I'm an engineering student and I'm working on a probability problem. I'm trying to find out the probability that two random diagonals of two circumferences intersect. I'm considering two circumferences with the same ray r and with their center points at distance c. In according to the value of a=c/r the problem assumes different formulations anyway in any case emerge a strange integral I'm not able to solve. The integral is the following:

$$\int \arcsin(a \sin{x}) dx $$

the value of the limits of integration is in according to the value of the parameter a ( for example one integral is integrated between $0$ and $\arccos (a/2)$ ). The software 'Mathematica' doesn't give me any results for this integral so I have tried to work on the integral expanding it with the definition of $\arcsin$, using by parts method and various substitutions with the hope to find out some known form that can be expressed in terms of some special function, but my efforts has been vain up to now. I have found some papers dealing with integrals involving $\ln{sin}$ that maybe have something to do with my problem.Probably I'm facing some hard mathematics that I'm not able to deal with or there is not a solution as the one I'm looking for. I'm waiting for some advice.

Thank you.

  • $\begingroup$ Hint: differentiate with respect to $a$. $\endgroup$ – J.G. Jul 3 '18 at 15:31
  • $\begingroup$ I have also tried to use the leibniz integration rule considering a as the variable, actually after having rewritten in some way this function, but no results I always end up with an integral i'm not able to solve. $\endgroup$ – marco Jul 3 '18 at 15:36

For any $z\in\mathbb{C}$ such that $|z|<1$ we have $$ \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}z^n = \frac{1}{\sqrt{1-z}},\qquad \sum_{n\geq 0}\frac{1}{(2n+1)4^n}\binom{2n}{n}z^{2n+1}=\arcsin(z)\tag{A} $$ hence, for instance, for any $a$ such that $|a|<1$ we have

$$ \int_{0}^{\pi/2}\arcsin(a\sin x)\,dx=\sum_{n\geq 0}\frac{a^{2n+1}}{(2n+1)4^n}\binom{2n}{n}\int_{0}^{\pi/2}\sin(x)^{2n+1}\,dx \tag{B}$$ and a small miracle happens, since by integration by parts $$\int_{0}^{\pi/2}\sin(x)^{2n+1}\,dx = \frac{4^n}{(2n+1)\binom{2n}{n}}\tag{C}$$ hence $$ \int_{0}^{\pi/2}\arcsin(a\sin x)\,dx = \sum_{n\geq 0}\frac{a^{2n+1}}{(2n+1)^2}=\color{red}{\frac{\text{Li}_2(a)-\text{Li}_2(-a)}{2}}.\tag{D} $$ Considering the limit of both sides of ($\text{D}$) as $a\to 1^-$ leads to a proof (due to Euler) of $\zeta(2)=\frac{\pi^2}{6}$. The procedure above can be adapted to the case $\int_{0}^{\color{red}{\pi/4}}\arcsin(a\sin x)\,dx$, involving incomplete Beta functions. On the other hand it is pretty difficult to provide an insightful answer to your question without knowing the integration bounds you have to deal with. Besides $$\begin{eqnarray*} \int \arcsin(a\sin x)\,dx &=& \int_{0}^{a}\underbrace{ \int\frac{\sin x}{\sqrt{1-b^2 \sin^2(x)}}\,dx}_{\text{logarithmic integral}}\,db\\&\stackrel{x\mapsto\arcsin z}{=}&\int_{0}^{a}C-\frac{1}{b}\log\left(b^2\sqrt{1-z^2}+b\sqrt{1-b^2 z^2}\right)\,db \end{eqnarray*}$$ there is no reason for expecting that the LHS always has a nice closed form (in terms of $\text{Li}_2$).
On the other hand $g(a)=\int_{0}^{\arccos(a/2)}\arcsin(a\sin x)\,dx $ is a smooth, decreasing function on $(\sqrt{2},2)$, which is well-approximated by $(2-a)-\frac{1}{2}(2-a)^3$ on such interval.

  • $\begingroup$ Thank you for your reply. Now I 'm actually coming back by train when I'll arrive at home I'll read you answer with the due attention. Polylogarithmic functions are something I come across reading the various papers mentioned in my topic. For the integration buonds I got different combinations. When $\sqrt {2}<a <2 $ the limits of integration are $0$ and $ \arccos {a/2}$. Is this case also tractable in a smart way? $\endgroup$ – marco Jul 3 '18 at 16:31
  • $\begingroup$ How did you commute $\int \sum$? $\endgroup$ – GFauxPas Jul 3 '18 at 17:08
  • $\begingroup$ @GFauxPas: we are dealing with holomorphic functions and absolutely convergent series, there is no issue in commuting $\int$ and $\sum$. $\endgroup$ – Jack D'Aurizio Jul 3 '18 at 17:09
  • $\begingroup$ Thank you jack. Tomorrow I will work on my problem following your advices. The next Step of my problem is to consider also 'a' as a random variable so I will try to join the various limits of integration to se if I can reduce the presence of these kind of integrals to an integral between $0$ and $\pi/2$. If something will remain I will try with the ' besides' part of your answer or with some approximation. $\endgroup$ – marco Jul 3 '18 at 17:59

If you want an exact solution, then I don't know how to approach this.

But since this is an engineering problem and not a pure math problem, then an answer that's close enough should be good enough.

If you do a numerical integration of this integral between 0 and $arccos(\frac{a}{2})$ and vary the parameter a from 0 to 2, you get a curve that looks like this

numerical integration

Which looks almost parabolic, but you can see there is a small asymmetry.

You can get a pretty good curve fit with

$$f(a)=0.00146 + 0.958 a - 0.338 a^2 - 0.0736 a^3$$

I'll look around and see if there is something nicer...

Looking around, $$f(a)=-a(a-2)(0.05a+0.5)$$ is a pretty good fit.

Here's what the curve for the data and the curve for this latest fit look like together... enter image description here

  • $\begingroup$ Thank you David.... $\endgroup$ – marco Jul 4 '18 at 6:08

Using $$\arcsin(t) = \sum^{\infty}_{n=0} \frac{(2n)!}{4^n (n!)^2 (2n+1)} t^{2n+1} \qquad \text{for}\qquad |t|\leq 1$$ we should have $$\arcsin(a \sin(x))=\sum^{\infty}_{n=0} \frac{(2n)!\,\,a^{2n+1}}{4^n\, (n!)^2 \,(2n+1)} \sin^{2n+1}(x)$$ and $$\int_0^\alpha \sin^{2n+1}(x)\,dx=-\cos (\alpha) \,\, _2F_1\left(\frac{1}{2},-n;\frac{3}{2};\cos ^2(\alpha)\right)+\frac{\sqrt{\pi }\,\, \Gamma (n+1)}{2\, \Gamma \left(n+\frac{3}{2}\right)}$$ where appears the Gaussian or ordinary hypergeometric function.


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