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I'm having big trouble finding the complex Fourier series coefficient of the following periodic function

$$\frac{a-b\cos\varphi}{\sqrt{a^2+b^2-2ab\cos\varphi}}$$

Mathematica is unable to compute it!!

Minas Michael Opethian
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    Even if the question gets closed here and you decide to post it elsewhere, it might help to format it properly. I have edited the formula from the linked image into your post. You can find pointers to some information about typing mathematical expressions on Stack Exchange sites here: How does one type mathematical formulas on this site? – Martin Sleziak Aug 15 '19 at 08:08
  • Thank you, it won't happen again! – Minas Michael Opethian Aug 15 '19 at 08:09
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    To start with: One constant is enough: use c:=a/b. – Pietro Majer Aug 15 '19 at 10:17
  • for $a=b$ the Fourier coefficients are $f_n=(2/\pi)(1-4n^2)^{-1}$. – Carlo Beenakker Aug 15 '19 at 10:22
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    Have a look at https://en.wikipedia.org/wiki/Legendre_polynomials and in particular the § "Applications of Legendre polynomials". The square root is expanded in terms of Legendre polynomials in $ \cos(\varphi) $. For the "pure" Fourier coefficient, though, I think you may have to use the link between Legendre and Chebychev polynomials (also on wikipedia)... – Synia Aug 15 '19 at 10:27
  • for $a\neq b$ there are no closed form expressions in terms of elementary functions; for example, for $a=2b$ one has $f_1=(12\pi)^{-1}\left[60 K(-8)-11 K\left(\frac{8}{9}\right)-12 E(-8)+15 E\left(\frac{8}{9}\right)\right]$, with $E$ and $K$ elliptic functions. – Carlo Beenakker Aug 15 '19 at 10:34

2 Answers2

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You might combine the Legendre expansion (wlog $a>b$) $$ \frac1{\sqrt{a^2+b^2-2ab\cos\varphi}}=\frac1a\sum_{n=0}^\infty\left(\frac ba\right)^nP_n(\cos\varphi) $$ with the Fourier expansion (Gradshteyn-Ryzhik 8.826) $$ P_n(\cos\varphi)=\\\frac{2^{n+2}n!}{\pi(2n+1)!!}\left(\sin(n+1)\varphi+\frac11\frac{n+1}{2n+3}\sin(n+3)\varphi+\frac11\frac32\frac{n+1}{2n+3}\frac{n+2}{2n+5}\sin(n+5)\varphi+\dots\right) $$ and product-to-sum formulas when multiplying by $a-b\cos\varphi$.

  • Thank you for your reply! What do you mean with "and product-to-sum formulas when multiplying by a−bcosφ"? – Minas Michael Opethian Aug 15 '19 at 11:25
  • @MinasMichaelOpethian Put another way: multiplying the Fourier series for $(a^2+b^2-2ab\cos\varphi)^{-1/2}$ by $a-b\cos\varphi$ (a finite Fourier series) amounts to convolving coefficients. –  Aug 15 '19 at 11:35
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$$\frac{1}{\pi} \int_0^{\pi} \frac{a-b\cos\varphi}{\sqrt{a^2+b^2-2ab\cos\varphi}} \cos n \varphi\, d\varphi = \, _3\tilde{F}_2\left(\tfrac{1}{2},\tfrac{1}{2},1;1-n,n+1;\tfrac{4 a b}{(a+b)^2}\right)$$ $$\qquad\qquad\qquad\qquad-\frac{ b}{a+b} \, _3\tilde{F}_2\left(\tfrac{1}{2},\tfrac{3}{2},2;2-n,n+2;\tfrac{4 a b}{(a+b)^2}\right),$$ with $\tilde{F}$ defined here.

Carlo Beenakker
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