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If I have to integrate a function in the region of the unit circle...Is there a way without even evaluating the integral I might get the result using the symmetry of a unit circle.

For example, if I'm integrating the following function over a unit circle. $$\int\int_R e^x dxdy$$ Can I solve without really solving the integral or get to a point where the integral might be not as difficult as this?

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  • $\begingroup$ For that one not really. Sometimes you could, though. Such as a function whose integral around each circle is just zero. $\endgroup$
    – Ian
    Sep 21, 2020 at 22:28
  • $\begingroup$ How could I evaluate something like $\int\int_R x^2 ydxdy$ without really integrating the integral where R is the same unit circle centered at the origin? $\endgroup$
    – Orpheus
    Sep 22, 2020 at 0:03
  • $\begingroup$ That one has odd symmetry through the $x$ axis, so it is zero. $\endgroup$
    – Ian
    Sep 22, 2020 at 0:05
  • $\begingroup$ so it depends on if the function has odd or even symmetry?.... So if it is odd about the x-axis(because in $dxdy$ we integrate with respect to x first and then with respect to y) then it is zero else if it is even it is not zero?... Is that right? $\endgroup$
    – Orpheus
    Sep 22, 2020 at 1:32
  • $\begingroup$ If the integrand is odd about any axis and you integrate over a region symmetric about that axis then you get zero. Besides that, usually all you can do with symmetry is sometimes chopping up the domain into several regions on which the integrals are equal and thus reducing the problem to integrating over just one of them. $\endgroup$
    – Ian
    Sep 22, 2020 at 1:44

1 Answer 1

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Using polar coordinates, the integral of $x^n$ over the unit circle is given by $$\int_0^{2\pi}\int_0^1 (r\cos\theta)^{n}r\,dr\,d\theta=\frac{1}{n+2}\int_0^{2\pi}\cos^{n}\theta\,d\theta=2\pi\frac{(2n-1)!!}{(2n)!!(n+2)}$$ for even $n$, and $0$ otherwise (using a table of definite integrals or establishing a recurrence relation by parts).

Hence the integral of $e^x$ yields

$$2\pi\sum_{k=0}^\infty\frac{(4k-1)!!}{(4k)!!(2k+2)(2k)!}$$

and this will work for other functions with a known Taylor development.

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