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I have tried to evaluate the following integral for the last few hours, and I did not succeed:

$$ \int\limits_{0}^{2 \pi} e^{\mathrm{i} \cdot n \cdot\mathrm{arcsin}(r \cdot\mathrm{sin}(\theta))} \cdot e^{\mathrm{i}\cdot m \cdot \mathrm{arcsin}(r \cdot \mathrm{cos}(\theta))} d \; \theta$$

for $0<r<1$. And also this other integral:

$$ \int\limits_{0}^{2 \pi} e^{\mathrm{i} \cdot n \cdot\mathrm{arctan}(t \cdot\mathrm{sin}(\theta))} \cdot e^{\mathrm{i}\cdot m \cdot \mathrm{arctan}(t \cdot \mathrm{cos}(\theta))} d \; \theta.$$

Here $m$ and $n$ are integers, and $t \in \mathbb{R}$ is scalar.

I am pretty sure that is nonzero, if and only if $n=m$, and indepedent of $r$ otherwise, but I cannot figure what substitution makes this easy to see.

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    $\begingroup$ Seems unusual, since $\arcsin(r\sin\theta)$ is often undefined. $\endgroup$ – André Nicolas May 15 '12 at 16:07
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    $\begingroup$ In order to inverse trigonometric functions to be real for all $\theta$ in the range of integration, you should further require $0< r \leqslant 1$. $\endgroup$ – Sasha May 15 '12 at 16:08
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    $\begingroup$ It is bad to change the question this way, without leaving information about what it was before, since now my answer appears irrelevant to your question, and the time I put into answering goes wasted. Please do not do this. $\endgroup$ – Sasha May 15 '12 at 16:40
  • $\begingroup$ Dear Sasha, I did not expect an answer so fast. I posted back the original integral. $\endgroup$ – Marc Palm May 15 '12 at 16:47
  • $\begingroup$ Please, think over the question before asking it - at this moment the whole thing has changed two times! $\endgroup$ – AD. May 15 '12 at 16:47
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This addresses the question in its original form, where $\arcsin$ was used.

First, let's massage the integral: $$ \begin{eqnarray} \mathcal{I} &=& \int_0^{2 \pi} \exp\left( i n \arcsin(r \sin(\theta)) + i m \arcsin(r \cos(\theta)) \right) \mathrm{d} \theta \\ &\stackrel{\theta \to 2\pi - \theta}{=}& \int_0^{2 \pi} \exp\left( -i n \arcsin(r \sin(\theta)) + i m \arcsin(r \cos(\theta)) \right) \mathrm{d} \theta \end{eqnarray} $$ Now averaging out both lines: $$ \begin{eqnarray} \mathcal{I} &=& \int_0^{2 \pi} \cos\left(n \arcsin(r \sin(\theta)) \right)\cdot \exp\left( i m \arcsin(r \cos(\theta)) \right) \mathrm{d} \theta \\ &=& 2 \int_0^{\pi} \cos\left(n \arcsin(r \sin(\theta)) \right)\cdot \cos\left( m \arcsin(r \cos(\theta)) \right) \mathrm{d} \theta \end{eqnarray} $$

Here is a counter-example to your claim. Let $r = \frac{1}{2}$, and $n=1$, $m=2$. Then the integrand is positive, and hence the integral does not vanish:

enter image description here

Added The above counterexample actually carries over to the case with $\arctan$ just the same, i.e. the integrand is positive.

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  • $\begingroup$ I do not understand the last identity of your integerals. $\endgroup$ – Marc Palm May 15 '12 at 18:20
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    $\begingroup$ @late_learner $\int_0^{2\pi} = \int_0^{\pi} + \int_{\pi}^{2\pi}$. In the second integral change variables, $\theta = \pi + \theta$, and use $\cos(\pi+\theta) = -\cos(\theta)$. Then combined two integrals into one, and simplify the integrand. $\endgroup$ – Sasha May 15 '12 at 18:22

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