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The limit $$\lim_{a \to \infty} \int_0^a \sin x \, \mathrm{d} x$$ does not exist. However, consider that $$ \lim_{\epsilon \to 0} \int_0^\infty e^{- \epsilon x} \sin x \, \mathrm{d} x = 1 \,.$$ Here I have 'regulated' the integral. What I discovered, and what strikes me as very surprising, is that if, instead of an exponential, I choose a different function $f(x, \epsilon)$ which tends pointwise to $1$ as $\epsilon$ goes to $0$ and tends to $0$ as $x$ goes to $\infty$, I get convergence to exactly the same limit. So if I choose $$ f(x, \epsilon) = \frac{1}{1 + \epsilon x^2} \quad \text{or} \quad \mathrm{sech}^2(\epsilon x) \quad \text{or} \quad (1 + 2 \epsilon x^2) e^{-\epsilon x^2}\,,$$ Then the integral of $f(x, \epsilon) \sin x$ from 0 to $\infty$ tends to $1$ as $\epsilon$ tends to $0$.

Why is this happening?

EDIT: I was initially satisfied with the responses given, but on further thought I don't think I follow the logic of tired's answer, which invokes the stationary phase approximation.

In particular, my understanding of the stationary phase approximation is that one looks for stationary points of the argument of the exponential since these correspond to the points where the oscillation is slowest – away from this point, the oscillations 'cancel out' because of how rapid they are. However, in this case the argument of the exponential has no stationary points. Further, whilst I can appreciate that (in the case where there are no stationary points) a 'boundary maximum point' would dominate the integral in the real case (that is, for which the argument of the exponential is real), I can't see that this would be relevant in the imaginary case.

I am looking for an answer that includes the following three points:

  • A proof that the limit of this integral is independent of regulator (for a suitable class of regulator).
  • Some intuition as to why we might expect this particular integral to be independent of regulator.
  • Information on whether there is some general theory about assigning, perhaps uniquely, a value to non-convergent integrals. In particular, I would like to know whether the fact that the integral at the top of this question is 'almost convergent' (in the sense that it is bounded for all $a$) makes it easier to unambiguously regulate.
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  • $\begingroup$ note that a gaussian regulator also works $\endgroup$
    – tired
    Oct 14, 2016 at 17:22
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    $\begingroup$ That probably depends on the fact that the averaged version of $$\int_{0}^{a}\sin(x)\,dx = 1-\cos(a),$$ i.e. $$\frac{1}{a}\int_{0}^{a}(1-\cos b)\,db = 1-\frac{\sin a}{a},$$ tends to $1$ as $a\to +\infty$. So my bet is on a side-effect of integration by parts. $\endgroup$ Oct 14, 2016 at 17:29
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    $\begingroup$ @JackD'Aurizio Cesaro... integral? $\endgroup$ Jan 27, 2017 at 0:23

3 Answers 3

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Here is a partial answer that seconds @Jack D'Aurizio's idea.


Let us focus on the class of locally integrable functions $f : [0,\infty) \to \Bbb{R}$ such that

  • $f(0) = 1 = \lim_{x \to 0^+} f(x)$, and
  • $ I(f, \epsilon) := \lim_{R\to\infty} \int_{0}^{R} f(\epsilon x) \sin x \, dx$ converges for all $\epsilon > 0$.

For appropriate choices of $f$, this covers many classical summability methods:

  1. If $f(x) = \mathbf{1}_{[0,1]}(x)$, then $I(f,\epsilon) = \int_{0}^{1/\epsilon} \sin x \, dx$ corresponds to the ordinary definition of improper integrability.

  2. If $f(x) = (1 - x)_+$, then $I(f, \epsilon) = \epsilon \int_{0}^{1/\epsilon} \int_{0}^{x} f(t) \, dtdx $ corredponds to the Cesaro summability.

  3. If $f(x) = e^{-x}$, then $I(f, \epsilon) = \int_{0}^{\infty} e^{-\epsilon x} \sin x \, dx$ corresponds to the Abel summability.

It also covers all the examples you have discussed. We can simply choose $f(x)$ to be

$$ \frac{1}{1+x^2}, \quad \operatorname{sech}^2 x, \quad (1 + 2x^2)e^{-x^2}, \quad \cdots. $$

Since we know that 1 fails to converge while 2 and 3 converge, we may suspect that regularity of $f$ is one factor which determines how regularly $I(f,\epsilon)$ behaves. Toward this direction, let us make the following assumption:

Assumption 1. We have $f(x) = \int_{x}^{\infty} \rho(t) \, dt$ for some $\rho$ which is integrable on $[0, \infty)$.

