9
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ The computation's still going to be narly no matter what you do,but I have a suggestion and I'm working through it. $\endgroup$ – Mathemagician1234 Mar 6 '15 at 3:09
  • $\begingroup$ This is a question with an astonishing answer, $\frac{64}{45}$. $\endgroup$ – Jack D'Aurizio Mar 6 '15 at 3:41
  • 1
    $\begingroup$ @JackD'Aurizio: I did not see your comment before, but I am glad it agrees with my answer. $\endgroup$ – robjohn Mar 9 '15 at 1:50
12
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Mathemagician1234 Mar 6 '15 at 3:30
  • $\begingroup$ @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. $\endgroup$ – Jack D'Aurizio Mar 6 '15 at 3:32
  • 1
    $\begingroup$ 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! $\endgroup$ – CameronJWhitehead Mar 6 '15 at 12:20
  • 1
    $\begingroup$ @Mathemagician1234 Yep, you're right, I'm from the UK, at a university where we did some complex analysis in second year :) $\endgroup$ – CameronJWhitehead Mar 6 '15 at 21:11
  • 1
    $\begingroup$ Your alternative way is beautiful. $\endgroup$ – CameronJWhitehead Mar 7 '15 at 23:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.