I am trying to solve the following: $$\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\sin(a\cos\phi)}{1+b\cos\phi}$$ with $-1 < b < 0$.

I started with $\cos\phi = \operatorname{Re}[z]$, but it led to nowhere as I had to find the residue at 0, which doesn't have a closed form.

  • $\begingroup$ Is there any reason to think that this has a closed form? $\endgroup$ – Gerald Edgar Jan 9 '18 at 22:48
  • $\begingroup$ a power series in $a$ has a closed form expression, is that helpful? $\endgroup$ – Carlo Beenakker Jan 9 '18 at 22:53
  • $\begingroup$ I thought of expanding it as a power series in $a$, but then I cannot reconcile the terms after I integrate them as a closed form. $\endgroup$ – John Jan 9 '18 at 22:57

Mathematica finds

Integrate[Sin[a*Cos[t]]*b^n*Cos[t]^n,{t, 0, 2*Pi},Assumptions->n \[Element] Integers&&n>0]

$$-\frac{1}{2} \pi a \left((-1)^n-1\right) b^n \Gamma \left(\frac{n}{2}+1\right) \, _1\tilde{F}_2\left(\frac{n+2}{2};\frac{3}{2},\frac{n+3}{2};-\frac{a^2}{4}\right) $$


Integrate[Cos[t]^n/(1 + b*Cos[t]), {t, 0, 2*Pi}, 
 Assumptions -> n \[Element] Integers && n >= 0 && b > -1 && b < 0]

$$-\left(2 \sqrt{\pi } (-i b)^{-n} \left(\cos \left(\frac{\pi n}{2}\right)+i \sin \left(\frac{\pi n}{2}\right)\right) \left((-i b)^n \cos \left(\frac{\pi n}{2}\right) \Gamma \left(\frac{n+1}{2}\right) \Gamma \left(\frac{n-1}{2}\right) \, _2F_1\left(\frac{1}{2},1;\frac{3-n}{2};\frac{1}{1-b^2}\right)+i b (-i b)^n \sin \left(\frac{\pi n}{2}\right) \Gamma \left(\frac{n}{2}\right)^2 \, _2F_1\left(\frac{1}{2},1;1-\frac{n}{2};\frac{1}{1-b^2}\right)+2 \sqrt{\pi -\pi b^2} \Gamma \left(\frac{n}{2}\right) \Gamma \left(\frac{n+1}{2}\right)\right)\right. $$

$$\left(\left(b^2-1\right) \Gamma \left(\frac{n}{2}\right) \Gamma \left(\frac{n+1}{2}\right)\right)^{-1}$$

Therefore, the integral under consideration can be expressed as a series.

  • $\begingroup$ Thank you. Do you think there is a closed form, though? $\endgroup$ – John Jan 10 '18 at 13:04
  • $\begingroup$ @ John: A series is not a closed-form expression. Mathematica fails with the integral under consideration even for $b= -\frac 1 2 $. $\endgroup$ – user64494 Jan 10 '18 at 13:20

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