Under this condition, let us show that $I(f, \epsilon)$ converges to $1$. Indeed, by the Fubini's theorem, we have

\begin{align*} \int_{0}^{R} f(\epsilon x) \sin x \, dx &= \int_{0}^{R} \left( \int_{\epsilon x}^{\infty} \rho(t) \, dt \right) \sin x \, dx \\ &= \int_{0}^{\infty} \left( \int_{0}^{(t/\epsilon) \wedge R} \sin x \, dx \right) \rho(t) \, dt \\ &= \int_{0}^{\infty} [ 1 - \cos ((t/\epsilon) \wedge R) ] \rho(t) \, dt. \end{align*}

Now we can apply the dominated convergence theorem to the last integral to find that $I(f, \epsilon)$ converges and is given by

$$ I(f, \epsilon) = \int_{0}^{\infty} ( 1 - \cos (t/\epsilon) ) \rho(t) \, dt. $$

Taking $\epsilon \to 0^+$, Riemann-Lebesgue lemma tells that the last integral converges to:

$$ \lim_{\epsilon \to 0^+} I(f, \epsilon) = \int_{0}^{\infty} \rho(t) \, dt = f(0^+) = 1. $$


This proof already hints us that this regularization captures the notion of the average of the antiderivative $\int_{0}^{x} \sin t \, dt = 1 - \cos x$. Indeed, with a bit of extra assumption, we can check that this regularization yields the integral of a nice function in Cesaro summability sense. Let us formalize this claim.

Assumption 2.

  1. $f$ satisfies Assumption 1 with $\rho : [0, \infty) \to \Bbb{R}$ which is of bounded variation and satisfies $\int_{[0,\infty)} x |d\rho(x)| < \infty$.
  2. $\varphi : [0,\infty) \to \Bbb{R}$ is locally integrable such that the antiderivative $\Phi(x) = \int_{0}^{x} \varphi(t) \, dt$ is bounded and has Cesaro-mean $\ell \in \Bbb{R}$, i.e., $\frac{1}{R} \int_{0}^{R} \Phi(x) \, dx \to \ell$ as $R \to \infty$.

Although this assumption seems technical, it loosely tells that $\rho$ does not oscillate heavily and that $\Phi$ is bounded and has an average.

As in the previous proof, we find that

$$ \int_{0}^{R} f(\epsilon x) \varphi(x) \, dx = \int_{0}^{\infty} \Phi((t/\epsilon)\wedge R) \rho(t) \, dt \xrightarrow[R\to\infty]{} \int_{0}^{\infty} \Phi(t/\epsilon) \rho(t) \, dt. $$

Now from the assumption that $\rho$ is of bounded variation, we easily find that $\rho(t) \to 0$ as $t \to \infty$. Thus by the Fubini's theorem again,

\begin{align*} \int_{0}^{\infty} \Phi(t/\epsilon) \rho(t) \, dt &= \int_{0}^{\infty} \Phi(t/\epsilon) \left( - \int_{(t,\infty)} d\rho(u) \right) \, dt \\ &= \int_{(0,\infty)} \left( \int_{0}^{u} \Phi(t/\epsilon) \, dt \right) \, d\rho(u) \\ &= \int_{(0,\infty)} \left( \frac{\epsilon}{u} \int_{0}^{u/\epsilon} \Phi(t') \, dt' \right) u \, d\rho(u) \end{align*}

where $t' = t/\epsilon$. Then by the assumption, we can invoke the dominated convergence theorem to find that the last integral converges to:

$$ \lim_{\epsilon \to 0^+} \int_{(0,\infty)} \left( \frac{\epsilon}{u} \int_{0}^{u/\epsilon} \Phi(t') \, dt' \right) u \, d\rho(u) = \int_{(0,\infty)} \ell u \, d\rho(u) = \ell. $$

Putting together, we have

$$ \lim_{\epsilon \to 0^+} \int_{0}^{\infty} f(\epsilon x) \varphi(x) \, dx = \ell = \lim_{R\to\infty} \frac{1}{R}\int_{0}^{R} \int_{0}^{x} \varphi(t) \, dtdx. \tag{*}$$

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  • $\begingroup$ Thank you very much for the reply. Quick, naive question: what do you mean by $\wedge$ when invoking Fubini's theorem? $\endgroup$
    – gj255
    Jan 23, 2017 at 18:46
  • $\begingroup$ @gj255, Oh, it is the minimum. For instance, $x \wedge y = \min\{x, y\}$. $\endgroup$ Jan 24, 2017 at 0:01
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    $\begingroup$ This is not a (+1), it is a $\large{(+1)}$. $\endgroup$ Jan 27, 2017 at 0:43
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Let $f(\epsilon x)$ a function which decays to zero (to be safe, i think it is sufficent if it can be also fastly oscillating) as $x\rightarrow \infty$ with $f(0)=1$. Furthermore uit should not contain any singular points. Then we have

