How to evaluate $\int\limits^{2\pi}_{0} \frac{\cos\left(u\right) - r}{\left(1 + r^{2} - 2r \cos\left(u\right)\right)^{\frac{3}{2}}} du$

Question

I am looking for a practical solution for $0<r<1$

$$\int\limits^{2\pi}_{0} \frac{\cos\left(u\right) - r}{\left(1 + r^{2} - 2r \cos\left(u\right)\right)^{\frac{3}{2}}} du$$

My attempt:

$y = 1 + r^2 - 2r \cos(u)$

$du=\frac{dy}{2r\sin(u)}$ with this integral becomes

$$\frac{1}{2r}\int\limits^{(r-1)^2}_{(r-1)^2}\frac{\cos(u) - r}{y^{3/2}\sin(u)}dy=0$$ but WolframAlpha finds diffrently (for example it finds $0.003141596...$ for $r=0.001$). I am not knowledgeable in Complex Analysis or Elliptic Integrals, so an elementary solution would be greatly appreciated, if possible. Thank you in advance.

Answer 1

You can try in the following lines: $$\int_0^{2\pi}\frac{\cos(u)-r}{(1+r^2-2r\cos(u))^{\frac{3}{2}}}du=\frac{1}{2r}\int_0^{2\pi}\frac{2r\cos(u)-2r^2-1+1}{(1+r^2-2r\cos(u))^{\frac{3}{2}}}du$$

then the right hand side is just evaluating $$\frac{1}{2r}\int_0^{2\pi}\frac{1-r^2}{(1+r^2-2r\cos(u))^{\frac{3}{2}}}du-\frac{1}{2r}\int_0^{2\pi}(1+r^2-2r\cos(u))^{-\frac{1}{2}}du$$

Now substitute $y=(1+r^2-2r\cos(u))^{\frac{1}{2}}$. This will yield $$\frac{dy}{r\sin(u)}=(1+r^2-2r\cos(u))^{-\frac{1}{2}}du$$

Replace the value of $r\sin(u)$ from the substitution equation in terms of $y$ and change the limits of the integration accordingly. I think you will be good to go.

Answer 2

Let $\rho=\dfrac{2\sqrt r}{1+r}$. By $(1)$ symmetry and $(2)$ substitution of $u\to\dfrac{\pi-u}2$, we have

$$\begin{align*} I(r) &= \int_0^{2\pi} \frac{\cos u - r}{\left(1+r^2-2r\cos u\right)^{3/2}} \, du \\ &\stackrel{(1)}= \frac2{\left(1+r^2\right)^{3/2}} \int_0^\pi \frac{\cos u-r}{\left(1-\frac{\rho^2}2\cos u\right)^{3/2}} \, du \\ &\stackrel{(2)}= -\frac4{\left(1+r^2\right)^{3/2}} \int_0^\tfrac\pi2 \frac{\cos(2u)+r}{\left(1+\frac{\rho^2}2\cos(2u)\right)^{3/2}} \, du \\ &= \frac8{(1+r)^3}\int_0^\tfrac\pi2\frac{\sin^2u}{\left(1-\rho^2\sin^2u\right)^{3/2}} \, du-\frac4{(1+r)^2}\int_0^\tfrac\pi2\frac{du}{\left(1-\rho^2\sin^2u\right)^{3/2}} \\ &\!\!\!\stackrel{(3),(4)}= \frac8{(1-r)^2(1+r)} \underbrace{\int_0^\tfrac\pi2 \frac{1-\sin^2u}{\sqrt{1-\rho^2\sin^2u}} \, du}_{I_1} -\frac4{(1-r)^2}\underbrace{\int_0^\tfrac\pi2\sqrt{1-\rho^2\sin^2u}\,du}_{I_2} \end{align*}$$

The reductions to $I_1$ and $I_2$ respectively follow from $(3)$ and $(4)$. $I_1$ can be further broken down into a linear combination of $E(\rho)$ and $K(\rho)$.

We ultimately find that

$$I(r)=\frac2{r(1-r)}E\left(\frac{2\sqrt r}{1+r}\right)-\frac2{r(1+r)}K\left(\frac{2\sqrt r}{1+r}\right)$$

or equivalently, applying identities listed here,

$$\begin{align*} K(k)&\stackrel{(127)}=\frac1{1+k}K\left(\frac{2\sqrt k}{1+k}\right) \\ E(k)&\stackrel{(139)}=\left(1+\sqrt{1-k^2}\right)E\left(\frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}}\right)-\sqrt{1-k^2}K(k) \end{align*}$$

$$\implies I(r) = \frac4{r\left(1-r^2\right)} E(r) - \frac4r K(r)$$

Answer 3

Too long for comments.

@user170231 gave the exact solution.

If, by practical, you mean that you wish a shortcut method for the numerical evaluation of $$I(r)= \frac4{r\left(1-r^2\right)} E(r) - \frac4r K(r)$$ you could build the series expansion around $r=0$ and have

$$I(r)=\frac{\pi r}{ \left(1-r^2\right)}\sum_{n=0}^\infty \frac {a_n}{b_n} r^{2n} $$ where the $b_n$ correspond to sequence $A278145$ in $OEIS$ and the first $a_n$ are $$\{1,1,3,25,245,1323,7623,184041,4601025,29548805,193947611,2591845347,\cdots \}$$

The summation will converge quite fast since, if $$c_n= \frac {a_n}{b_n}\qquad \implies \qquad \frac {c_{n+1}}{c_n} \sim 1-\frac 2 n$$