Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is the equation:

\begin{eqnarray*} f\left(e\right) & = & -r\left(e\right)\left(1+\int_{E}\cos\left(s\left(e\right)-s\left(e'\right)\right)de'-\int_{I}\cos\left(s\left(e\right)-s\left(i\right)\right)di\right) \end{eqnarray*}

All functions are one-to-one and range between $-1$ and $1$. The functions $r$ and $s$ evaluate to constants not to other functions. However, they are not linear but more like a hash. The subscripts on the integrals denote the domains.

Edit

I began by expanding the integrals to isolate terms. For example, what follows is an expansion of the middle term of the right-hand side.

\begin{eqnarray*} \int_{E}\cos\left(s\left(e\right)-s\left(e'\right)\right)de'=\cos\left(s\left(e\right)\right)\int_{E}\cos\left(s\left(e'\right)\right)de'-\sin\left(e\right)\int_{E}\sin\left(s\left(e'\right)\right)de' \end{eqnarray*}

How can I evaluate these integrals?

share|improve this question
    
Where is the differential? What is the unknown? What do you call functions which evaluate to constants? Etc. –  Did Feb 20 '13 at 20:54
    
What are you trying to do? What does "functions evaluate to constants" mean? What are $E$ and $I$? How is this a functional integral if $r,s$ are functions but the parameters being integrated over, $e'$ and $i$, are presumably numbers? Perhaps you're doing a functional integral, and the quantity being integrated is the evaluation of varying functions at given arguments? –  anon Feb 20 '13 at 20:56
    
The differentials are $de'$ and $di$. The variables over which to integrate are $e'$ and $i$. By "functions which evaluate to constants", I meant a univariate function I guess. (Like the computer science idea of evaluating to a literal instead of another function.) –  mac389 Feb 20 '13 at 20:57
    
$f$ depends only on $e$. What are the relations b/w $f$ and $s$ besides of the equation you wrote? –  Ilya Feb 20 '13 at 21:03
1  
Precisely: $\int\limits_I\cos(s(e)-s(i))di=\int\limits_I\cos(s(e)-s(w))dw=\int\limits_I\cos‌​(s(e)-s(\alpha))d\alpha=...$ –  Did Feb 20 '13 at 21:37

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.