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.

Is there an application for, or an industry that needs repeated calculations of similar integrals?

Let me begin by explaining this a little. I began a little project, mostly for fun and learning, to "build" integrals. Suppose we have an integral consisting of elementary functions. The idea is, we can construct another integral from knowledge of the first one, using much less effort than would otherwise be required. So, for example, if we have:

$$\int_a^b{x^2 \sin{(x+5)}dx}$$

we can quickly calculate another similar integral such as:

$$\int_a^b{(x^2+3x+5)x^2 \sin{(x+5)}dx}$$

The idea is that there is a way to quickly "attach" products (of elementary functions, for example) to existing integrals with special preparations.

I'm wondering if there is any chance that a method like this has a use in the real world.

Again, the idea is mainly a way to quickly get a similar integral.

share|improve this question
    
Specifically how do you "quickly calculate" the second integral given the first in your example? Sounds like it's just by-parts integration to me. –  anon Sep 11 '11 at 23:23
    
I probably should have added the method - I use differentiation under the integral. I have to calculate one integral first, adding a special function such as $\displaystyle e^{y \sin{x}}$ or $\sin{(y x)}$, which I can then use to extend the integral as needed. Afterwords, there are ways of eliminating the extra function. But once the special integral is determined, it seems relatively easy to calculate another similar integral. I guess this still may be just a curiousity, but I'd like to know. –  Matt Groff Sep 11 '11 at 23:28
    
A. Rich has written Rubi (rule based integrator) see his webpage where much of similar ideas have been developed. Rules are provided as plain text, as Mathematica code, and as Maple code. –  Sasha Sep 11 '11 at 23:29
add comment

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.