Considering $\arccos(x)=-i\ln(x+\sqrt{x^2-1})$, which has branch points at $1$, $-1$, and infinity, it seems natural to attempt evaluating $$\int\limits^{1}_{-1}\arccos(x)dx$$ using the residue theorem. I tried and it didnt seem to work; can this actually be done, and if so how? If not, why?
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What did you try? – sammy gerbil Apr 22 '20 at 15:18
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You don't need the residue theorem, you can integrate by parts. – Alessio K Apr 22 '20 at 15:22
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@sammygerbil I tried a dogbone since I couldn't reason out how to work with the logarithm. Thinking about it now, it seems the logarithm could be the problem. – Spador Yedi Apr 22 '20 at 15:51
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@Wrench I am aware, I just wanted to see if this could be done with the residue theorem so I could apply a similar method to similar integrands. – Spador Yedi Apr 22 '20 at 15:51
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For $\arccos(z)$, the points $z=1, z=-1$ are branch points, no good for computing via the residue theorem. When this happens, we may sometimes evaluate by finding the residues outside the contour. But there the point $z=\infty$ is also a branch point, so we still cannot use the residue theorem.
However, $\arccos z - i\log z$ (with the appropriate branches of $\arccos$ and $\log$) is single-valued at $z=\infty$, with residue $0$, so for a contour $\gamma$ that surrounds both $1$ and $-1$ once in the clockwise direction,
we get $\oint_\gamma \arccos z\;dz = \oint_\gamma i\log z\;dz = 2 \pi$. The limit of this branch of $\arccos z$ approaching the real axis from below is $\arccos x$, but the limit approaching from above is $-\arccos x$. So $$ 2\pi = \int_1^{-1} (-\arccos x)\;dx + \int_{-1}^1 \arccos x\;dx = 2\int_{-1}^1 \arccos x $$ Therefore our answer is $\pi$.
GEdgar
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Oh, I see now. My error was taking the residue at infinity. Thank you! – Spador Yedi Apr 22 '20 at 17:18
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Does the arccosine function have branch points at $0$ and $\infty$? – Mark Viola May 06 '20 at 15:43
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The arccosine has branch points at $1,-1,\infty$. But it is regular at $0$. – GEdgar May 06 '20 at 21:17