Integrate $\int_0^1\int_0^\pi \frac{u}{\sqrt{1+u^2-2u \cos\phi}} d\phi du$

Question

I was trying to solve this integral while solving an electromagnetics problem in physics.

$$\int_0^1\int_0^\pi \frac{u}{\sqrt{1+u^2-2u \cos\phi}} d\phi \ du$$

My approach

My idea was to first make this integral into an single-variable one by solving the inner integral. $$\int_0^\pi \frac{u}{\sqrt{1+u^2-2u \cos\phi}} d\phi\\ =\frac{u}{1+u}\int_0^\pi\frac{1}{\sqrt{1-\frac{2u}{(1+u)^2}(1+\cos\phi)}}d\phi\\ =\frac{u}{1+u}\int_0^\pi\frac{1}{\sqrt{1-\frac{4u}{(1+u)^2}\cos^2\frac\phi2}}d\phi\\ =\frac{2u}{1+u}\int_0^\frac\pi2\frac{1}{\sqrt{1-\frac{4u}{(1+u)^2}\sin^2\frac\phi2}}d\left(\frac\phi2\right)\\ =\frac{2u}{1+u}K\left(\frac{2\sqrt u}{1+u}\right)$$ ($K$ is the complete elliptic integral of the first kind)

I managed to make the integral into an single-variable one, but the elliptic integral showed up and I couldn't proceed. $$=\int_0^1\frac{2u}{1+u}K\left(\frac{2\sqrt u}{1+u}\right)du$$ I tried to use partial integration or substitution as the Weierstrass substitution, but it didn't solved the question. According to Wolfram Alpha, the value of the integral is 2.00000... approximately. Therefore I thought there would be some way to calculate the value of the integral without numeric methods.

Any help is appreciated. Thanks for reading!

Answer 1

With $K$ as the Complete Elliptic Integral of the First Kind with Paramater $k$, $$K:=K(k)=\int_0^{\pi/2}\frac{1}{\sqrt{1-k^2\sin^2t}}dt$$

Also using Landen's Transformation:

$$K\left(\frac{2\sqrt{x}}{1+x}\right)=(1+x)K(x)$$

Here is a Collection of Proofs for the Transformation.

This transformation is also related to a larger family of Hypergeometric Transformations (Quadratic Transformation of Gauss).

The Integral in question is then reduced to:

$$I=2\int_0^1kKdk$$

$$=-2\int_0^{\pi/2}\csc^2(t)\left[\int_0^1\frac{d}{dk}\sqrt{1-k^2\sin^2t}\ dk\right]dt$$ $$=-2\int_0^{\pi/2}\frac{\cos t-1}{\sin^2 t}dt$$

Therefore, $$I=2$$

Answer 2

Thanks to @john_barber for the idea. noting $$\frac{u}{\sqrt{1+u^2-2u\cos\phi}}= \sum_{k=0}^\infty u^{k+1} P_k(\cos\phi)$$ where : $$P_k(x) = 2^k \sum_{i = 0}^k x^i {k\choose i}{\frac{k+i-1}{2}\choose k} $$

thus for the final expression we have : $$ \frac{u}{\sqrt{1+u^2-2u\cos\phi}}= \sum_{k=0}^\infty \sum_{i = 0}^k u^{k+1} 2^k \cos^i\phi {k\choose i}{\frac{k+i-1}{2}\choose k} $$ knowing that the said sum always converges we can do the following operation : $$ \int_0^\pi \frac{u}{\sqrt{1+u^2-2u\cos\phi}} d\phi = \int_0^\pi \sum_{k=0}^\infty \sum_{i = 0}^k u^{k+1} 2^k \cos^i\phi {k\choose i}{\frac{k+i-1}{2}\choose k}d\phi = $$ $$\sum_{k=0}^\infty \sum_{i = 0}^k u^{k+1} 2^k {k\choose i}{\frac{k+i-1}{2}\choose k}\int_0^\pi\cos^i\phi d\phi $$ now using the reduction formula : $$ \int_0^\pi\cos^i\phi d\phi = \frac{1}{i}\cos^{n-1}\phi \sin{\phi}|_0^\pi + \frac{i-1}{i}\int_0^\pi \cos^{i-2}\phi d\phi = \frac{i-1}{i}\int_0^\pi \cos^{i-2}\phi d\phi = $$$$ \frac{i-1}{i} \frac{i-3}{i-2}\frac{i-5}{i-4}...\frac{\pi}{2} = \frac{(i-1)!!}{i!!}\pi $$ note that for odd i, the integral vanishes. now we are solving : $$ \int_0^1\int_0^\pi \frac{u}{\sqrt{1+u^2-2u\cos\phi}} d\phi du = \int_0^1\int_0^\pi \sum_{k=0}^\infty \sum_{i = 0}^k u^{k+1} 2^k \cos^i\phi {k\choose i}{\frac{k+i-1}{2}\choose k}d\phi du= $$ $$\int_0^1\sum_{k=0}^\infty \sum_{i = 0}^k u^{k+1} 2^k {k\choose i}{\frac{k+i-1}{2}\choose k}\int_0^\pi\cos^i\phi d\phi du = $$ $$\int_0^1\sum_{k=0}^\infty \sum_{i = 0, 2|i}^k u^{k+1} 2^k {k\choose i}{\frac{k+i-1}{2}\choose k}\frac{(i-1)!!}{i!!}\pi du = \pi\sum_{k=0}^\infty \sum_{i = 0, 2|i}^k \frac{2^k}{k+2}{k\choose i}{\frac{k+i-1}{2}\choose k}\frac{(i-1)!!}{i!!} = 2$$

