Transform complex exponential integral to real Question: Transform
$J_n (x)=\frac{1}{2 \pi} \int_{\theta=-\pi}^\pi e^{i(n\theta-x\sin\theta)}$
On answer sheet:
$J_n (x)=\frac{1}{\pi} \int_{\theta=0}^\pi \cos(n\theta-x\sin\theta)$
But how? I am not very good at complex analysis, thanks for any help!
 A: By definition, the real part:
$$e^{i(n\theta-x\sin\theta)}=\cos(n\theta-x\sin\theta)+i\sin(n\theta-x\sin\theta)\implies$$
$$\implies \text{Re}\,e^{i(n\theta-x\sin\theta)}=\cos(n\theta-x\sin\theta)$$
and as this last one is an even function, we get
$$\frac1{2\pi}\int_{-\pi}^\pi\cos(n\theta-x\sin\theta)=2\left(\frac1{2\pi}\int_0^\pi\cos(n\theta-x\sin\theta)\right)=\frac1{\pi}\int_0^\pi\cos(n\theta-x\sin\theta)$$
A: The imaginary part is $$\dfrac{1}{2\pi} \int_{-\pi}^\pi \sin(n\theta - x \sin\theta)\; d\theta$$ but that is $0$ because the integrand is an odd function.
The real part is  $$\dfrac{1}{2\pi} \int_{-\pi}^\pi \cos(n\theta - x \sin\theta)\; d\theta$$
Here the integrand is even, so you can write this as $\dfrac{1}{\pi} \int_0^\pi \ldots$.
A: $$e^{i(n\theta-x\sin{\theta})} = \cos(n\theta-x\sin{\theta}) +i \sin(n\theta-x\sin{\theta})$$
Let,
$$f(\theta) = \cos(n\theta-x\sin{\theta}) +i \sin(n\theta-x\sin{\theta})$$
Therefore,
$$f(-\theta) = \cos(-n\theta+x\sin{\theta}) +i \sin(-n\theta+x\sin{\theta})$$
$$\implies f(-\theta) = \cos(n\theta-x\sin{\theta}) -i \sin(n\theta-x\sin{\theta})$$
Let,
$$f(\theta) = a(\theta) + ib(\theta)$$
Therefore,
$$f(-\theta) = a(\theta) - ib(\theta)$$
Hence we can say that $a(\theta)$ is an even function as $a(\theta) = a(-\theta)$ and $b(\theta)$ is an odd function since $b(-\theta) = -b(\theta)$
Hence integral of an odd function over a symmetric domain is zero whereas for an even function we can thus write,
$$J_n(\theta) = 2(\frac{1}{2\pi}\int_0^{\pi}\cos(n\theta-x\sin{\theta}) d\theta)$$
$$\therefore J_n(\theta) = \frac{1}{\pi}\int_0^{\pi}\cos(n\theta-x\sin{\theta}) d\theta$$
A: $e^{i t}=\cos t +i\sin t$ . So $$(2 \pi)^{-1}\int_{-\pi}^{\pi}\exp (i(n \theta-x\sin \theta))\;d\theta=(2 \pi)^{-1}\int_{-\pi}^{\pi}(f(y)+i g(y))dy$$ $$ \text {where }\quad  f(y)=\cos (y-x\sin y)$$ $$ \quad \text {and }\quad  g(y)=\sin (y-x\sin y).$$ $$\text {Observe that }\quad  f(-y)=f(y) \quad \text {and }\; g(-y)=-g(y)\quad  \text {for all }\; y,$$ $$\text {Therefore }\quad  \int_{-\pi}^{\pi}i g(y)\;dy=0$$ $$\text { and }\; \int_{-\pi}^{\pi}f(y)\;dy=2\int_0^{\pi} f(y)\;dy.$$
