How do I evaluate integrals of the form $\int_{0}^{2\pi}\arccos(-a-b\cos x)dx$

Question

Are there any closed-form/analytical expressions for integrals of the form: $$\int_{0}^{2\pi}\arccos(-a-b\cos x)dx$$

or

$$\int_{0}^{2\pi}\sqrt{1 - (-a-b\cos x)^2}(-a-b \cos x) dx$$

where $a>0$, $b>0$ and $a+b<1$? Such integrals come up when trying to use the method of averaging to obtain the slow-flow equations for multi-degree-of-freedom dynamical systems with discontinuous forcing.

I tried to proceed with the first one as follows:

$$2\int_{0}^{\pi}\left(\pi-\arccos(a+b \cos x) \right) dx $$

Then, after a substitution $y=\arccos(a+b \cos x)$, we have:

$$ I = 2\pi^2 - \dfrac{2}{b}\int_{\arccos(a+b)}^{\arccos(a-b)} \dfrac{y \sin y}{\sqrt{1-\left(\frac{-a+\cos y}{b} \right)^2} }dy $$

A further substitution $\cos y = z$, brings the integral term to:

$$\bar{I} = -\int_{a+b}^{a-b} \dfrac{\arccos z}{\sqrt{1 - \left(\dfrac{z-a}{b} \right)^2 }} dz$$

The expressions appear to have a relation with some elliptic integral but is shifted by a parameter $\frac{a}{b}$. I have attempted to solve it in Mathematica but it is unable to perform the integration.

If anybody can provide an answer, that would be of great help. A step-by-step derivation would be ideal. But any suggestions/tips are also welcome.

Thanks in advance!

Answer 1

Not sure if a closed-form expression exists, but maybe a series might help.
We start with the formula $$ \arccos (-a-b\cos(x))=\frac{\pi}{2}-\sum_{n=0}^{\infty}\frac{(2n)!}{2^{2n}(n!)^2}\frac{(-a-b\cos(x))^{2n+1}}{2n+1} $$ which converges for $x\in [-1, 1].$
By the assumptions on $a, b$ we have that $-a-b\cos(x)\in [-1, 1] \quad \forall x\in [0, 2\pi].$
By the general binomial formula, we then have $$ (-a-b\cos(x))^{2n+1} = \sum_{k=0}^{2n+1} \begin{pmatrix} 2n+1\\k\end{pmatrix}(-a)^{2n+1-k}(-b)^k\cos(x)^k $$

Note that using the recursion formula we get $$I_k:=\int_{0}^{2\pi}\cos(x)^k dx = \frac{\sqrt{\pi } \left((-1)^k+1\right)^2 \Gamma \left(\frac{k+1}{2}\right)}{k \Gamma \left(\frac{k}{2}\right)} = \begin{cases} 0, \quad k \text{ is odd}\\ \frac{ 4\sqrt{\pi } \Gamma \left(\frac{k+1}{2}\right)}{k \Gamma \left(\frac{k}{2}\right)}, \quad k \text{ is even}\end{cases}$$

Therefore after re-indexing the sum for $k\to 2j$ and noticing that $I_{2n+1}=0$ one obtains in terms of Hypergeometric functions $$ \begin{aligned} \int_0^{2\pi}(-a-b\cos(x))^{2n+1} dx &= \sum_{j=0}^{n} \begin{pmatrix} 2n+1\\2j\end{pmatrix}(-a)^{2n+1-2j}(-b)^{2j} I_{2j} \\&= -2 \pi a^{2 n+1} \, _2F_1\left(-n-\frac{1}{2},-n;1;\frac{b^2}{a^2}\right) =: J_n(a,b) \end{aligned} $$

Finally this gives $$ \boxed{\int_0^{2\pi}\arccos(-a-b\cos x)dx = \pi^2 -\sum_{n=0}^{\infty}\frac{(2n)!}{2^{2n}(n!)^2}\frac{J_n(a,b)}{2n+1}} $$

---

Numerical test for $a=1/3, b=1/3$, the actual value is around $12.08099847999267$

and the formula $$ \pi^2 -\sum_{n=0}^{N}\frac{(2n)!}{2^{2n}(n!)^2}\frac{J_n(\frac{1}{3},\frac{1}{3})}{2n+1} $$ gives

\begin{array}{|c|c|} \hline \textbf{N} & \textbf{Value} \\ \hline 5 & 12.080\color{red}{86707985155151} \\ 10 & 12.08099\color{red}{7709047028028} \\ 100 & 12.0809984\color{red}{800892542} \\ \hline \end{array}

---

2nd Integral

Here we make use of the binomial series, so we have

$$ \sqrt{1-x^2} \cdot x = \sum_{k=0}^{\infty} \begin{pmatrix}\frac 12 \\ k\end{pmatrix} (-1)^kx^{2k+1} $$ for $|x|<1.$ Therefore we get $$ \boxed{\begin{aligned} \int_{0}^{2\pi}\sqrt{1-(-a-b\cos(x))^2}(-a-b\cos(x)) dx &= \sum_{k=0}^{\infty} \begin{pmatrix}\frac 12 \\ k\end{pmatrix} (-1)^k \int_0^{2\pi}(-a-b\cos(x))^{2k+1}dx \\ &=\sum_{k=0}^{\infty} \begin{pmatrix}\frac 12 \\ k\end{pmatrix} (-1)^k J_k(a,b) \end{aligned}} $$ where $J_k(a,b)$ has been defined before in the 1st Integral.

Answer 2

For the particular case where $a=b$.

More than inspired by @Idividedbyzero's answer,

$$\cos ^{-1}\big(-a (1+\cos (x))\big)=\frac{\pi}{2}+\sum_{n=0}^{\infty}\frac{(2n)!}{2^{2n}(n!)^2} a^{2n+1}\frac{(1+\cos(x))^{2n+1}}{2n+1}$$ Using $$\int_0^{2\pi}(1+\cos(x))^{2n+1}\,dx= 4^{n+1}\,\sqrt{\pi }\,\,\frac{\Gamma \left(2 n+\frac{3}{2}\right)}{\Gamma (2 n+2)}$$ the integral is $$\pi^2+\pi\,\sum_{n=0}^{\infty}\frac{2^{-4 n}\, \Gamma (4 n+3)}{(2 n+1)^3\, \Gamma (n+1)^2\, \Gamma (2 n+1)} a^{2n+1}$$ that is to say $$\large\color{blue}{\pi^2+2 \pi\, a \,\, _4F_3\left(\frac{1}{2},\frac{1}{2}, \frac{3}{4},\frac{5}{4};1,\frac{3}{2},\frac{3}{2};4 a^2\right)}$$ If $b_n$ is the summand $$\frac {b_{n+1}}{b_n}=a^2\,\,\frac{(2 n+1)^2 (4 n+3) (4 n+5)}{4 (n+1)^2 (2 n+3)^2}=4a^2 \Bigg(1-\frac 2n +\frac{63}{16 n^2}+O\left(\frac{1}{n^3}\right) \Bigg)$$ making the convergence quite slow as @Idividedbyzero already showed in the example. Moreover, we cannot use $a>\frac 12$ if we stay in the real domain.

Edit

Notice that $$\int (1+c \cos (x))^{2 n+1}\,dx$$ is given in terms of the Appell hypergeometric function of two variables (have a look here). This would lead to a monster.