I am trying to do the following integration using Mathematica, but failed miserably.

Integrate[Exp[I q Cos[θ]]/(a - b Sin[θ]), {θ, 0, 2 π}]

I need some help or a hint.

  • 3
    $\begingroup$ Are you certain that this has an analytic form for all a, b, and q? $\endgroup$ – Jason B. Sep 6 '18 at 20:59
  • $\begingroup$ I am not very sure whether there is any analytical form but I need to do another integral after this on q, and I do not want to do numerical integration if the integral can be done analytically. Thanks. $\endgroup$ – ark Sep 7 '18 at 8:52
  Integrate[Exp[I q Cos[θ]]/(a - b Sin[θ]), {θ, 0, 2 π}]

Whenever one faces hard integral, it is good to break it down to smaller parts to try to find which is the problem part within the original integral.

\begin{align*} I & =\int\frac{e^{iq\cos\theta}}{a-b\sin\theta}d\theta\\ I & =\int\frac{\cos\left( q\cos\theta\right) +i\sin\left( q\cos \theta\right) }{a-b\sin\theta}d\theta\\ I & =\int\frac{\cos\left( q\cos\theta\right) }{a-b\sin\theta}d\theta +i\int\frac{\sin\left( q\cos\theta\right) }{a-b\sin\theta}d\theta\\ I & =I_{1}+I_{2}% \end{align*}

Looking at $I_{2}$

$$ I_{2}=\int\frac{\cos\left( q\cos\theta\right) }{a-b\sin\theta}d\theta $$

This seems not to have antiderivative, at least Mathematica can't solve it. Even this $$ I_{2}=\int\cos\left( q\cos\theta\right) d\theta $$

Mathematica can't be solved. i.e. there is no function whose derivative is $\cos\left( q\cos\theta\right) $. Even this $ \int\cos\left( \cos\theta\right) d\theta $ can't be solved.

So before trying to solve $\int\frac{e^{iq\cos\theta}}{a-b\sin\theta} d\theta\,$, one needs to try to solve $\int\cos\left( q\cos\theta\right) d\theta$.

I tried Rubi also and it can't solve $\int\cos\left(q\cos\theta\right) d\theta$ nor $\int\cos\left(\cos\theta\right) d\theta$

So I would say, your original integral has no analytical anti-derivative.

  • $\begingroup$ MMA is able to find a closed-form antiderivative for the definite integral: Integrate[Cos[q Cos[t]], {t, 0, 2*Pi}] = ConditionalExpression[2 \[Pi] BesselJ[0, Abs[q]], q \[Element] Reals]; but not for Integrate[Cos[q Cos[t]]/(a - b Sin[t]), {t, 0, 2*Pi}] $\endgroup$ – theorist Sep 8 '18 at 7:48

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