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Intuitively, it is rather obvious that

$$\lim_{l\to\infty}\sum_{n=-\infty}^{\infty}f(n\Delta x)\Delta x = \int_{-\infty}^{\infty}f(x)dx \tag{1}$$

where $\Delta x = \frac{1}{l}$, assuming $f$ is integrable and the limit exists.

The fact that this equality is true is the core part of deriving Fourier transform from Fourier series, see page 4, eq. 4.7 in this document. Or maybe we cannot consider this derivation as formal, as it was never intended to be formal, but I thought n mathematics there's no place for informal thinking.

My question is how can we prove it's true from the definitions and properties of improper integral, definite integral and limits?

I've listed the important definitions below in case you would like to refer to some of these in your answers.

Oh, and please ignore mrf's answer - it doesn't refer to my question anymore, I've reformulated it.


If function $f$ is integrable on $[a,b]$, then: $$\int_{a}^{b}f(x)dx=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i)\Delta x \tag{2}$$ where $\Delta x = \frac{b-a}{n}$ and $x_i = a+i\Delta x$.

Improper integral definitions

$$\int_{a}^{\infty}f(x)dx=\lim_{t\to\infty}\int_{a}^{t}f(x)dx \tag{3}$$

$$\int_{-\infty}^{b}f(x)dx=\lim_{t\to-\infty}\int_{t}^{b}f(x)dx \tag{4}$$

$$\int_{-\infty}^{\infty}f(x)dx=\int_{a}^{\infty}f(x)dx + \int_{-\infty}^{a}f(x)dx \tag{5}$$

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  • $\begingroup$ They are equal if and only if the function is actually Riemann integrable. But it is possible for limits of sums over two sequences of partitions to be different. In this case your limit will exist while the integral will not. $\endgroup$ – Ian Feb 11 '15 at 20:07
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    $\begingroup$ I've got a crude heuristic argument suggesting that (1) is true if $f$ is integrable on all finite intervals and $f(x) = O(1/x^2)$ for large $\left\lvert{x}\right\rvert$. I'll try to work it into a proper proof tomorrow - meanwhile, bed! $\endgroup$ – Calum Gilhooley Feb 16 '15 at 0:18
  • $\begingroup$ Often to avoid these sorts of complications, the Fourier transform is applied to functions with compact support and Schwartz functions (rapidly decaying functions). In these cases the integration becomes much easier. In the former case, it is actually a finite integral. $\endgroup$ – Joel Feb 16 '15 at 20:57
  • $\begingroup$ The Wikipedia article Fourier transform (which a quick inspection suggests is really excellent) refers to the textbook by Stein & Shakarchi, Fourier Analysis: An introduction (2003), which mentions the concept of "functions of moderate decrease", which are defined in MathSE question #519238 (8 Oct 2013) as those which are $O(1/(1 + \left\lvert{x}\right\rvert^{1 + \epsilon}))$ for some $\epsilon > 0$ (presumably this is for large $x$). I hazard the wild guess that a more thoroughgoing and expert analysis than mine (below) would show your result to be true for all such functions $f$. $\endgroup$ – Calum Gilhooley Feb 17 '15 at 23:50
  • $\begingroup$ Indeed, that seems equivalent to the combination of boundedness, for small $x$, and being $O(1/\left\lvert{x}\right\rvert^{1 + \epsilon})$, for large $x$, which (in combination with integrability) I already thought might be a sufficient condition. $\endgroup$ – Calum Gilhooley Feb 18 '15 at 0:28
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This is a possibly overcomplicated proof of the conjectured identity: \begin{equation} \lim_{\delta \to 0+} \sum_{n = -\infty}^\infty f(n\delta)\delta = \int_{-\infty}^\infty f \tag{1}\label{eq:A} \end{equation} for the improper Riemann integral of a function $f: \mathbb{R} \to \mathbb{R}$, based on these hypotheses about $f$:

  1. $f$ is integrable on all finite intervals of $\mathbb{R}$;
  2. $f(x) = O\!\left(\frac{1}{x^2}\right)$, for large $\left\lvert{x}\right\rvert$. $\newcommand{\abs}[1]{\left\lvert#1\right\rvert}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\Z}{\mathbb{Z}}$ $\renewcommand{\phi}{\varphi}$ $\newcommand{\floor}[1]{\left\lfloor#1\right\rfloor}$

By hypothesis, there exist $M, A > 0$ such that $\abs{f(x)} \leqslant M/x^2$ for $\abs{x} \geqslant A$.

$\int_0^\infty f$ exists, by the Cauchy convergence criterion, because, if $A \leqslant a \leqslant b$, $$ \bigg\lvert\int_0^b f - \int_0^a f\bigg\rvert = \bigg\lvert\int_a^b f\bigg\rvert \leqslant \int_a^b \abs{f} \leqslant \int_a^b \frac{M}{x^2}\,dx = M\left(\frac{1}{a} - \frac{1}{b}\right) < \frac{M}{a} $$ and this tends to $0$ as $a$ tends to $\infty$. Similarly for $\int_{-\infty}^0 f$; therefore $\int_{-\infty}^\infty f$ exists.

