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- anonymous

Find a continuous f on the interval (0, inf) such that the integral of f(x) on (0, inf) exists, but the limit as x-> infinity =/= 0

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- anonymous

- katieb

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- anonymous

cosine, sine, lots. actually any function that can be dissolved vie fourier series into plain cos and sine will fit these parameters. integral will be zero. Did you want a non-zero integral?

- anonymous

Maybe you like that answer. Otherwise, if you want an answer with a non-zero integral simply make your function of the form sine(x) + g(x); where g(x) is any function with an existing integral. It doesn't matter whether the limit of g(x) approaches zero when x-> infinity, because sine(x) +g(x) will still have a non-convergent limit (the limit will not exist). If you want the integral of your answer to be X, simply choose a g(x) that has integral X. The integral like the derivative is a linear operator and therefore obeys the first law of linearity ( int[f(x)+g(x)] = int[g(x)]+ int[f(x)]. A less clever way to do this that a lay mathematician may think would work is simply attaching the function g(x) to the function sine(x) (or cos (x)) like making a composite function of the form: from 0 to x inclusive, g(x), and from x to infinity exclusive, sine(x). However, this will not work as at the point that sin(x) is joined to g(x), there will be a cusp where the left limit will not equal the right limit. Right, that is all you will ever have to know about limit theory as a context to this problem. Though it is a fascinating area to specialize in and learn more about. A family of sinusoidal functions is also a perfect place to use a meta-limit that approaches sine of cos. But that is a relatively new (very new) concept that hasn't quite caught on yet. Stick to the conventional epsilon delta proof of limits if you want to verify that I am right about all this for your professor. I've done all this in my head, if you want the delta epsilon proof, I can supply this.

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