The standard way to define an improper integral of the form $\int_a^\infty f(t)dt$ is as follows. We first define the Riemann integral $\int_a^xf(t)dt$ for each $x>a$ in the standard way, i.e. using partitions of the interval $[a,x]$ and limits of Riemann sums and all that. Then we define $\int_a^\infty f(t)dt$ as $\lim_{x\rightarrow\infty}\int_a^xf(t)dt$.

But my question is, why can’t we define improper integrals in an analogous fashion to how we define Riemann integrals on closed intervals? That is, why can’t we take partitions of the interval $[a,\infty)$, take Riemann sums which would be infinite series, and then take the limit of those Riemann sums either under the refinement partial order or as the mesh of the partition goes to $0$?

Is the issue that the Riemann sum of a given partition of $[a,\infty)$ may not be a convergent series?

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    $\begingroup$ Riemann sums are always finite sums by partitioning a closed interval $[a,b]$ into intervals of length $\delta>0$. If you partition the interval $[a,\infty)$ into partitions of length $\delta$, then you would have an infinite partition, and thus an infinite sum. This might indeed not converge, for example take $f(x)=e^{-x}$. Then $\int_0^\infty f(x)dx=1$ but $$\lim_{\delta\rightarrow 0}\sum_{n=1}^\infty\delta\cdot f(n\delta)=\lim_{\delta\rightarrow 0}\delta\sum_{n=1}^\infty e^{-\delta n}$$ Using the power series of the exponential function, you can see that this does not converge. $\endgroup$ – Pink Panther Mar 31 at 15:42
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    $\begingroup$ Sometimes it works, other times it does not. It is a common problem with interchanging limits generally. For counterexamples see here. $\endgroup$ – RRL Mar 31 at 15:48
  • $\begingroup$ @RRL Hmm, but are there partitions where the subintervals are of unequal width for which the Riemann sum in this case is a convergent series? In the case maybe we just need to restrict ourselves to limits of convergent Riemann sums. $\endgroup$ – Keshav Srinivasan Mar 31 at 15:52
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    $\begingroup$ @PinkPanther: $$\lim_{\delta\rightarrow 0}\delta\sum_{n=1}^\infty e^{-\delta n} = \lim_{\delta\rightarrow 0} \delta \frac{e^{-\delta}}{1 - e^{-\delta}} = \lim_{\delta\rightarrow 0} \frac{\delta }{e^{\delta}-1} = 1 $$ $\endgroup$ – RRL Mar 31 at 16:17
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    $\begingroup$ @PinkPanther: No problem. There are other counterexamples of course. $\endgroup$ – RRL Mar 31 at 20:27

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