Also, it is clear that all the sums on the left hand side of $\eqref{eq:A}$ converge, by comparison with $\sum 1/n^2$. But we need precise information on this convergence.

For $\delta > 0$, let $S(\delta) = \sum_{n = -\infty}^\infty f(n\delta)\delta$. There exists a positive real number $N(\delta)$ (which we may take as large as we please) such that: \begin{equation} \bigg\lvert S(\delta) - \!\!\!\!\sum_{\abs{n} \leqslant N} f(n\delta)\delta \bigg\rvert < -\frac{1}{\log\delta} \ \text{ for all } N \geqslant N(\delta). \tag{2}\label{eq:B} \end{equation} The expression $-1/(\log\delta)$ is chosen so that it tends only slowly to $0$ as $\delta \to 0$; any other similarly slowly shrinking function would have done instead.

The size of $N(\delta)$ turns out to be critical, so we estimate it carefully. Certainly we require $N(\delta)\delta > A$ for all $\delta$, in order to use our hypothesis on $f$. Also: $$ \lim_{\delta \to 0+} N(\delta)\delta = +\infty. $$ Both these properties are guaranteed by the choice of $N(\delta)$ that follows, so long as $\delta$ is small enough. (Certainly $\delta < 1$, otherwise $\eqref{eq:B}$ is ill-defined.) If $N\delta \geqslant A$: \begin{gather*} \bigg\lvert \sum_{\abs{n} > N} f(n\delta)\delta \bigg\rvert \leqslant \sum_{\abs{n} > N} \abs{f(n\delta)}\delta \leqslant \frac{2M}{\delta}\!\!\!\sum_{n=N+1}^\infty \frac{1}{n^2} < \frac{2M}{\delta}\!\!\!\sum_{n=N+1}^\infty \frac{1}{n(n - 1)} = \frac{2M}{N\delta}. \end{gather*}

Accordingly, we define $N(\delta)$ by the equation $N(\delta)\delta = -2M\log\delta$.

Consider the change of variables $\phi: (-1, 1) \to \R$, where: $$ \phi(y) = \frac{1}{1 - y} - \frac{1}{1 + y} = \frac{2y}{1 - y^2}, \ \ \phi'(y) = \frac{2(1 + y^2)}{(1 - y^2)^2} \ \ \ (-1 < y < 1). $$

If we write $y_n = \phi^{-1}(n\delta)$ for all $n \in \Z$, then by Taylor's Theorem: \begin{gather*} \sum_{\abs{n} \leqslant N(\delta)} f(n\delta)\delta = \!\!\!\!\sum_{\abs{n} \leqslant N(\delta)} f(\phi(y_n))(\phi(y_{n+1}) - \phi(y_n)) = \\ \sum_{\abs{n} \leqslant N(\delta)} f(\phi(y_n))\phi'(y_n)(y_{n+1} - y_n) + \!\!\!\!\sum_{\abs{n} \leqslant N(\delta)} f(\phi(y_n))\frac{\phi''(y_n^*)}{2}(y_{n+1} - y_n)^2, \end{gather*} for some $y_n^*$ such that $y_n < y_n^* < y_{n+1}$ ($\abs{n} \leqslant N(\delta)$).

The first of these two subexpressions is 'almost' a Riemann sum for the integral $\int_{-1}^{1} f(\phi(y))\phi'(y)\,dy$. Note that the integrand remains bounded at the endpoints, because $\phi'(y) \sim \phi(y)^2$ as $y \to \pm 1$, and therefore, for $y$ close to $\pm 1$, $$ \abs{f(\phi(y))\phi'(y)} \leqslant \frac{M\phi'(y)}{\phi(y)^2} \sim M. $$

Define a function $F: (-1, 1) \to \R$, and numbers $c(\delta), d(\delta) \in (-1, 1)$, by: \begin{align*} F(y) & = f(\phi(y))\phi'(y) \ \ (-1 < y < 1), \\ c(\delta) & = \phi^{-1}(-N(\delta)\delta), \\ d(\delta) & = \phi^{-1}((N(\delta) + 1)\delta)). \end{align*} Then $\lim_{\delta \to 0+} c(\delta) = -1$, $\lim_{\delta \to 0+} d(\delta) = 1$, and, because $F$ is bounded at $\pm 1$, \begin{equation} \int_{-\infty}^\infty f = \lim_{\delta \to 0+} \int_{-N(\delta)\delta}^{(N(\delta) + 1)\delta} f = \lim_{\delta \to 0+} \int_{c(\delta)}^{d(\delta)} F = \int_{-1}^1 F. \tag{3}\label{eq:D} \end{equation}

The desired conclusion $\eqref{eq:A}$ follows from $\eqref{eq:B}$, $\eqref{eq:D}$, and: \begin{equation} \lim_{\delta \to 0+} \sum_{\abs{n} \leqslant N(\delta)} f(n\delta)\delta = \int_{-1}^1 F, \tag{4}\label{eq:F} \end{equation} which we now prove.

