2}=\frac{|z-ia|e^{i\theta_1}|z+ia|e^{i\theta_2}}{|z|^2e^{i2\theta}}=\frac{|z^2+a^2|}{|z|^2}e^{i(\theta_1+\theta_2-2\theta)}=\Big|\frac{z^2+a^2}{z^2}\Big|e^{i(\theta_1+\theta_2-2\theta)}$$

Segment C1

Now, as our segment $C_1$ approaches arbitrarily closer to the vertical branch, we have that: $$\theta \to \frac{\pi}{2}, \theta_1 \to \frac{3\pi}{2}, \theta_2 \to \frac{\pi}{2}$$

Then:

$$\frac{z^2+a^2}{z^2}=\Big|\frac{z^2+a^2}{z^2}\Big|e^{i(\frac{3\pi}{2}+\frac{\pi}{2}-2\frac{\pi}{2})}=\Big|\frac{z^2+a^2}{z^2}\Big|e^{i\pi}$$

Therefore:

$$\log\Big(\frac{z^2+a^2}{z^2}\Big)=\log\Big|\frac{z^2+a^2}{z^2}\Big|+i\pi$$

Segment $C_2$:

The reasoning in this case is analogous to the previous one. With the difference now we have that, when the segment $C_2$ approaches the branch, the angles tend to:

$$\theta \to \frac{\pi}{2}, \theta_1 \to \frac{-\pi}{2}, \theta_2 \to \frac{\pi}{2}$$

Segment C2

Then:

$$\frac{z^2+a^2}{z^2}=\Big|\frac{z^2+a^2}{z^2}\Big|e^{i(-\frac{\pi}{2}+\frac{\pi}{2}-2\frac{\pi}{2})}=\Big|\frac{z^2+a^2}{z^2}\Big|e^{-i\pi}$$

Therefore:

$$\log\Big(\frac{z^2+a^2}{z^2}\Big)=\log\Big|\frac{z^2+a^2}{z^2}\Big|-i\pi$$

And that way we have determined the angles corresponding to a composite function at the path segments running along the branch cut.

ppmbb
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