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.

Are there any well known techniques to solve a problem of the following form: $$\int_a^b f(x,\alpha) dx = g(\alpha),$$ where $a,b\in\mathbb{R}$ are fixed, $f$ and $g$ are known functions, $\alpha\in\mathbb{C}$ is the unknown variable, and the expression is not an identity. Put another way, given the above expression are there techniques available to find the values of $\alpha$ for which the expression holds true, assuming we know from empirical study that there do exist such $\alpha$ ?

share|improve this question
2  
Can this be any easier than finding all solutions to $p(x)=q(x)$ for known functions $p,q$? And can that be easier than finding solutions to $h(x)=0$ for known function $h$? –  Gerry Myerson Aug 26 '12 at 10:38

1 Answer 1

It's a question slightly strange, but under certain not-too-tight conditions, we have $$\frac{d}{d\alpha}\left(\int_a^bf(x,\alpha)dx\right)=\int_a^b\frac{d}{d\alpha}\left(f(x,\alpha)\right)dx$$ So if you know both functions you could check whether $\,g'(\alpha)\,$ equals the above...

share|improve this answer
1  
It seems to be you are trying to find the arguments where two functions agree by finding the arguments where their derivatives agree. –  Gerry Myerson Aug 26 '12 at 10:36
    
Well, there'll be equality up to an added constant...The question looks to me too general, but perhaps I'm missing something. –  DonAntonio Aug 26 '12 at 10:46
1  
We assume the expression is not an identity so it is not true for all values of $\alpha$. However, suppose it is true for some values of $\alpha$. I'm trying to find which values it is true for (in closed form, not just numerical points). So, I wondered if there were any standard techniques for doing this. –  pbs Aug 26 '12 at 12:10

Your Answer

 
discard

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.