My intuition was to try and see if the series is a Riemann Sum of a function and then see what happens but I can't really see which function fits here.
Thanks!
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1Your instinct was good. Look at $\int \frac{dz}{z^2+1}$. For a given increment, let's call it $x$, what are the Riemann sums? – lulu Aug 01 '15 at 18:43
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Right, and this just appeared today: http://math.stackexchange.com/questions/1381185/convergence-integrals-and-limits-question – zhw. Aug 01 '15 at 18:47
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@lulu I am slightly confused: don't the usual theorems about Riemann sums only apply for integrals over a compact interval? (i.e., this one would have to be done "manually") – Clement C. Aug 01 '15 at 18:49
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@ClementC. If the function is positive and decreasing, you always get the "infinite Riemann sum" result. See the link I gave above. – zhw. Aug 01 '15 at 18:51
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@zhw -- I read it (that is what I call "manually"): it doesn't appeal to a theorem about Riemann integrals and sums, since in particular (unless I'm missing something) there is no proper Riemann integral over $[0,\infty)$. – Clement C. Aug 01 '15 at 18:53
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@ClementC. I didn't look at the link, but I don't see the problem. Given the integrand, however one may want to define the improper integral it lies between the right and left Riemann sums, no? But those just differ by the first rectangle which, in the limit, does not matter. Am I missing something? – lulu Aug 01 '15 at 19:00
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My point (again, I may be wrong, it's been a while) is not that the difference between left and Riemann sum does not go to zero; my issue is that for improper integrals, convergence of a "Riemann sum" to the improper integral is not guaranteed by any theorem -- it's on a case-by-case basis. (The integral over $[0,\infty)$ in question does not even exist in the setting of Riemann integration.) – Clement C. Aug 01 '15 at 19:13
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(This is assuming you are referring to an integral over $[0,\infty)$, of course... over a compact, I guess everything would be fine.) – Clement C. Aug 01 '15 at 19:22
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@ClementC. I assure you, it has been longer for me so it might well be me who is confused. Are you concerned about uniformity? I could see a problem arising in the absence of that. But, in the case of a decreasing positive integrand, I think uniformity is quite clear, no? For a fixed cut off, the tail decreases as x increases. So...truncate the integral, The (finite) partial sums are then simultaneously close to the limit of the truncated integrals and to the infinite sum. – lulu Aug 01 '15 at 19:37
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@ClementC. Do you believe you can find a decreasing, positive function f for which the improper integral exists (as a limit of the finite integral, say) but is not equal to the limit of the Riemann sums? – lulu Aug 01 '15 at 19:39
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As an aside, $\displaystyle\sum_{n=-\infty}^\infty\frac1{n^2x^2+1}=\frac\pi x~\coth\frac\pi x~.~$ This can be shown by differentiating the natural logarithm of Euler's infinite product formula for the sine function. – Lucian Aug 02 '15 at 05:06
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@lulu: I wouldn't bet on it, no... it does seem more than plausible. – Clement C. Aug 02 '15 at 06:27
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@ClementC. But wait, you said you read what was at the link I gave, and it is only plausible? Really it's quite elementary. – zhw. Aug 03 '15 at 00:02
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@zhw: OK, to reiterate what is my position: I believe in the result (for positive, decreasing functions on $[0,\infty)$ for which the integral on $[0,\infty)$ converges), and the proof you linked. I haven't thought about it for arbitrary, possibly weirder domains and integration domains (the latter for the improper integral), but my point is that a proof (like the one you gave) is needed: it does not merely follow by theorems from Riemann integration. – Clement C. Aug 03 '15 at 04:52
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I'll replace $x$ by $\Delta x.$ Let $f(x) = 1/(x^2+1).$ Then we want fo find
$$\lim_{\Delta x\to 0^+}\sum_{n=1}^{\infty} f(n\Delta x)\Delta x.$$
The above sum can be viewed as an "infinite Riemann sum" of areas of rectangles of base $\Delta x$ and heights $f(n\Delta x).$ Because $f$ is positive and decreasing, the limit is what you would expect, namely
$$\int_0^\infty f(x)\,dx = \int_0^\infty \frac{dx}{x^2 + 1}= \pi/2.$$
See my link in the comments for the justification of this limit; it's quite simple.
zhw.
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Notice, $$\lim_{x\to 0}\sum_{n=1}^{\infty}\frac{x}{n^2x^2+1}=\lim_{x\to 0}\sum_{n=1}^{\infty}\frac{x}{(nx)^2+1}$$ Let $nx=u\implies u\in[0, \infty)\ as \ x\to 0, n\to \infty $ $$=\int_{0}^{\infty}\frac{du}{u^2+1}$$ $$=[\tan^{-1}(u)]_{0}^{\infty}$$ $$=\tan^{-1}(\infty)-\tan^{-1}(0)=\frac{\pi}{2}$$
Harish Chandra Rajpoot
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May want to explain a bit more on the transition from the series to the integral. But (+1) for the clever substitution. – John Molokach Aug 01 '15 at 20:36
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2Since for any $x$, $nx\to \infty$ as $n\to \infty,$ why do you think $0\le u \le 1$? – zhw. Aug 01 '15 at 21:12