of course further simplification is possible and is left as an exercise to the reader. (comment if in need of help.)

Answer 3

Consider a triangle with side lengths $1$, $u$ and $ \sqrt{1-2u \cos\phi+u^2} $, respectively, and let $\theta$ be the angle opposite of the side $u$. The sine rule states that
$$\frac{\sin\phi}{\sqrt{1-2u \cos\phi+u^2} } = \frac{\sin\theta}{u}=\frac{\sin(\phi+\theta)}1$$ Then, change the variable from $u$ to $\theta$ via $u=\frac{\sin\theta}{\sin(\phi+\theta)}$ along with $du =\frac{\sin\phi}{\sin^2(\phi+\theta)}d\theta$, which simplifies the double integral considerably \begin{align} &\int_0^1\int_0^\pi \frac{u}{\sqrt{1+u^2-2u \cos\phi}} d\phi \ du\\ =& \int_0^\pi\int_0^{\frac{\pi-\phi}2}\frac{\sin\theta}{\sin^2(\phi+\theta)}d\theta\ d\phi\overset{\theta \to \frac{\pi-\theta}2} =\int_0^\pi\int_{\phi}^\pi \frac{\cos\frac{\theta}2}{2\cos^2(\frac{\theta}2-\phi)}d\theta\ d\phi\\ = &\int_0^\pi\int_0^{\theta} \frac{\cos\frac{\theta}2}{2\cos^2(\frac{\theta}2-\phi)}d\phi\ d\theta =\int_0^\pi\sin\frac{\theta}2\ d\theta=2 \end{align}

Answer 4

Denote $(c,s)=(\cos\phi,\sin\phi)$. Integrate w.r.t. $u$ first by completing the square and $(*)$ substituting $u=c+sv$ :

$$\newcommand{arsinh}{\operatorname{arsinh}} \begin{align*} I &= \int_0^\pi \int_0^1 \frac{u}{\sqrt{(u-c)^2+s^2}}\,du\,d\phi \\ &= \int_0^\pi \left[\int_{-\tfrac cs}^{\tfrac{1-c}s} \frac{c}{\sqrt{1+v^2}} \, dv + J(\phi)\right] \, d\phi & (*) \\ &= \int_0^\pi \left(c \left(\arsinh\frac{1-c}s-\arsinh\frac cs\right) + \frac{\sin\phi}{\cos\frac\phi2} - 1\right) \, d\phi \\ &= \int_0^\pi \left(\cos\phi[\color{blue}{\arsinh(\csc\phi-\cot\phi)} + \color{red}{\arsinh(\cot\phi)}] + \frac{\sin\phi}{\cos\frac\phi2} - 1\right) \, d\phi \\ &= \int_0^\pi \left(-\frac12 \color{blue}{\frac{\sin\phi}{\cos\frac\phi2}} + \color{red}{1} + \frac{\sin\phi}{\cos\frac\phi2} - 1\right) \, d\phi & \text{by parts} \\ &= 2 \int_0^\tfrac\pi2 \sin\phi \, d\phi = \boxed{2} & \phi\to2\phi \end{align*}$$

where

$$\begin{align*} J(\phi) &= \int_{-\tfrac cs}^\tfrac{1-c}s \frac{sv}{\sqrt{1+v^2}}\,dv \\ &= \sqrt{s^2+(1-c)^2} - \sqrt{s^2+c^2} \\ &= \sqrt2\sqrt{1-c}-1 \\ &= \frac{s}{\sqrt{\frac{1+c}2}} - 1 \end{align*}$$