We obtained, above, a lengthy expression of the form: $$ \sum_{\abs{n} \leqslant N(\delta)} f(n\delta)\delta = I(\delta) + J(\delta), $$ remarking at the time that $I(\delta)$ is 'almost' a Riemann sum. In fact (here we temporarily denote the integer $\floor{N(\delta)}$ by '$N$' for readability), the expression $$ F(y_{-N})(1 + y_{-N}) + I(\delta) + F(y_{N+1})(1 - y_{N+1}) $$ is a Riemann sum for the partition $(-1, y_{-N}, y_{-N+1}, \ldots, 0, \ldots, y_N, y_{N+1}, 1)$, tagged with values $(y_{-N}, y_{-N}, y_{-N+1}, \ldots, 0, \ldots, y_N, y_{N+1})$. By the continuity of $\phi$, the maximum of the interval lengths $y_{n+1} - y_n$ tends to $0$ with $\delta$; and we have already remarked that $y_{N+1}$ tends to $1$ and $y_{-N}$ to $-1$; and $F$ is bounded.

From the facts just mentioned, it follows that: $$ \lim_{\delta \to 0+} I(\delta) = \int_{-1}^1 F, $$ and so the proof of $\eqref{eq:F}$, and therefore of $\eqref{eq:A}$, reduces to: $$ \lim_{\delta \to 0+} J(\delta) = 0, $$ or in full: $$ \lim_{\delta \to 0+} \sum_{\abs{n} \leqslant N(\delta)} f(\phi(y_n))\frac{\phi''(y_n^*)}{2}(y_{n+1} - y_n)^2 = 0. $$

By our hypotheses, $f$ is integrable, and therefore bounded, on the interval $[-A, A + \delta]$, therefore the factor $f(\phi(y_n))\phi''(y_n^*)$ is bounded for $n \in \Z$ such that: $$ -\phi^{-1}(A) \leqslant y_n < y_n^* < y_{n+1} \leqslant \phi^{-1}(A + \delta), $$ or equivalently, $$ -A \leqslant n\delta < \phi(y_n^*) < (n + 1)\delta \leqslant A + \delta. $$ Such terms therefore contribute at most a fixed multiple of $\sum_n (y_{n+1} - y_n)^2$ to the absolute value of the summation; and because $\lim_{\delta \to 0+} \max_n (y_{n+1} - y_n) = 0$, and $\sum_n (y_{n+1} - y_n) < 2$, this part of the sum tends to $0$ in the limit as $\delta \to 0$.

What now remains to be proved is: \begin{equation} \lim_{\delta \to 0+} \sum_{A/\delta \leqslant \abs{n} \leqslant N(\delta)} f(\phi(y_n))\frac{\phi''(y_n^*)}{2}(y_{n+1} - y_n)^2 = 0. \tag{5}\label{eq:H} \end{equation} For such $n$, we have: $$ \abs{f(\phi(y_n))\frac{\phi''(y_n^*)}{2}} \leqslant \frac{M\abs{\phi''(y_n^*)}}{2\phi(y_n)^2} = \frac{M\abs{\phi''(y_n^*)}}{2n^2\delta^2}. $$ Note that: $$ \phi''(y) = \frac{4(3y + y^3)}{(1 - y^2)^3} \ \ \ (-1 < y < 1). $$ By taking $A$ large enough, we can assume that all values of the argument $y$ under consideration satisfy $1/\sqrt{2} \leqslant \abs{y} < 1$, so that $\abs{y} \leqslant 2\abs{y^3}$, and therefore: $$ \abs{\phi''(y)} \leqslant \frac{28\abs{y}^3}{(1 - y^2)^3} = \frac{7\abs{\phi(y)}^3}{2}. $$ We also have the inequality: $$ \abs{\phi(y_n^*)} \leqslant (\abs{n} + 1)\delta. $$ Putting it all together: $$ \abs{f(\phi(y_n))\frac{\phi''(y_n^*)}{2}} \leqslant \frac{7M\abs{\phi(y_n^*)}^3}{4n^2\delta^2} \leqslant \frac{7M\delta(\abs{n} + 1)^3}{4n^2} = \frac{7M\abs{n}\delta}{4}\left(1 + \frac{1}{\abs{n}}\right)^3. $$ Taking $A$ large enough, we can assume $\abs{n} \geqslant 22$, therefore $\left(1 + \frac{1}{\abs{n}}\right)^3 < \frac{8}{7}$, and: $$ \abs{f(\phi(y_n))\frac{\phi''(y_n^*)}{2}} \leqslant 2M\abs{n}\delta \leqslant 2MN(\delta)\delta = -4M^2\log\delta. $$ Having been careful with our estimates so far, we can afford to be sloppy now! We have $\phi'(y) \geqslant 2$, for all $y$, therefore $y_{n+1} - y_n \leqslant \delta/2$, for all $n$. This and the fact that $\sum_n (y_{n+1} - y_n) < 2$ together imply that the sum in $\eqref{eq:H}$ is bounded above by $-4M^2\delta\log\delta$, which does tend to $0$ with $\delta$. This completes the proof.

