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So I'm pretty confident that the answer is infinity, because the sum of all real numbers on $[0, 1]$ is greater than the harmonic series. However, I don't understand why the following integral can't be used to solve this problem: $$\int^1_0x\,dx$$ Wouldn't this integral add up infinitely small $x$ values to get the correct answer? But the answer I get when I integrate is $0.5$ I'm still pretty new to integration, so I apologize if it's something extremely simple.

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    $\begingroup$ It does not add up all the real values $x$ in $[0,1]$, instead it adds up all the real values times a very very small number d$x$ (of course this is not a rigorous statement). $\endgroup$
    – Levent
    Dec 11, 2017 at 23:10
  • $\begingroup$ @Levent Ohhhh that makes so much more sense. Thank you! $\endgroup$
    – Badr B
    Dec 11, 2017 at 23:13
  • $\begingroup$ By the way if you try to add an uncountable number of positive numbers you will get $\infty$. Hence we don't define an uncountable sum. $\endgroup$
    – Paramanand Singh
    Dec 14, 2017 at 5:00

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Adding all real numbers in any interval is not quite the same as integrating $xdx$ on that interval. You can think of integration as finding the area under the curve (here the curve $y=x$) between the limits 0 and 1.

I am not quite sure how one would formulate the problem of adding all real numbers in any interval as an integration problem. The area you are looking for is that of an infinite number of rectangles with very small widths but unit height. Hope this gives you some insights on how to think about integration.

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    $\begingroup$ Thank you for the explanation! I figured that integration was just finding the area under some graph, but I never really understood how to apply it. I'm mostly interested in using integration to calculate infinite sums, but as you can see, I need some practice with that haha. $\endgroup$
    – Badr B
    Dec 11, 2017 at 23:47
  • $\begingroup$ Integrating with respect to the counting measure on the reals is basically just summing them, FWIW. $\endgroup$ Dec 12, 2017 at 0:53
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No uncountable sum of positive terms can be convergent in the usual sense. To see this, partition the positive reals as $$[1,\infty), [\frac 12,1), [\frac 13, \frac 12),\cdots$$

As there are only countably many slots, one of them must contain infinitely many of your summands and the sum of those alone will diverge.

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By doing an integration you are summing up the values of the function (here $x$) times some width ($dx$). Basically this yields the area under the curve.

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    $\begingroup$ Yeah that's what threw me off. I was assuming that $dx=1$ which isn't the case here. Thank you for the quick answer! $\endgroup$
    – Badr B
    Dec 11, 2017 at 23:15

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