I am doing some analytical work that includes the integral of $e^{i(n\, t - x \sin t)}$. I know the result of this integral is a Bessel function.

$$J_n(x)=\frac1{2\pi}\int_{-\pi}^\pi e^{i(x\sin\tau-n\tau)}\mathrm d\tau$$

However, when making the calculation, Mathematica does not seem to know the identity shown above and does not use it. How can I inform Mathematica of the identity so it will use it and give me results expressed in terms of Bessel functions?

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    $\begingroup$ For specific integer values of $n$, Mathematica is certainly aware of this identity (try e.g. Integrate[Exp[I (x Sin[t] - 3 t)], {t, -π, π}]/(2 π)). In general, however, this isn't true, and the result comes out in terms of Anger and Weber functions (in Mathematica, AngerJ[] and WeberE[]). $\endgroup$ – J. M.'s torpor Nov 13 '19 at 21:53
  • $\begingroup$ Thank you very much! In my case, n can be an integer. However, it is not a given value. I tried this: Integrate[E^(I (-n t + x Sin[t])), {t, -\[Pi], \[Pi]}, Assumptions -> n \[Element] Integers]/(2 \[Pi]) that assumes n is integer, but it does not work. $\endgroup$ – yeliya201 Nov 14 '19 at 13:20
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    $\begingroup$ Yes, it seems Mathematica does not know what to do for generic n. $\endgroup$ – J. M.'s torpor Nov 15 '19 at 7:07

One method is to set the function as

J[n_, x_] := (1/(2*Pi))*Integrate[Exp[I*(x*Sin[u] - n*u)], {u, -Pi, Pi}];

As a simple use case the script

J[n_, x_] := (1/(2*Pi))*Integrate[Exp[I*(x*Sin[u] - n*u)], {u, -Pi, Pi}];
Sum[J[n, 0], {n, 0, Infinity}]

provides the result of 1, the desired result.

Since the Bessel function of the first kind is known to be useful in the calculation, based upon the description by the OP, then it would be more useful to use the series expansion, or other forms, built into Mathematica for calculations and then convert the result to the desired form.

  • $\begingroup$ Thank you very much! I think my problem is a little different. I hope Mathematica can know that the result of the integral contains a Bessel function and give the result with J[n,x]. Maybe I have not fully understood you, sorry! $\endgroup$ – yeliya201 Nov 14 '19 at 13:27

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