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  • $\begingroup$ The definition of the functions $c(\delta)$, $d(\delta)$ is a hangover from an earlier version of the proof, and they really need have nothing to do with $N(\delta)$. For clarity's sake, I'll edit this part of the proof later (and then I'll delete this comment). $\endgroup$ – Calum Gilhooley Feb 17 '15 at 15:20
  • $\begingroup$ I admire the work you've put into this answer. We probably should ask why for a few hundred years nobody has ever questioned the formulation of fourier transfom from fourier series. Maybe people who are expert in this area know a simpler explanation, but they don't have time to share it with us. I wonder what would happen if I asked it on mathoverflow... By the way, was $O(\frac{1}{x^2})$ a pure guess? $\endgroup$ – user4205580 Feb 17 '15 at 15:46
  • $\begingroup$ The condition $f(x) = O(1/x^2)$ dropped out naturally when I tried to prove the result using the change-of-variables substitution $\varphi$. That instantly made the result plausible, but it got quite beastly when I tried to prove it yesterday, and then it turned around and bit me (with the size of $N(\delta)$) even after I thought I'd got it tamed! $\endgroup$ – Calum Gilhooley Feb 17 '15 at 17:11
  • $\begingroup$ Tomorrow I may try arguing similarly for $f(x) = O(1/\left\lvert{x}\right\rvert^{1 + \epsilon})$, using the change of variables formula $\psi(y) = \log\frac{1 + y}{1 - y}$. $\endgroup$ – Calum Gilhooley Feb 18 '15 at 1:49
  • $\begingroup$ (Red face.) Actually, I think that should have been $\psi(y) = (1 - y)^{-1/\epsilon} - (1 + y)^{-1/\epsilon}$ (with hindsight, the obvious generalisation from $\epsilon = 1$). Try again tomorrow. Have written some of it up today. $\endgroup$ – Calum Gilhooley Feb 19 '15 at 0:55
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I think this is a counterexample.

Let $g$ be any non-negative function on $\mathbb{R}$ such that the improper Riemann integral $\int_{-\infty}^\infty g$ exists. Let $S = \{ q + (p/q) : q, p = 1, 2, \ldots \}$, and let $f(x)$ be equal to $g(x)$ except on $S$, where $f(x) = 1$.

Since $S$ has only finitely many points in any finite interval, $f$ is continuous and equal to $g$ except on the countable set $S$, therefore $\int_{-\infty}^\infty f$ exists, and is equal to $\int_{-\infty}^\infty g$.

But for $q = 1, 2, \ldots$, the sum $\sum_{n = -\infty}^\infty f(n/q)$ diverges to $+\infty$, therefore the limit on the left hand side of (1) does not exist.

Update: it gets worse, I'm afraid.

One might still reasonably hope that, if $f$ is continuous (which rules out this counterexample as it stands), and $f$ is non-negative, and $\int_{-\infty}^\infty f$ exists, then all the infinite sums on the left hand side of (1) exist, so there's a fair chance that (1) holds under these (arguably not too restrictive) conditions.

Define the countable set $S \subset \mathbb{R}$, in the same way as before. Because $S$ has only finitely many points in any finite interval, it can be arranged as a strictly increasing sequence, $s_1 < s_2 < \ldots$.

Choose a convergent series $\sum_{k=1}^\infty t_k$ such that $t_k > 0$ and $s_k + t_k \leqslant s_{k+1} - t_{k+1}$ ($k = 1, 2, \ldots$).

Let $h: (-1, 1) \to \mathbb{R}$ be a "bump function", such as: $$ h(y) = e^{y^2/(y^2 - 1)} \qquad (-1 < y < 1). $$

Define: $$ f(s_k + yt_k) = \frac{h(y)}{s_k} \qquad (k = 1, 2, \ldots; \ -1 < y < 1), $$ and let $f$ have the value $0$ everywhere outside the pairwise disjoint open intervals $(s_k - t_k, s_k + t_k)$.

