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Is there a complex version of integration-by-part? I saw someone used it but didn't find it in textbook. I tested integrals $\int_{\mathcal{C}}\frac{\log(x+1)}{x-2}\mathrm{d}x$ and $\int_{\mathcal{C}}\frac{\log(x-2)}{x+1}\mathrm{d}x$, where $\mathcal{C}$ encloses both -1 and 2. But the results do not match. Is it because they are not equal at the first place or I chose the wrong branch cut?

user88809
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1 Answers1

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Integration by parts is just the product rule and the Fundamental Theorem of Calculus. But you need well-defined analytic functions on your contour, which you don't have here.

Ted Shifrin
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  • Not everyone would agree in your answer. Take a look at https://math.stackexchange.com/a/505865/622884 for example. – Dr Potato Jan 03 '22 at 11:57
  • @DrPotato Looks to me as if Daniel said exactly the same thing, using the word “holomorphic” instead of “analytic.” – Ted Shifrin Jan 03 '22 at 14:10
  • (+1) This looks like a good answer. There is no analytic extension of $\log(x+1)$ onto a contour that has a non-zero winding number around $-1$. The same goes for $\log(x-2)$ and $2$. – robjohn Jan 03 '22 at 18:37
  • Yes Ted you're right. So is there any alternative for contour integration by parts when the contour encloses some poles or singularities? – Dr Potato Jan 05 '22 at 14:29
  • @DrPotato For each integrand to be a well-defined holomorphic/analytic function on the closed contour, every function appearing must have the sum of residues equal to $0$ inside. – Ted Shifrin Jan 05 '22 at 18:14