I've stumble into this integral on a paper (page 5, right column)

$\displaystyle\int_{\tfrac {\cos(k)}{\sqrt{1+\cos^2(k)}} \leq~y}~ \dfrac{\mathrm{d}k}{4\pi}$

with solution

$\displaystyle{1-\frac{1}{\pi}\arccos \left(\frac{k}{\sqrt{1-k^2}}\right)}$.

What type of integral is this? How would you solve it? How do the authors of this paper solve this?

  • $\begingroup$ @Marcos: It may look funny and weird, but re-write this on paper nicely space out and see if you notice anything. As user6312 pointed out, there is no integrand, so you can think of it being the integral of one as such: $\int \! 1\,\mathrm{d}k~$ with a factor of $\dfrac {1}{4\pi}$ multiplied outside. Then it just comes down to the inequality constraint. $\endgroup$ – night owl Jun 30 '11 at 2:44
  • $\begingroup$ yes, I just did that, and is very easy. The problem, as I wrote in my commment below, is the notation of the integral $\endgroup$ – Marcos Villagra Jun 30 '11 at 2:49
  • $\begingroup$ is it ok to write integrals like this? $\endgroup$ – Marcos Villagra Jun 30 '11 at 2:50
  • $\begingroup$ @night owl: thank you very much for that info!! $\endgroup$ – Marcos Villagra Jun 30 '11 at 4:34
  • $\begingroup$ @Marcos: Yes, it is quite often used when dealing with integrals that have constraints on them. A way to think about if you are familiar with vector calculus is that you may want to find the area of some region and it can be looked at as so: $\iint\limits_{D}~$ where D is the region of constraint that is of interest. $\endgroup$ – night owl Jun 30 '11 at 7:21

This is as easy an integral as could be imagined, since no integration is necessary (the integrand is constant). All we need to do is to solve the inequality $$\frac{\cos k}{\sqrt{1+\cos^2 k}} \le y$$ for $k$ in terms of $y$.

The result is, however, unfortunately marred by a bad typo. In the answer, the $k$'s should be $y$'s.

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  • $\begingroup$ I see, but is it normal to use that kind of notation?. That's what completely misguided me. $\endgroup$ – Marcos Villagra Jun 30 '11 at 2:47
  • $\begingroup$ I mean, the cosine on the integral there? $\endgroup$ – Marcos Villagra Jun 30 '11 at 2:48
  • $\begingroup$ @Marcos Villagra: Fairly common, though this one is a bit of an extreme case. It would have been better to define a region $D$ and do it in night owl style. An alternative that has become popular is to define $I(k)$ to be $0$ outside the region, $1$ inside, and write the integral as $\int_k \frac{I(k)}{4\pi}dk$. What is intended in the paper is clear from the context, but would have been a lot clearer without the typo! That's one of the problems with a non-refereed paper. $\endgroup$ – André Nicolas Jun 30 '11 at 3:19
  • $\begingroup$ thank you very much, now is crystal clear! $\endgroup$ – Marcos Villagra Jun 30 '11 at 4:34

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