C/++/# Value of x that maximises the integral in a numerical integrator

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CAF123

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Consider an integral of form $$\int_a^b dx f(x) g(x).$$ Is it possible to tell a numerical integrator to spit out the value of ##x \in [a,b]## that maximises the value of ##f(x)g(x)##? I'm mostly interested in incorporating this into some code I have for adaptive integrator gsl_qags in C++.

Thanks!
 

CAF123

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This link directs me simply to google -> in any case I suppose the gsl_qags will use some stepwise increment over the integration domain. Is it maybe simpler to print out the values sampled in the integration domain together with the value of the integrand? Then I can see the maximum by inspection. Thanks.
 

CAF123

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@DrClaude Thanks, sorry I see I was not careful with my expression, I meant the value of the integral in my above post.
 

DrClaude

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Sorry, but I don't understand. Lets start from the beginning: what function do you want to maximise, and with respect to what parameter/variable?
 
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I meant the value of the integral in my above post.
That doesn't make sense. You originally asked:

Is it possible to tell a numerical integrator to spit out the value of ##x \in [a,b]## that maximises the value of ##f(x)g(x)##?
If you meant maximizes the integral, then you would be asking for a particular value of ##x## that maximizes an integral over a range of values of ##x##. That doesn't make sense.
 

Ray Vickson

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That doesn't make sense. You originally asked:



If you meant maximizes the integral, then you would be asking for a particular value of ##x## that maximizes an integral over a range of values of ##x##. That doesn't make sense.
I suppose it might make sense if he/she wants---for some unknown reason--- to spit up the integral as
$$\int_a^b f(x) g(x) \, dx = \int_a^m f(x) g(x) \, dx + \int_m^b f(x) g(x) \, dx,$$
where ##m = \text{arg max}_{x \in [a,b]} \, f(x) g(x).##

Even then the problem may be ill-posed, because for some nasty functions ##f,g## the integrand could have hundreds of values of ##m##--that is, hundreds of points that maximize the integrand.
 

rcgldr

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If looking for a maximum x for f(x)g(x), and assuming that f(x)g(x) has a maximum, then the maximum value will occur at one of the roots of the derivative of f(x)g(x).

If looking for the maximum of the indefinite integral of f(x)g(x), and assuming that there is a maximum, then the maximum value will occur at one of the roots of f(x)g(x).

If looking for a maximum integral range, and if the entire range of the integral results in "positive" areas, then the maximum is the entire range. Otherwise you're looking for a range with a mix of "positive" and "negative" areas that has the maximum "positive" area, which seems complicated.
 
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If looking for the maximum of the indefinite integral of f(x)g(x), and assuming that there is a maximum, then the maximum value will occur at one of the roots of f(x)g(x).
This doesn't make sense. You don't maximize an integral over a range of values of ##x## by picking one value of ##x##. See my response to the OP in post #7.

And in fact, you can't maximize an indefinite integral at all since it's indefinite. You might mean maximize the function of ##x## that is the antiderivative of the function ##f(x) g(x)##, but that's not quite the same thing.
 

rcgldr

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you can't maximize an indefinite integral at all since it's indefinite. You might mean maximize the function of ##x## that is the antiderivative of the function ##f(x) g(x)##, but that's not quite the same thing.
I meant antiderivative. "In calculus, an antiderivative, primitive function, primitive integral or indefinite integral ... "

https://en.wikipedia.org/wiki/Antiderivative
 

Ray Vickson

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I meant antiderivative. "In calculus, an antiderivative, primitive function, primitive integral or indefinite integral ... "

The OP started this all by stating a definite integral ##\int_a^b f(x) g(x) \, dx.##
 
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I am closing this thread because there is no point in speculating about what the OP meant. The OP needs to tell us, and that hasn't happened.

@CAF123 if you want to clarify what you meant, please PM me; if I get sufficient clarification I can reopen the thread for discussion.
 

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