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

Question

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

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

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.)