$$ F(\epsilon)=\int_0^{\infty}\sin(x)f(x\epsilon)\, \mathrm dx= \frac{1}{\epsilon}\int_0^{\infty}\sin\left({\frac{x}{\epsilon}}\right)f(x)\, \mathrm dx=\frac{1}{\epsilon}\Im\int_0^{\infty}\exp\left({\frac{i x}{\epsilon}}\right)f(x) \, \mathrm dx $$

if $\epsilon$ goes to zero this is a classic example for a integral which might be tackled by the method of stationary phase and the dominating contribution will be from the boundary point at $x=0$ (see formula 2.22 in the given reference)

$$ F(\epsilon)\underset{\epsilon \to 0}{\sim}\frac{1}{\epsilon}\Im\left(\epsilon\frac{f(0)}{-i} \right)+\mathcal{O}(\epsilon)=1+\mathcal{O}(\epsilon) $$

This is because heuristically the heaviy oscillations tend to cancel out each other (but to get them you need an additional small parameter $\epsilon$)

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  • $\begingroup$ Thank you for the answer. The only thing I don't quite understand is why the dominant contribution comes from $x = 0$. If the function $f(x)$ had a maximum away from $0$, e.g. $f(x) = (1 + 2x^2)\exp(-x^2)$, would the limit be different? $\endgroup$
    – gj255
    Oct 16, 2016 at 21:24
  • $\begingroup$ heuristically as $\epsilon \rightarrow 0$ any maximum of $f(x \epsilon)$ will shifted so close to the origin that it is well approximated as $f(0)$ so there shouldn't be a change in the leading order expansion $\endgroup$
    – tired
    Oct 17, 2016 at 11:15
  • $\begingroup$ OK, but we've moved the $\epsilon$ out of $f$ and into the exponential. I can't see why the point $0$ is special. $\endgroup$
    – gj255
    Oct 17, 2016 at 12:03
  • $\begingroup$ because $x=0$ is the only point where$\exp(i x /\epsilon)$ doesn't osicllate and no cancelations occur $\endgroup$
    – tired
    Oct 17, 2016 at 12:06
  • $\begingroup$ What do you mean, doesn't oscillate? $\mathrm{d}/\mathrm{d}x \, \exp (i x/ \epsilon) |_{x = a} = i/\epsilon \exp(i a / \epsilon)$. This has the same magnitude for all $a$, so it seems to me as though this exponential oscillates equally fast everywhere. $\endgroup$
    – gj255
    Oct 17, 2016 at 12:11
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Suppose $f\in C^1([0,\infty),$ $\lim_{x\to \infty} f(x) = 0,$ and both $f,f' \in L^1([0,\infty)).$ Then

$$\tag 1 \lim_{\epsilon\to 0} \int_0^\infty f(\epsilon x) \sin x\, dx = f(0).$$

Proof: Integrating by parts shows the integral equals

$$f(\epsilon x)(-\cos x)\big |_0^\infty + \int_0^\infty \epsilon f'(\epsilon x) \cos x\, dx = f(0) + \int_0^\infty f'(y) \cos (y/\epsilon)\, dy.$$

As $\epsilon \to 0,$ the last integral $\to 0$ by the Riemann-Lebesgue lemma. This gives $(1).$

Remarks: 1. In the cases $f(x) = e^{-x}, 1/(1+x^2), \mathrm {sech}^2 x, (1+2x^2)e^{-x^2},$ $(1)$ gives the results you mentioned (although with $\sqrt \epsilon$ in place of $\epsilon$ for the second and fourth functions).

  1. The hypothesis $\lim_{x\to \infty} f(x) = 0$ is not really needed, because it's implied by the other hypotheses. This is because $f'\in L^1$ implies $f$ is uniformly continuous, and a uniformly continuous function in $L^1([0,\infty))$ must vanish at $\infty.$ I included the $f(x)\to 0$ hypothesis to keep the proof simple.

$(1)$ can be generalized: Keep the hypotheses on $f$ as above, and assume $G\in C^1([0,\infty)),$ with both $G,G'$ bounded. Then

$$\tag 2 \lim_{\epsilon\to 0} \int_0^\infty f(\epsilon x) G'(x)\, dx = -G(0)f(0).$$

For example, taking $G(x) = - \cos x$ gives $(1).$ The proof of $(2)$ is very much like that of $(1);$ we need an analogue of Riemann-Lebesgue, but that's straightforward.

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  • $\begingroup$ Thanks for the answer. My intuition for the step invoking the Riemann-Lebesgue lemma is that the cosine function oscillates so rapidly in this limit that, provided $f$ is continuous, the integral over several oscillations is essentially just the integral of the cosine multiplied by a constant, which is zero. Thus the integral over the whole interval is zero. Is this intuition correct? Can I ask also whether this step has any connection to the stationary phase approximation? $\endgroup$
    – gj255
    Jan 27, 2017 at 12:36

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