Undergrad Understanding the "branch cut argument"

[TMO]

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.

 

[Orodruin]

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]

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

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

instead of

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

 

[Orodruin]

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

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

instead of

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

Not so, the ln(x) part of C1 will cancel that of C3.

 

[stevendaryl]

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]

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 $\pm 2\pi i$.

Indeed.