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I would like to ask why, or how the expression

$$J{_n}(x)=\frac{i^{-n}}{2\pi}\int_{\frac{-3\pi}{2}}^{\frac{\pi}{2}}e^{i(x cos \phi+n\phi)}d\phi $$

is the same or leads to the following:

$$J{_n}(x)=\frac{i^{-n}}{\pi}\int_{0}^{\pi}e^{i(x cos \phi)}cos{(n\phi)}d\phi $$

I have tried substitutions and all sorts of simplifications with the symmetry of cosine and sine. Please help me and include all the possible details.

Thank you very much!

Heber
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    If $f$ is periodic with period $T$, then $\int_{t_0}^{t_0+T} f(t) dt$ does not depend on $t_0$. Therefore, we can change the integration range to $[0, 2\pi]$. Next, $\cos (2 \pi - \phi) = \cos \phi$, $\sin n (2\pi - \phi) = -\sin n \phi$, $\cos n (2\pi - \phi) = \cos n \phi$. Therefore, the change of variables $\phi = 2\pi - \phi$ on $[\pi, 2\pi]$ gives the answer as an integral over $[0, \pi]$ with the contributions from $\sin n \phi$ cancelled out and the contribution from $\cos n \phi$ doubled. – Maxim Jun 21 '18 at 17:41
  • @Maxim: Thank you very much for your hint. I was able to finally see! – Heber Jun 27 '18 at 16:50

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