I'm trying to derive an equation for which I have a relationship between a constant and its integral -- namely, $$W = 2 \int_0^\infty f(L)\,dL$$ -- but also have another definition for $W$ -- $$W = -L\ln(1-P)$$ $f(L)$ itself is $f(L) = P_d$, a value related to $P$ but not the same ($P$ is the value of $P_d$ integrated from $0$ to $L$ somehow, though my sources don't actually give the $P_d$ equation). So what I need to do is solve $$2\int_0^\infty f(L)\,dL = -L\ln(1-P)$$ so that $P_d$ is the $y$ value and $L$ is the $x$ value, but I'm flubbing this one as I keep coming up with $P_d = \frac{W}{2L}$, which is clearly wrong (partly because it should go to $P_d = 1$ at $L = 0$ rather than infinity). I'm an experimentalist and out of my depth so long after taking calculus!
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1 Answers
The first thing you should note is that $\int_0^\infty f(L)\,dL$ is a constant. It is just a number. The only thing that it depends upon is the function $f$. If you change what function you are integrating, you will get a different value. But it does not depend on a value of the variable $L$. The $L$ in $\int_0^\infty f(L)\,dL$ is a dummy variable - it only exists to support the integral notation, and it does not exist outside that integration. You could equally well use any other variable:
$$\int_0^\infty f(L)\,dL = \int_0^\infty f(q)\,dq = \int_0^\infty f({\frak L})\,d{\frak L}$$ And since $L$ is also a variable in use outside this expression, you definitely should use a different dummy variable. Then you don't get confused thinking that the variable $L$ inside the notation means the same thing as the variable $L$ outside the notation.
But since $\int_0^\infty f(q)\,dq$ is just a constant, that means $W$ is a constant. It doesn't change with $L$. So the equation you need to solve is $$W = -L\ln(1 - P)\\-\dfrac WL = \ln(1 - P)\\e^{-W/L} = e^{\ln(1-P)}\\e^{-W/L} = 1 - P\\P = 1 - e^{-W/L}$$ Now you say that $P = \int_0^L P_d(q)\,dq$. By the fundamental theorem of calculus, that would mean $P_d$ is the derivative of $P$: $$P_d = \frac{dP}{dL} = -\frac{W}{L^2}e^{-W/L}$$
But alas, this cannot be true. If $P = \int_0^L P_d(q)\,dq$, then $P = 0$ when $L = 0$, but instead $P=1 - e^{-W/L}$ approaches $1$ as $L$ approaches $0$ from above.
So unfortunately something you've said here has been garbled too much to be able to figure out the answer.
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Dang it, lost my reply when I logged in. I didn't explain well, since P actually does go to 1 when L (for lateral distance, BTW) goes to 0. P goes to 0 at infinite L. I think the negative exponential is the right track, since a naive fit of a negative exponential to my data is a really good fit (R square > 0.95). My problem now is teasing W out of the resulting constant, so I'll try what you derived. – Ken Chiacchia Jan 23 '21 at 17:42
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But as I pointed out, this violates your claim that "$P$ is the value of $P_d$ integrated from $0$ to $L$ somehow". If $P(L) = \int_0^L$ of anything, then by necessity, $P(0) = 0$, not $1$. Something you've said in the OP is wrong. What it is, I cannot say, but the claims you've made are contradictory. Whether the solution for $P$ I gave is appropriate depends on where the error is. – Paul Sinclair Jan 23 '21 at 17:48
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Huh. I'll have to think about it. These are physical quantities, so the error has to be in my presentation somehow. Thanks. – Ken Chiacchia Jan 26 '21 at 14:36
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The statement that P is an integration of Pd is intuitive, so that's the most likely place I'm wrong. – Ken Chiacchia Jan 26 '21 at 14:37
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P being the probability of detection in an area defined by the detector's path out to lateral distance L (on either side of the detector); Pd is the probability of detection at distance L exactly. – Ken Chiacchia Jan 26 '21 at 14:38
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Probability doesn't quite work that way for continuous variables. The probability of detection at any single point is $0$ because there are infinite number of points, and if they had non-zero probability each, the probability for all of them together would be infinite, not $1$. But the probability of detection on intervals is non-zero. The probability density $P_d$ is the probability of detection in small intervals around $L$ divided by the width of the interval - more accurately, it is the limit as the width of the interval goes to $0$. In particular $$P_d = \frac{d,P([0,L])}{dL}$$ – Paul Sinclair Jan 26 '21 at 14:48
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I see two possibilities: either $W = -L\ln(1-P)$ should be $W=-\ln(1-P)/L$, in which case $P = \text{Prob}([0,L]) = 1 - e^{-WL}$, with $P(0) = 0, P(\infty) = 1$ ($P$ is the probability the result will be $\le L$). Or the $W$ formula is correct as is, but $P = \text{Prob}([L,\infty)) = 1-e^{-W/L}$ as I had ($P$ is the probability that the result will be $\ge L$). So $P(0) = 1, P(\infty) = 0$. In this case $P_d = -\frac{dP}{dL}$ and $$P = \int_L^\infty P_d(q),dq$$. – Paul Sinclair Jan 26 '21 at 21:50