Observe that for $l = 1, 2, \ldots$, we have $n/l \in S$ and $f(n/l) = l/n$ for all $n > l^2$, and therefore: $$ \sum_{n = -\infty}^\infty f\left(\frac{n}{l}\right) \geqslant \sum_{n = l^2 + 1}^\infty f\left(\frac{n}{l}\right) = l\sum_{n = l^2 + 1}^\infty \frac{1}{n} = +\infty. $$

Thus: $f$ is smooth everywhere on $\mathbb{R}$; $f(x) \geqslant 0$ for all $x \in \mathbb{R}$; $f(x) \to 0$ as $x \to \pm \infty$; the improper Riemann integral $\int_{-\infty}^\infty f$ exists (it is bounded above by $2\sum_{k=1}^\infty t_k/s_k$, and therefore by $2\sum_{k=1}^\infty t_k$); yet, the inner series on the left hand side of (1) diverges to $+\infty$ for all positive integral values of $l$.

So, even though the parameter $\Delta x$ on the left hand side of (1) may assume any strictly positive value, the series expression under the outer limit sign becomes undefined for arbitrarily small values of $\Delta x$, so the limit itself is not well defined, even for this quite "reasonable" function $f$.

Further update: essentially the same construction, and same argument, with these minor changes: $$ f(s_k + yt_k) = \frac{h(y)}{s_k\log s_k}, \\ \sum_{n = -\infty}^\infty f\left(\frac{n}{l}\right) \geqslant l\sum_{n = l^2 + 1}^\infty \frac{1}{n\log n} = +\infty, $$ shows that (1) fails even for smooth $f$ such that $\int_{-\infty}^\infty f$ exists and $f(x) = O\left(\frac{1}{\left\lvert{x}\right\rvert\log\left\lvert{x}\right\rvert}\right)$ for large $\left\lvert{x}\right\rvert$.

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  • $\begingroup$ Oh, I see you've added the condition that the limit exists - as a comment on another answer. That changes things. I think it would have been better to have edited the question to incorporate the new condition, and you could still do that. $\endgroup$ – Calum Gilhooley Feb 13 '15 at 13:59
  • $\begingroup$ If the index variable $l$ in your limit expression is intended to be an integer, my counterexample still works - or at least, it can perhaps be tweaked so that it works! Were you intending to allow $l$ to assume any strictly positive real value? $\endgroup$ – Calum Gilhooley Feb 13 '15 at 14:09
  • $\begingroup$ Variable $l$ is only used in definition of $\Delta x$. It's a real number. The equality has to be true (at least for continuous functions) as it is the core part of derivation of fourier transform from fourier series - see page 4 of this document $\endgroup$ – user4205580 Feb 13 '15 at 14:15
  • $\begingroup$ Even assuming that eq. 4.7 in that PDF is OK (I've no reason to doubt it, ignorant as I am on the subject!), and assuming your conjecture is true, I think what my counterexample still shows is that the LHS of (1) may fail to be defined even when the RHS is defined, and so the LHS cannot be used as an alternative definition of the improper Riemann integral. Your question is still worth thinking about, however, and my 'answer' doesn't dispose of it, but should be considered more as an extended comment. $\endgroup$ – Calum Gilhooley Feb 13 '15 at 14:43
  • $\begingroup$ If (1) is wrong, then so is every derivation of fourier transform you can come across... Not a great news. $\endgroup$ – user4205580 Feb 15 '15 at 7:54
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$\newcommand{\abs}[1]{\left\lvert#1\right\rvert}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\Z}{\mathbb{Z}}$ $\renewcommand{\phi}{\varphi}$ $\newcommand{\floor}[1]{\left\lfloor#1\right\rfloor}$

See this Meta thread for the reason why this answer has been edited after 4 years. The content has not changed.

The counterexample in my first answer (as recently modified) shows that conjecture $\eqref{eq:A}$ fails even for some relatively well-behaved functions $f$ such that $f(x) = O(1/(\abs{x}\log\abs{x}))$ for large $\abs{x}$; whereas my second answer shows that $\eqref{eq:A}$ is true for all functions $f$, integrable on finite intervals, such that $f(x) = O(1/x^2)$.

We now close most, although not all, of the remaining gap between these estimates, by showing that conjecture $\eqref{eq:A}$ is true for all functions $f$, integrable on finite intervals, such that $f(x) = O(1/\abs{x}^{1 + \epsilon})$ for large $\abs{x}$, for some $\epsilon > 0$ (i.e. $f$ is a "function of moderate decrease").

To save space, I'll refer frequently to my second answer, and elaborate the new argument only in the places where it differs significantly from the old one, which I'll no longer bother to update (not even in places where it is untidy!) - except, of course, to correct any remaining errors.

Talking about closing gaps, and correcting errors, the following lemma (copied, with only minor notational alterations, from the answer I posted yesterday to the question "Improper Riemann integral of bounded function is proper integral") is needed to plug the subtle logical gap in my first answer:

Suppose that (i) the function $g: [a, b] \to \R$ is bounded, and (ii) the improper Riemann integral $\int_{a+}^b g = \lim_{\epsilon \to 0+} \int_{a + \epsilon}^b g$ exists. Then the proper Riemann integral $\int_a^b g$ also exists, and it is equal to the improper integral $\int_{a+}^b g$.

