Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Seems unusual, since $\arcsin(r\sin\theta)$ is often undefined. – André Nicolas May 15 '12 at 16:07
In order to inverse trigonometric functions to be real for all $\theta$ in the range of integration, you should further require $0< r \leqslant 1$. – Sasha May 15 '12 at 16:08
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. – Sasha May 15 '12 at 16:40
Dear Sasha, I did not expect an answer so fast. I posted back the original integral. – May 15 '12 at 16:47
Please, think over the question before asking it - at this moment the whole thing has changed two times! – AD. May 15 '12 at 16:47
up vote 1 down vote accepted

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.

share|cite|improve this answer
I do not understand the last identity of your integerals. – May 15 '12 at 18:20
@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. – Sasha May 15 '12 at 18:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.