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.
Best Answer
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.