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?
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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.
answered Aug 4, 2013 at 7:36
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$\begingroup$ Not everyone would agree in your answer. Take a look at math.stackexchange.com/a/505865/622884 for example. $\endgroup$ Commented Jan 3, 2022 at 11:57
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$\begingroup$ @DrPotato Looks to me as if Daniel said exactly the same thing, using the word “holomorphic” instead of “analytic.” $\endgroup$ Commented Jan 3, 2022 at 14:10
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$\begingroup$ (+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 $2$. $\endgroup$– robjohn ♦Commented Jan 3, 2022 at 18:37
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$\begingroup$ Yes Ted you're right. So is there any alternative for contour integration by parts when the contour encloses some poles or singularities? $\endgroup$ Commented Jan 5, 2022 at 14:29
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$\begingroup$ @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. $\endgroup$ Commented Jan 5, 2022 at 18:14