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(Apologies for any awkwardness. This is my very first post.) This is a question about how to get the sign right for the classic integral dealt with here Keyhole Contour with Square Root Branch Cut on Imaginary Axis

I get the same answer as Ron Gordon https://math.stackexchange.com/users/53268/ron-gordon offers in that reference except for the signs. I find -i instead of +i and vice versa in the denominators, leading to:

$$I_2 = i \int_{\infty}^m dy \, \frac{i y \, e^{-r y}}{-i \sqrt{y^2-m^2}} $$

$$I_3 = i \int_m^{\infty} dy \, \frac{i y \, e^{-r y}}{+i \sqrt{y^2-m^2}} $$

This is where my (likely wrong) signs are coming from. With the same contour setup as in the reference above, I have $-\pi/2<arg\left( z+i\right )<3\pi/2$ and $\pi/2<arg\left( z-i\right )<5\pi/2$.

So, on the $I_2$ side of the cut, the arg of the denominator is made up of $\pi/2$ from $\left( z+i\right )$ plus $5\pi/2$ from $\left( z-i\right )$, the whole thing times $1/2$ because of square roots. That produces a total arg of $3\pi/2$ in the denominator, which is the factor $e^{i3\pi/2}=-i$ in my (most likely wrong) $I_2$ above.

Similarly, on the $I_3$ side of the cut, the arg of the denominator is made up of $\pi/2$ again from $\left( z+i\right )$ plus $\pi/2$ from $\left( z-i\right )$, the sum times $1/2$ again. That produces a total arg of $\pi/2$ in the denominator, which is the factor $e^{i\pi/2}=+i$ in my (most likely wrong) $I_3$ above.

Where did I go wrong? Much obliged.

Alred
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  • IDK what restrictions the OP from that link put on their $I_1$, which is $\int_{-R}^R \frac{z e^{i r z}}{\sqrt{z^2+m^2}}$, to make it converge as $R \to \infty$, but you can write $\sqrt{x^{2}+m^{2}}=\exp\left(\frac{1}{2}\log\left(x^{2}+m^{2}\right)\right)$ as $$\exp\left(\frac{1}{2}\ln\left|x^{2}+m^{2}\right|+\frac{i}{2}\arg\left(x+im\right)+\frac{i}{2}\arg\left(x-im\right)\right)$$ but that equality may be false depending on the interval the $\arg$s live in. To recover the integral on the real axis, you would want $\arg\left(x^2+m^2\right) = 0$. – Accelerator Jul 01 '24 at 04:39
  • Specifically, $\arg(x+im) + \arg(x-im) = 0 \iff \arg(x-im) = -\arg(x+im)$. If $\arg(z-im) \in \left(\frac{\pi}{2},\frac{5\pi}{2}\right)$, then $\arg(z+im) \in \left(-\frac{5\pi}{2}, -\frac{\pi}{2}\right)$. The contour I have in mind is this one here which is pretty much the same as the one from the link. I believe one of your $\arg$s is incorrect as well. I used the intervals in this comment and I was able to get exactly what Dr. Gordon got in their answer. – Accelerator Jul 01 '24 at 04:40
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    Yes, you are absolutely correct. I put in the fix and it worked. Terrific, thank you. – Alred Jul 01 '24 at 06:14

1 Answers1

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Question answered thanks to comment by Accelerator https://math.stackexchange.com/users/1065595/accelerator

Once we impose a constraint on the argument intervals so that $arg\left( z+i\right )$ and $arg\left( z-i\right )$ cancel on the real axis to make the function there real, as the contour integration scheme requires, then the previous (incorrect) combination of intervals could be changed to, eg, $-5\pi/2<arg\left( z+i\right )<-\pi/2$ and $\pi/2<arg\left( z-i\right )<5\pi/2$.

Then the rest of the reasoning reproduces correctly the Ron Gordon answer. Many thanks to Accelerator.

Alred
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