Proof:

Let $M$ be any upper bound of $\{\abs{g(x)}: a \leqslant x \leqslant b\}$ such that $2M(b - a) > 1$. For $n = 1, 2, \ldots$, let $\epsilon_n = \frac{1}{2nM}$, and let $P_n$ be a partition of $[a + \epsilon_n, b]$ on which the upper and lower Darboux sums of $g$ both differ from $\int_{a + \epsilon_n}^b g$ by less than $\frac{1}{2n}$. The upper and lower Darboux sums of $g$ on $\{a\} \cup P_n$ both differ from $\int_{a + \epsilon_n}^b g$ by less than $\frac{1}{n}$, so $g$ has a sequence of upper Darboux sums over $[a, b]$ that converges to $\int_{a+}^b g$, and also a sequence of lower Darboux sums over $[a, b]$ that converges to $\int_{a+}^b g$. Hence, $g$ is Riemann integrable on $[a, b]$, and $\int_a^b g = \int_{a+}^b g$. Q.E.D.

The change of variables $x \mapsto a + b - x$ yields the corollary that the existence of $\int_a^{b-} g = \lim_{\epsilon \to 0+} \int_a^{b - \epsilon} g$ implies the existence of $\int_a^b g$, with the same value.

Now for the proof of the main result.

By hypothesis, there exist $M, A > 0$ such that $\abs{f(x)} \leqslant M/\abs{x}^{1 + \epsilon}$ for $\abs{x} \geqslant A$.

The simple proof that $\int_{-\infty}^\infty f$ exists goes through, almost exactly as before.

Define a smooth monotone bijection $\phi: (-1, 1) \to \R$, where, for $y \in (-1, 1)$, \begin{gather*} \phi(y) = \frac{1}{(1 - y)^{\frac{1}{\epsilon}}} - \frac{1}{(1 + y)^{\frac{1}{\epsilon}}}, \\ \phi'(y) = \frac{1}{\epsilon}\left[ \frac{1}{(1 - y)^{\frac{1}{\epsilon} + 1}} + \frac{1}{(1 + y)^{\frac{1}{\epsilon} + 1}} \right], \\ \phi''(y) = \frac{1}{\epsilon}\left(\frac{1}{\epsilon} + 1\right)\left[ \frac{1}{(1 - y)^{\frac{1}{\epsilon} + 2}} - \frac{1}{(1 + y)^{\frac{1}{\epsilon} + 2}} \right]. \end{gather*}

Define $F: [-1, 1] \to \R$ much as before, by putting $F(y) = f(\phi(y))\phi'(y)$ ($-1 < y < 1$), and assigning arbitrary values to $F(-1)$ and $F(1)$. As $y \to \pm 1$, \begin{gather*} \phi(y) \sim \pm \frac{1}{(1 \mp y)^{\frac{1}{\epsilon}}}, \ \ \phi'(y) \sim \frac{1}{\epsilon}\frac{1}{(1 \mp y)^{\frac{1}{\epsilon} + 1}}, \end{gather*} therefore $$ \abs{F(y)} = \abs{f(\phi(y))\phi'(y)} \leqslant \frac{M\phi'(y)}{\abs{\phi(y)}^{1 + \epsilon}} \sim \frac{M}{\epsilon}. $$ So $F$ is bounded on $[-1, 1]$.

By the theorem on change of variable in a Riemann integral (see e.g. Rudin, Principles of Mathematical Analysis, Theorem 6.19), $F$ is Riemann-integrable on any closed subinterval $[c, d]$ of $(-1, 1)$, and $\int_c^d F = \int_{\phi(c)}^{\phi(d)} f$. Therefore, the improper Riemann integral $\int_{(-1)+}^{1-} F$ exists and equals the improper Riemann integral $\int_{-\infty}^\infty f$. But $F$ is bounded on $[-1, 1]$, so the lemma and corollary above imply that the proper Riemann integral $\int_{-1}^1 F$ exists and equals $\int_{-\infty}^\infty f$.

In order to use this result to prove the conjecture $\eqref{eq:A}$, we now have to prove: \begin{equation} \lim_{\delta \to 0+} S(\delta) = \int_{-1}^1 F, \tag{6}\label{eq:J} \end{equation} where, as before: $$ S(\delta) = \sum_{n = -\infty}^\infty f(n\delta)\delta. $$

