# Understanding the "branch cut argument"

• I

## Main Question or Discussion Point

http://web.williams.edu/Mathematics/sjmiller/public_html/302/coursenotes/Trapper_MethodsContourIntegrals.pdf

See Type 5 Integrals. I don't understand why J is equal to the original real integral multiplied by a factor of $2\pi i$. I think the $2\pi i$ comes from the fact that as you go around $C_2$ you end up $2\pi i$ greater than when you started. But why does the difference between $C_1, C_3$ correspond to the real integral? Is it because these two segments touch the real line? I don't quite get it.

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Orodruin
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Did you read the part where the log is expanded according to $\log z = \log r + i\theta + i 2\pi n$? Along both curves $r \to x$. What does this mean for the integrals? What does it mean for the difference between the integrals?

stevendaryl
Staff Emeritus
I think I'm missing something simple. It seems to me that the contour described amounts to computing the integral:

$\int_0^\infty \frac{ln(x)}{1+x^3}$

$\int_0^\infty \frac{1}{1+x^3}$

Orodruin
Staff Emeritus
Homework Helper
Gold Member
I think I'm missing something simple. It seems to me that the contour described amounts to computing the integral:

$\int_0^\infty \frac{ln(x)}{1+x^3}$

$\int_0^\infty \frac{1}{1+x^3}$
Not so, the ln(x) part of C1 will cancel that of C3.

stevendaryl
Staff Emeritus
Not so, the ln(x) part of C1 will cancel that of C3.
Oh! Right above the real axis, you have $ln(|x|)$ and right below the real axis, you have $ln(|x|) + 2\pi i$. So when you subtract them, all that's left is $\pm 2\pi i$.

Orodruin
Staff Emeritus
Oh! Right above the real axis, you have $ln(|x|)$ and right below the real axis, you have $ln(|x|) + 2\pi i$. So when you subtract them, all that's left is $\pm 2\pi i$.