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August 25

Comparison of two integrals.

Hi there,

I want to know how two integrals given below are related. Let's assume m is a natural number, perhaps m = 40; then n = 360/m and n = 9; Function f(φ) is smooth, differentiable in its domain. The domain is a small fraction of circumference: {0,360/m} or {0,n}. Then the integral (1) will have the integrand as product of the function f(φ) and one whole period of cosine function cos(mφ) where function cos(mφ) is in fact defined on the whole circumference:

$\int \limits _{0}^{n}$ f(φ)cos(mφ) dφ (1)

Then I "stretch" the function and its domain m times to the extent that it will coincide with the whole circumference. The cosine function or rather the portion thereof defined initially on {0,n} interval will be stretched as well. The amplitude of the function f(φ) will be preserved, the same can be said of the cosine function. Then I will take the integral:

$\int \limits _{0}^{360}$ f(ω)cos(ω) d(ω) (2)

I want to know how the integrals (1) and (2) are related. My hunch is that I will have to multiply the integral (1) by m to get integral (2) but not sure about it. Not a homework :-) It is a practical problem I am facing in some computations I need to perform. Will appreciate any insight. Thank you. --AboutFace 22 (talk) 20:56, 25 August 2014 (UTC)[reply]

This is a simple change of variables (except for the "rescaling of f"). If one takes u = mx, du = m dx,

\int _{0}^{n}f(x)\cos(mx)dx={\frac {1}{m}}\int _{0}^{360}f({\frac {u}{m}})\cos(u)du

. I'm not sure about what you mean by "stretch" here, but your hunch is rather close, except that I think your second integral is not exactly what you're talking about (you shouldn't use the same letter f to denote two different functions - you should write it as

\int _{0}^{360}g(u)\cos(u)du

with g(u)=f(u/m)).--Jasper Deng (talk) 00:18, 27 August 2014 (UTC)[reply]

Thank you much for your note. You are correct. I should have changed the letter for the function. Actually I have been thinking about doing this for two days but somehow other things intevened and I missed the opportunity. Sure it is a different function after the "dilation." So, could you say with certainty that the two integrals are related in such a way that I can divide the value of the second integral by m to get the value of the first integral? It seems the formula you wrote says this clearly, right? Your "rather close" makes me a little bit nervous, though :-) --AboutFace 22 (talk) 13:28, 27 August 2014 (UTC)[reply]

@AboutFace 22: I know the expression in my previous comment to be correct. The reason why I say "rather close" is that you have not made "dilation" well defined; if you mean it as I did above, then that is correct. I think you know this, but m need not be a natural number. The equality is valid for all real nonzero m.--Jasper Deng (talk) 19:02, 27 August 2014 (UTC)[reply]

Mr. Deng, thank you. I know that my definitions are lame, I also know that any real non-zero number could do the same job as the whole number m. In my application though, m is a natural number, that is why I stated that. I am completely satisfied now. I appreciate your help very much. --AboutFace 22 (talk) 01:06, 28 August 2014 (UTC)[reply]