# How do I calculate the triple integral $\int_{0}^{2\pi}\int_0^1 \int_0^1 xy\sqrt{x^2 + y^2 -2xy\cos(\theta)} dx \text{ }dy \text{ } d\theta$?

I have tried a couple substitutions which didn't pan out, and I tried differentiating under the integral sign on $$I(\theta) = \int_0^1 \int_0^1 xy\sqrt{x^2 + y^2 -2xy\cos(\theta)} dx \text{ }dy$$ to get $$\frac{dI}{d\theta} = -\sin(\theta) \int_0^1 \int_0^1 \frac{x^2y^2}{\sqrt{x^2 + y^2 -2xy\cos(\theta)}} dx \text{ }dy$$ but nothing has worked so far. Can any of you help me?

• The computation's still going to be narly no matter what you do,but I have a suggestion and I'm working through it. – Mathemagician1234 Mar 6 '15 at 3:09
• This is a question with an astonishing answer, $\frac{64}{45}$. – Jack D'Aurizio Mar 6 '15 at 3:41
• @JackD'Aurizio: I did not see your comment before, but I am glad it agrees with my answer. – robjohn Mar 9 '15 at 1:50

By symmetry, we have: $$J=\int_{0}^{2\pi}I(\theta)\,d\theta = 2\int_{0}^{2\pi}\iint_{D}xy\sqrt{x-ye^{i\theta}}\sqrt{x-ye^{-i\theta}}\,dx\,dy\,d\theta$$ where $D$ is the region: $$D = \{(x,y)\in[0,1]^2 : y\leq x\},$$ so: $$\begin{eqnarray*} J &=& 2\int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{x}xy\sqrt{x-ye^{i\theta}}\sqrt{x-ye^{-i\theta}}\,dy\,dx\,d\theta\\&=&2\int_{0}^{2\pi}\int_{0}^{1}x^4\int_{0}^{1}t\sqrt{1-te^{i\theta}}\sqrt{1-te^{-i\theta}}\,dt\,dx\,d\theta\\&=&\frac{2}{5}\int_{0}^{1}t\int_{0}^{2\pi}\sqrt{1-te^{i\theta}}\sqrt{1-te^{-i\theta}}\,d\theta\,dt.\end{eqnarray*}$$ Since for any $|z|<1$ we have: $$\sqrt{1-z}=\sum_{n\geq 0}\binom{\frac{1}{2}}{n}(-1)^n z^n$$ it happens that: $$\int_{0}^{2\pi}\sqrt{1-te^{i\theta}}\sqrt{1-te^{-i\theta}}\,d\theta =2\pi\sum_{n\geq 0}\binom{\frac{1}{2}}{n}^2 t^{2n}$$ so: $$J = \frac{4\pi}{5}\int_{0}^{1}\sum_{n\geq 0}\binom{\frac{1}{2}}{n}^2 t^{2n+1}\,dt = \frac{4\pi}{5}\sum_{n\geq 0}\binom{\frac{1}{2}}{n}^2\frac{1}{2n+2}=\frac{4\pi}{5}\cdot\frac{16}{9\pi}=\color{red}{\frac{64}{45}}.$$ To prove the last identity, notice that: $$\sum_{n\geq 0}\binom{\frac{1}{2}}{n}^2\frac{1}{2n+2}=\sum_{n\geq 0}\frac{\frac{1}{16^n}\binom{2n}{n}^2}{(2n-1)^2(2n+2)}$$ can be computed from: $$\sum_{n\geq 0}\frac{\frac{1}{4^n}\binom{2n}{n}}{(2n-1)^2}x^{2n}=\sqrt{1-x^2}+x\arcsin x,$$ $$\sum_{n\geq 0}\frac{\frac{1}{4^n}\binom{2n}{n}}{(2n+2)}x^{2n}=\frac{1-\sqrt{1-x^2}}{x^2}$$ through a contour integral, as already guessed by the OP.

This can be tackled also in an alternative and very elegant way.

If we consider the extremely special function:

$$B(\lambda)=\sum_{n\geq 0}\left(\frac{\binom{2n}{n}}{(2n-1)4^n}\right)^2\lambda^{2n}=\frac{1}{2\pi}\int_{0}^{2\pi}\sqrt{1+\lambda^2-2\lambda\cos\theta}\,d\theta=\phantom{}_2 F_1\left(-\frac{1}{2},-\frac{1}{2};1;\lambda^2\right)$$ it is easy to check from the power series representation that $B$ satisfies the ODE: $$\lambda B = (\lambda^2+1) B' + (\lambda-\lambda^3) B''$$ but from that differential equation it is also easy to check, by integration by parts, that: $$\int_{0}^{1} \lambda\, B(\lambda)\,d\lambda = \frac{16}{9\pi}$$ since $B(1)=\frac{4}{\pi}=2\cdot B'(1)$. The previous integral is everything we need to solve our problem.

• Ok,that's bizarre. I was trying to convert the expression to spherical coordinates,but that probably would have resulted in a much more complicated integrand. – Mathemagician1234 Mar 6 '15 at 3:30
• @Mathemagician1234: I was working on the integral of the elliptic function given by integrating wrt $\theta$ first, then I noticed that by writing the square root as a series everything simplifies nicely by orthogonality. I agree, bizarre. – Jack D'Aurizio Mar 6 '15 at 3:32
• Thanks very much for this, but could you give me a hint for summing that last infinite series to get $\frac{16}{9\pi}$? from the looks of it, my only idea is to use a contour integral? Also @Mathemagician1234, I'm in my third year of university, so I understand the method, but I've dropped all the analysis courses and calculus courses, so I'm not really too keen on integration any more. This isn't homework though, don't worry! – CameronJWhitehead Mar 6 '15 at 12:20
• @Mathemagician1234 Yep, you're right, I'm from the UK, at a university where we did some complex analysis in second year :) – CameronJWhitehead Mar 6 '15 at 21:11
• Your alternative way is beautiful. – CameronJWhitehead Mar 7 '15 at 23:10