# Integral sinc(a sin theta) knowing sinc, sinc^2 etc.

1. Dec 10, 2012

### elizevin

1. The problem statement, all variables and given/known data
Solve main lobe of radio telescope if the power diagram is given as:
P$_{n}$($\vartheta$, $\varphi$)=sinc$^{2}$(a*sin$\vartheta$)

2. Relevant equations
Ω$_{m}$=$\int$$\int$P$_{n}$ sin$\vartheta$ d$\vartheta$d$\varphi$

3. The attempt at a solution
Purely math question - but I have problem solving the integral. I can solve sinc or sinc$^{n}$ etc. Hint is to solve integral in Descartes coordinates

x=asin$\vartheta$cos$\varphi$
y=asin$\vartheta$sin$\varphi$

Does anybody have an idea how to solve this integral?

2. Dec 10, 2012

### BruceW

hey man, welcome to physicsforums! Since they gave that hint, then try to make use of it.

3. Dec 10, 2012

### elizevin

Why thank you very much, and thank you for your helpful suggestion, but I already tried that, didn't work, so I came here asking for help. I can't seem to transform integral in purely Descartes coordinates, there is always one spheric coordinate left. I need fresh idea.

Last edited: Dec 10, 2012
4. Dec 10, 2012

### haruspex

Your integrand doesn't seem to depend on φ. So the specific substitution for x and y doesn't look useful. Is that exactly the hint, or did it just suggest Cartesian co-ordinates without specifying the equations?

5. Dec 11, 2012

### elizevin

TA just told us to think about those substitutions and similar problem of sinc x *sinc y which are easy to solve (where x and y can be expressed in spheric coordinates).

6. Dec 11, 2012

### BruceW

Your integrand does not depend on phi, so what happens if you integrate over phi?

7. Dec 11, 2012

### elizevin

Phi is not a problem, I'll have 2*pi constant before integral over theta, problem is this integral which I don't know how to solve: (sinc(a*sin theta))^2

8. Dec 11, 2012

### BruceW

mm. That's a very tricky integral. Are you sure you wrote down the problem correctly? hopefully not :)

9. Dec 11, 2012

### elizevin

Unfortunately I'm sure. To make matters worse, boundaries are the first nulls of sinc, and not infinity, so even if it's perfectly normal sinc integral I would have a problem.

10. Dec 14, 2012

### BruceW

Maybe stepping back and looking at it a bit differently would be a good idea. Am I right in thinking that the power diagram expresses the power (per area) flowing radially inwards toward the telescope? and 'a' is the radius?

And the aim of the problem is to integrate the power diagram over a surface surrounding the telescope, thereby giving the total energy (per time) reaching the telescope? So your method so far was to try to integrate over a unit sphere, right? So maybe try integrating over another shape which takes advantage of the symmetry around the z axis.