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Hello everyone! I need to calculate the following integral:

$\int\int_{\sqrt{x^2+y^2}\leq a}\cos^2(v_x x+v_y y)dxdy$

First step, I convert to polar coordinates:

$\int_0^a\int_0^{2\pi}\cos^2(v_x \rho \cos\theta+v_y \rho\sin\theta)\rho d \rho d\theta$

Then...what? Any suggestions?

Mathematica manages to find a closed form solution, but it is very ugly (4-5 terms involving Hypergeometric Regularised functions and Struve functions). I am wondering if a more elegant solution exists, given the apparent simplicity of the integral.

Notice that I can also use $|\cos(v_x x+v_y y)|$ instead of $\cos^2(v_x x+v_y y)$, but my guess is that it will only get more complicated. If needed, I can also change the domain of integration, as long as the integral does not change with rotation of the vector $[v_x,v_y]$ (a condition that is satisfied by my integral above).

Cheers!

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2 Answers 2

I am not sure that it is a proper question, but one has to excercise in integration once in a while, so:

First of all, choose coordinates so that vector $v$ is directed along $x$-axis. Then define $\Omega\subset\mathbb{R}^2$ to be the disc $||r||< a$, and function $f(r)=\cos(2v\cdot r)$. $$ I=\int_\Omega \cos^2(v\cdot r)d^2r=\frac{\pi a^2}{2}+\frac{1}{2}\int_\Omega f(r)d^2r $$ Notice that $4v^2f=-\Delta f$, so by some divergence theorem, $$ \int_\Omega f(r)d^2r=-\frac{1}{4v^2}\int_{0}^{2\pi}a\frac{\partial f}{\partial \rho}d\varphi=\frac{a}{2v}\int_{0}^{2\pi}\cos\varphi\sin (\lambda\cos\varphi)d\varphi $$ for $\lambda=2av$. The last integral is know to be $2\pi J_1(\lambda)$, so, as noted by Michael, we have: $$ I=\frac{\pi a^2}{2}\left(1+\frac{J_1(2av)}{av} \right) $$

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I used Mathematica, doing first the theta integration. The hypergeometric function that arises is actually a Bessel function, but Mathematica does not find that on its own. The final result I get is $${\pi a^2\over 2}(1+{J_1(2a\sqrt{v_x^2+v_y^2})\over a\sqrt{v_x^2+v_y^2}}).$$

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