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How can I obtain $$I_n(x)= \frac{1}{2\pi}\int_{0}^{2\pi}e^{in\theta}e^{x \cos\theta}\,d\theta ,$$ from the integral $$J_n (x) =\frac{1}{2\pi} \int_{0}^{2\pi}e^{-in\theta}e^{ix \sin\theta}\,d\theta .$$

I started with $I_n(x) = i^{-n} J_n(ix)$, but when I change the variable accordingly in the integral of $J_n$ above, I cannot get something that resembles the final result. I have tried manipulating the trigonometric identities in several ways, but couldn't get it, either.

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1 Answer 1

up vote 2 down vote accepted

Hint: try the substitution $$\theta \mapsto -\pi/2- \theta.$$

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