3
$\begingroup$

The title says it all, really - I am looking for

$$\int \arccos\left(\frac{\cos(x)}{r}\right) \, \mathrm{d}x$$

where $0<r<1$ and $x$ is in a domain where the integrand is real. It came up in a research problem involving billiard dynamics, but this is probably not relevant. I tried the usual techniques, Gradstein & Ryzhik and Mathematica without success. I am mostly seeking an exact answer, though it could involve obscure special functions.

Edit: Following Axoren, I am trying to expand the integrand in a series. Putting the following in mathematica

$Assumptions={r>0,r<1,x<Pi/2,x>ArcCos[r]};
a=Series[ArcCos[Cos[x]/r],{x,Pi/2,10}];
Integrate[a,x]

yields

$$\frac{1}{2} \pi \left(x-\frac{\pi }{2}\right)+\frac{\left(x-\frac{\pi }{2}\right)^2}{2 r}-\frac{\left(r^2-1\right) \left(x-\frac{\pi }{2}\right)^4}{24 r^3}+\frac{\left(r^4-10 r^2+9\right) \left(x-\frac{\pi }{2}\right)^6}{720 r^5}-\frac{\left(r^6-91 r^4+315 r^2-225\right) \left(x-\frac{\pi }{2}\right)^8}{40320 r^7}+\frac{\left(r^8-820 r^6+8694 r^4-18900 r^2+11025\right) \left(x-\frac{\pi }{2}\right)^{10}}{3628800 r^9}+O\left(\left(x-\frac{\pi }{2}\right)^{12}\right) $$

which gives me some hope of a closed form solution.

Edit 2: O.L.'s comment leads to the following exact solution, which checks numerically: $$ \int\chi(x)dx=\frac{1}{2}\left[{\rm Li}_2\left(\frac{e^{i(x+\chi(x))}}{r}\right) +{\rm Li}_2\left(\frac{e^{-i(x+\chi(x))}}{r}\right)+\pi(x+\chi(x))-\chi(x)^2 \right]+C$$ where $\chi(x)=\arccos(\cos(x)/r)$. (Thanks also for the comments at the other question regarding tags). Further remark: This expression involves the real part of the dilogarithm at complex argument. The mathematica pages state (without source or proof) a formula for the real part in terms of real arguments: See

http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog2/19/01/

$\endgroup$
  • $\begingroup$ @Integrator I meant it looks better now than the previous one, specially for the tittle format (you include some words, not $\LaTeX$ only). $\endgroup$ – Venus Dec 12 '14 at 10:38
  • $\begingroup$ I wouldn't be surprised if this was one of the many cases where there is no closed form. You could resort to a more qualitative study though. $\endgroup$ – Myself Dec 12 '14 at 10:58
  • 1
    $\begingroup$ Although this is not an exact duplicate, your integral trivially reduces to the indefinite integral from my old question, which can be computed in terms of dilogarithms. $\endgroup$ – Start wearing purple Dec 12 '14 at 22:34
1
$\begingroup$

I can give you a closed form for $|\cos(x)| \le r$. Since $r$ will always be less than $\frac \pi 2$, can just say the following:

$$ \int_{\arccos(-r)}^{\arccos(r)} \arccos \left(\frac{\cos(x)} r \right)\ dx $$

However, even that's not going to be pretty.

First, an identity:

$$ \arccos(x) = \frac \pi 2 - \sum_{n=0}^{\infty} \frac{\binom{2n}{n} x^{2n + 1}}{4^n (2n + 1)} \quad \mbox{for } |x| \le 1 $$

Now, we apply it to the integral.

$$ \int_{\arccos(-r)}^{\arccos(r)} \frac \pi 2 dx - \int_{\arccos(-r)}^{\arccos(r)}\sum_{n=0}^{\infty} \frac{\binom{2n}{n} \left(\frac{\cos(x)}{r}\right)^{2n + 1}}{4^n (2n + 1)}dx $$

$$ \int_{\arccos(-r)}^{\arccos(r)}\frac{\pi}{2} - \int_{\arccos(-r)}^{\arccos(r)}\sum_{n=0}^{\infty} \frac{\binom{2n}{n} \left(\frac{\cos(x)}{r}\right)^{2n + 1}}{4^n (2n + 1)} dx $$

$$ \int_{\arccos(-r)}^{\arccos(r)}\frac{\pi}{2} - \int_{\arccos(-r)}^{\arccos(r)}\sum_{n=0}^{\infty} \frac{\binom{2n}{n} \cos^{2n + 1}(x)}{4^n (2n + 1)r^{2n + 1}} dx $$

$$ \int_{\arccos(-r)}^{\arccos(r)}\frac{\pi}{2} - \sum_{n=0}^{\infty} \frac{\binom{2n}{n} \int_{\arccos(-r)}^{\arccos(r)}{\cos^{2n + 1}(x)\ dx}}{4^n (2n + 1)r^{2n + 1}} $$

Since at this point, $r$ is free, we can't simplify that inner integral expression until it's bound. So, all together, we have:

$$ \int_a^b \arccos \left(\frac{\cos(x)} r \right)\ dx = \left[\frac{\pi x}{2}\right]_a^b - \sum_{n=0}^{\infty} \frac{\binom{2n}{n} \int_{a}^{b} \cos^{2n + 1}(x)\ dx} {4^n (2n + 1)r^{2n + 1}} \quad \mbox{for } \arccos(-r) \le a \le b \le \arccos(r) $$

That's actually prettier than I expected it to be. However, it's not exactly what you're looking for, it's the closest I expect you to get. You can follow the same process so long as you can bound $x \in [\arccos(-r), \arccos(r)]$ when you use the result from it.

$\endgroup$
  • $\begingroup$ Thanks very much for this. But I'm having trouble with the first line, which expands around $x=0$. For any $0<r<1$ as in the question, the integrand is not real for $x=0$. So I wonder whether your final sum converges. Using series expansions is a good idea, though. $\endgroup$ – user200629 Dec 12 '14 at 14:04
  • $\begingroup$ The identity I used required the the argument to $\arccos x$ be $|x| \le 1$. Since the argument was going to be $\frac{\cos(x)}{r}$, it limited the domain to $\frac{|\cos(x)|}{r} \le 1$. I forgot the restriction on $r$ at this point and can easily include it. $\endgroup$ – Axoren Dec 12 '14 at 20:16
  • $\begingroup$ @Carl why is the integrand not real for $x = 0$? I know that $r$ must not be zero, by why musn't $x$? $\endgroup$ – Axoren Dec 12 '14 at 20:39
  • $\begingroup$ $x=0$ so $\cos(x)=1$, so $\cos(x)/r>1$, then you try to take the $\arccos$. $\endgroup$ – user200629 Dec 12 '14 at 20:42
  • $\begingroup$ Oh, I see. You're right. There are somethings I missed about the relationship between $cos(x)$ and $r$. Careless on my part. $\endgroup$ – Axoren Dec 12 '14 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy