I have had a really bad cold for the last few days and I feel like my head is full of molasses. Here is what I have.
There is a function, L(x), that only has value (i.e. is non-zero) between x=a and x=d. This function is dependent on a bunch of different unknown things (temperature, pressure, wind, etc…), but L(x) always has the following property:
\int_b^c L(x)dx = \alpha\int_a^d L(x) dx
Where \alpha is a constant and a\le b <c\le d.
There is another weighting function, W(x), which is known and only has value between x=b and x=c.
What I need to prove is that given that W(x) is known, prove that
\int_b^c W(x) L(x)dx = \beta\int_a^d L(x) dx
where \beta is another constant.
I am %99.9 sure this is correct, but I just can’t think how to prove it.
What I am hoping to do is to measure \beta and W(x) in the lab and then use this as a calibration to determine \int L(x) dx in any condition going forward as I can always measure \int W(x) L(x)dx.
Thanks. I feel like this should be easy, but my brain is wrapped in cotton.
What does “L(x) always has the following property” mean? In particular I’m asking about always. Does that mean for all a, b, c, d that satisfy the inequalities?
Sorry, I can see how this is vague. As I said, the function L(x) only has value from a to d. These are fixed values and are unimportant, you can replace a and d with \pm\infty if you want. So rephrased,
For any constants b and c, \int _b^c L(x)dx /\int_{-\infty}^\infty L(x)dx = \alpha , where \alpha is a constant.
L(X) is a set of spectral emission bands in a material (x is actually wavelength). The emission bands can change shape or brightness depending on a bunch of different variables (for instance, Doppler broadening), but the ratio of energy in a particular emission line (or sum of lines) to the total emission is always the same. In fact the ratio of energy in any particular emission line to another emission line is also a constant for the band system, at least it is in the conditions I am interested in.
So, \alpha is constant for any given value of b and c, even if all of the other variables (temperature etc.) change?
If I’m understanding correctly, I’d use the fact that the integral from b to c is a constant fraction of the total integral, to say that the integral from x to x+\Delta x is constant, then take the limit as \Delta x goes to 0, such that on that interval, W(x) is a constant, and so can be taken out of the interval.
That is exactly it @Chronos, thanks. Unfortunately, you also just showed me where I am wrong. If a spectral feature is on the edge of my filter function and undergoes non-symmetric broadening, the integral ratio will change. Damn.
There is something wrong here. Using your second formulation, let \int_{-\infty}^{\infty}L(x)dx=\gamma and then it says that independent of b,c, \int_b^cL(x)=\alpha\gamma. It is easy to see that this forces L(x) to be 0 (almost everywhere, anyway).
Yeah, it’s a little confusing. His L function isn’t just a function of x, but also of some other unlisted variables. When he says that alpha is a constant, he means independent of those other variables, not independent of the choice of b and c.