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I was trying to understand contour integral which have or those involving branch cuts. Like $$I=\int_0^\infty\frac{dx}{x^3+1}$$ because the integrand is not even we cannot extend the integration to the whole real axis and then halve the result. However, suppose we look at the contour integral $$J=\oint_C\frac{\ln z \:dz}{x^3+1}$$ around the Keyhole contour. Everything seems nice until now. But then I get to know that "There is a connection between $J$ and the original definite integral $I$" $$J=2\pi iI$$ I get the partial intuition from one of M.SE answer, but couldn't understand the logic.

\begin{align}&\bbox[10px,#ffd]{\oint_C\frac{\ln z}{1+x^3}dz}=2\pi i\sum\text{Res}f(z)=\int_0^\infty\frac{\ln z}{1+z^3}dz+\int_\infty^0 \frac{\overbrace{\ln z+2\pi i}^\star}{1+z^3}dz=-2\pi i\int_0^\infty\frac{1}{1+z^3}dz\end{align}

Here I don't understand the $\star$. I think $\ln z=\ln r+i\theta+2\pi in$ used, but aren't we evaluate the integral on same branch? Then why to consider $2\pi i$?

Some contour integral which have branch cuts use wedge-shaped contour. Where $\Gamma = \Gamma_1 + \Gamma_2 + \Gamma_3$ going in a straight line $\Gamma_1$ from $0$ to $R$ on the real axis, then a circular arc $\Gamma_2$, then a straight line $\Gamma_3$ back to $0$. But I couldn't understand which angle I should use for $\Gamma_3$? Like for $f(z) = \frac{z^n}{1+z^m}$, it was suggested to use a wedge-shaped contour of angle $\frac{2\pi}{m}$.

Is there any trick or rule of thumb which help to decide the angle?

WhyMeasureTheory
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1 Answers1

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Here the holomorphic branch of the logarithm is chosen as $$ \ln(z) = \ln(|z|) + i \arg(z) \quad \text{with } 0 < \arg(z) < 2 \pi \, , $$ so that the function $f$ is holomorphic in $\Bbb C \setminus [0, \infty)$, and in particular on and inside the contour $C$.

The keyhole contour has two horizontal segments in some distance $\epsilon > 0$ from the real axis.

When integrating along the “upper” horizontal segment of $C$, $\arg(z)$ is close to zero, and in the limit $ \epsilon \to 0$ that contribution becomes $\int_r^R\frac{\ln x}{1+x^3} \, dx$, where $r$ and $R$ are the inner and outer radius of the contour, respectively.

On the “lower” horizontal segment of $C$, $\arg(z)$ is close to $2 \pi$, and in the limit $\epsilon \to 0$ that contribution becomes $\int_R^r\frac{\ln x + 2 \pi i}{1+x^3} \, dx = - \int_r^R\frac{\ln x + 2 \pi i}{1+x^3} \, dx$.

Then result then follows with $r \to 0$ and $R \to \infty$.

Martin R
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  • Aha, I see. I was forgot to consider $\arg(z)$. Thanks(+1) @MartinR. Could you say something on my second question? – WhyMeasureTheory Sep 19 '21 at 13:18
  • @WhyMeasureTheory: I don't think there is a general answer to that. Often you try to use some symmetries, e.g. $z^m$ is invariant under a rotation of $z$ by the angle $2\pi/m$, and each “wedge” contains exactly one pole of the function. – Martin R Sep 19 '21 at 13:22
  • "$z^m$ is invariant under a rotation of $z$ by the angle $2\pi/m$" How symmetries help us to decide the angle. It will be great help for me if you explain it a little bit @MartinR – WhyMeasureTheory Sep 19 '21 at 13:26
  • @WhyMeasureTheory: It gives you a simply relation between the integrals along the straight lines. – Martin R Sep 19 '21 at 13:28
  • I was really screwed up with the symmetries thing. Like Here they used Fresnel contour where the angle is $\frac{\pi}{4}$. I didn't see any relation $\sin(x^2)$ with that. It will be great help if you use an example and break down that thing, please. It bother me a lot @MartinR – WhyMeasureTheory Sep 20 '21 at 16:06
  • @WhyMeasureTheory: I have tried to answer your concrete question about $\int_0^\infty\frac{dx}{x^3+1}$, but I cannot give you a general recipe which works for all kinds of integrands like $\frac{z^n}{1+z^m}$, $\sin(x^2)$, and whatever. – Martin R Sep 20 '21 at 16:15