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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 $\eqref{eq:A}$.