The positive real number $N(\delta)$ is again defined so as to satisfy the inequality: \begin{equation} \bigg\lvert S(\delta) - \!\!\!\!\sum_{\abs{n} \leqslant N} f(n\delta)\delta \bigg\rvert < -\frac{1}{\log\delta} \ \text{ for all } N \geqslant N(\delta). \tag{2}\label{eq:M} \end{equation} We cannot be quite so precise about the value of $N(\delta)$ this time: it depends on some new constants, whose values we do not attempt to estimate. As before, we require $N(\delta)\delta > A$, and $\lim_{\delta \to 0+} N(\delta)\delta = +\infty$. It is "well known" (for instance from Apostol, Introduction to Analytic Number Theory, Theorem 3.2(c)) that: $$ \sum_{n > N} \frac{1}{n^{1 + \epsilon}} = O\left(\frac{1}{N^\epsilon}\right). $$ That is to say, there exist real $K, B > 0$ such that: $$ \sum_{n > N} \frac{1}{n^{1 + \epsilon}} < \frac{K}{N^\epsilon} \text{ for all } N \geqslant B. $$ (In our first proof, we had $K = B = \epsilon = 1$.) If $N \geqslant B$, and $N\delta \geqslant A$, then: \begin{gather*} \bigg\lvert \sum_{\abs{n} > N} f(n\delta)\delta \bigg\rvert \leqslant \sum_{\abs{n} > N} \abs{f(n\delta)}\delta \leqslant \frac{2M}{\delta^\epsilon}\!\!\!\sum_{n=N+1}^\infty \frac{1}{n^{1 + \epsilon}} < \frac{2MK}{(N\delta)^\epsilon}. \end{gather*} Accordingly, we define $N(\delta)$ by the equation: $$ (N(\delta)\delta)^\epsilon = -2MK\log\delta. $$

From $\eqref{eq:B}$ and $\eqref{eq:J}$, what we now have to prove is, the same as before: \begin{equation} \lim_{\delta \to 0+} \sum_{\abs{n} \leqslant N(\delta)} f(n\delta)\delta = \int_{-1}^1 F. \tag{4}\label{eq:K} \end{equation}

Exactly as before, we use Taylor's Theorem to get an expression of the form: $$ \sum_{\abs{n} \leqslant N(\delta)} f(n\delta)\delta = I(\delta) + J(\delta). $$

The proof that $$ \lim_{\delta \to 0+} I(\delta) = \int_{-1}^1 F $$ is also exactly the same as before. We are reduced, as before, to proving that: $$ \lim_{\delta \to 0+} J(\delta) = 0, $$ or in full: $$ \lim_{\delta \to 0+} \sum_{\abs{n} \leqslant N(\delta)} f(\phi(y_n))\frac{\phi''(y_n^*)}{2}(y_{n+1} - y_n)^2 = 0. $$

The next part of the argument has changed slightly from the earlier version, which is why it has to be repeated in detail:

By our hypotheses, $f$ is integrable, and therefore bounded, on the interval $[-A - \delta, A + 2\delta]$, therefore the factor $f(\phi(y_n))\phi''(y_n^*)$ is bounded for $n$ such that: $$ -\phi^{-1}(A + \delta) \leqslant y_n < y_n^* < y_{n+1} \leqslant \phi^{-1}(A + 2\delta), $$ or equivalently, $$ -A - \delta \leqslant n\delta < \phi(y_n^*) < (n + 1)\delta \leqslant A + 2\delta. $$ Such terms therefore contribute at most a fixed multiple of $\sum_n (y_{n+1} - y_n)^2$ to the absolute value of the summation; and because $\lim_{\delta \to 0+} \max_n (y_{n+1} - y_n) = 0$, and $\sum_n (y_{n+1} - y_n) < 2$, this part of the sum tends to $0$ in the limit as $\delta \to 0$.

What now remains to be proved is: \begin{equation} \lim_{\delta \to 0+} \sum_{(A + \delta)/\delta \leqslant \abs{n} \leqslant N(\delta)} f(\phi(y_n))\frac{\phi''(y_n^*)}{2}(y_{n+1} - y_n)^2 = 0. \tag{$5'$}\label{eq:L} \end{equation} For such $n$, we have $\abs{\phi(y_n)} = \abs{n\delta} \geqslant A$, therefore: $$ \abs{f(\phi(y_n))\frac{\phi''(y_n^*)}{2}} \leqslant \frac{M\abs{\phi''(y_n^*)}}{2\abs{\phi(y_n)}^{1 + \epsilon}}. $$ We also have $\abs{\phi(y_n^*)} \geqslant A$. (That was the reason for the finicky change in the argument: we replaced $A$ with $A + \delta$, in order to get this inequality.) We can suppose that $A \geqslant 1$, therefore $\abs{\phi(y_n^*)} \geqslant 1$.

We now estimate $\abs{\phi''(y_n^*)}$ in terms of $\abs{\phi(y_n^*)}$.

I claim that if $\abs{\phi(y)} \geqslant 1$, then: $$ \abs{\phi''(y)} \leqslant \frac{(1 + \epsilon)2^{1 + 2\epsilon}}{\epsilon^2} \abs{\phi(y)}^{1 + 2\epsilon}. $$ (Obviously this is a much poorer bound than we got before for the case $\epsilon = 1$, but the multiplying constant doesn't matter.)

