Can this integral be written in terms of Bessel functions?

Question

I have seen in the literature that $$J_n(z)=\frac{1}{\pi i^n}\int_0^{\pi}e^{i z \cos{\theta}}\cos(n\theta)d\theta.$$ I have to deal with the following expression $$f_n(z)=\int_0^{\pi}e^{i z \cos{\theta}}\sin(n\theta)d\theta.$$ My question is: can the function $f_n(z)$ be written in terms of Bessel functions or modified Bessel functions?

Answer 1

Bessel function not needed. The indefinite integral $$ W_n(z) = \int e^{i z \cos\theta}\sin(n\theta)\;d\theta $$ can be done for any fixed positive integer $n$.

$$ W_1(z) = \frac{2\sin z}{z} \\ W_2(z) = \frac{-4 i \cos z}{z} + \frac{4 i \sin z}{z^2} \\ W_3(z) = \frac{6 \sin z}{z} + \frac{16\cos z}{z^2} + \frac{-16\sin z}{z^3} $$

It is interesting that the cosine integral is not elementary, while the sine integral is!

Explanation...
Expansion of $\sin(n\theta)$ using the Chebyshev polynomial $U_n$ is: $$ \sin(n\theta) = \sin \theta \;U_{n-1}(\cos\theta) $$ Then change variable $x=\cos\theta$ to get $$ \int e^{i z \cos\theta}\sin(n\theta)\;d\theta = - \int e^{i z x} U_{n-1}(x)\;dx \\ \int_0^\pi e^{i z \cos\theta}\sin(n\theta)\;d\theta = - \int_{-1}^1 e^{i z x} U_{n-1}(x)\;dx $$ and $U_{n-1}(x)$ is a polynomial of degree $n-1$.