Understanding the "branch cut argument"

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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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  • #2
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?
 
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stevendaryl
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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}##
 
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Orodruin
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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.
 
  • #5
stevendaryl
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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##.
 
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Orodruin
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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.
 

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