Proof: $\phi$ and $\phi''$ are odd functions, and $\phi'$ is an even function, so our previous expressions for these functions can be rewritten as: \begin{gather*} \abs{\phi(y)} = \frac{1}{(1 - \abs{y})^{\frac{1}{\epsilon}}} - \frac{1}{(1 + \abs{y})^{\frac{1}{\epsilon}}}, \\ \phi'(y) = \frac{1}{\epsilon}\left[ \frac{1}{(1 - \abs{y})^{\frac{1}{\epsilon} + 1}} + \frac{1}{(1 + \abs{y})^{\frac{1}{\epsilon} + 1}} \right] \geqslant \frac{2}{\epsilon}, \\ \abs{\phi''(y)} = \frac{1}{\epsilon}\left(\frac{1}{\epsilon} + 1\right)\left[ \frac{1}{(1 - \abs{y})^{\frac{1}{\epsilon} + 2}} - \frac{1}{(1 + \abs{y})^{\frac{1}{\epsilon} + 2}} \right]. \end{gather*} (The separate inequality for $\phi'(y)$ will be used shortly.) Therefore: \begin{gather*} \frac{(1 + \abs{y})^{\frac{1}{\epsilon}}} {(1 - \abs{y})^{\frac{1}{\epsilon}}} - 1 = (1 + \abs{y})^{\frac{1}{\epsilon}}\abs{\phi(y)} \geqslant 1, \ \ \therefore\ \frac{(1 + \abs{y})^{\frac{1}{\epsilon}}} {(1 - \abs{y})^{\frac{1}{\epsilon}}} \geqslant 2, \\ \therefore\ \frac{1}{(1 + \abs{y})^{\frac{1}{\epsilon}}} \leqslant \frac{1}{2(1 - \abs{y})^{\frac{1}{\epsilon}}}, \ \ \therefore\ \abs{\phi(y)} \geqslant \frac{1}{2(1 - \abs{y})^{\frac{1}{\epsilon}}}, \\ \therefore\ \abs{\phi''(y)} \leqslant \frac{1}{\epsilon}\left(\frac{1}{\epsilon} + 1\right) \frac{1}{(1 - \abs{y})^{\frac{1}{\epsilon} + 2}} \leqslant \frac{1}{\epsilon}\left(\frac{1}{\epsilon} + 1\right) (2\abs{\phi(y)})^{1 + 2\epsilon}, \end{gather*} as required.

Now, $\abs{n} \geqslant 1 + A/\delta \geqslant 2$, therefore $1 + 1/\abs{n} \leqslant 3/2$, and: \begin{gather*} \abs{f(\phi(y_n))\frac{\phi''(y_n^*)}{2}} \leqslant \frac{(1 + \epsilon)2^{1 + 2\epsilon}M\abs{\phi(y_n^*)}^{1 + 2\epsilon}} {2\epsilon^2\abs{\phi(y_n)}^{1 + \epsilon}} \\ \leqslant \frac{(1 + \epsilon)2^{1 + 2\epsilon}M((\abs{n} + 1)\delta)^{1 + 2\epsilon}} {2\epsilon^2(\abs{n}\delta)^{1 + \epsilon}} = \frac{(1 + \epsilon)2^{1 + 2\epsilon}M(\abs{n}\delta)^\epsilon} {2\epsilon^2} \left(1 + \frac{1}{\abs{n}}\right)^{1 + 2\epsilon} \\ \leqslant \frac{(1 + \epsilon)3^{1 + 2\epsilon}M(N(\delta)\delta)^\epsilon} {2\epsilon^2} = -\frac{(1 + \epsilon)3^{1 + 2\epsilon}M^2K\log\delta}{\epsilon^2}. \end{gather*} But, as was noted a moment ago: $$ \phi'(y) \geqslant \frac{2}{\epsilon} \ \ (\abs{y} < 1), $$ therefore: $$ y_{n + 1} - y_n = \phi^{-1}((n + 1)\delta) - \phi^{-1}(n\delta) \leqslant \frac{\epsilon\delta}{2} \ \ (n \in \Z). $$ By our previous argument, it follows that the sum in $\eqref{eq:L}$ is bounded above by $$ -\frac{(1 + \epsilon)3^{1 + 2\epsilon}M^2K\delta\log\delta}{2\epsilon} $$ which tends to $0$ with $\delta$, as required. This completes the proof of $\eqref{eq:A}$.

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  • $\begingroup$ I wonder if this argument can be generalised, to show that the improper Riemann integral over $\mathbb{R}$ of any function of moderate decrease $f: \mathbb{R} \to \mathbb{R}$ is the limit of a net of separately convergent infinite "Riemann sums" or "Darboux sums", each defined over an infinite partition of $\mathbb{R}$? If so, it's probably not a difficult modification to make ($\delta$ becomes the mesh of a partition that is no longer necessarily uniform), but this isn't the place to write it up, now that the stated question has been pretty much answered (barring errors). $\endgroup$ – Calum Gilhooley Feb 23 '15 at 1:55

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