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Logically and by method 1 the limit should be undefined, but with some juggling it comes out to be $1$.

  • Method 1. $\displaystyle \lim_{k\to\infty} \int_0^k \sin(x) \, dx = -\lim_{k\to\infty} (\cos(k)-1) = \text{not defined}$.

  • Method 2. Let $I = \int e^{-tx}\sin(x) \, dx$ and $J=\int e^{-tx}\cos(x) \, dx$. Using integration by parts,

    \begin{align*} I &= -e^{-tx}\cos x - tJ, \tag{i} \\ J &= e^{-tx}\sin x + tI \tag{ii} \end{align*}

    from $\text{(i)}$ and $\text{(ii)}$,

    $$ I = -e^{-tx} \left[ \frac{\cos x + t\sin x}{1+t^2} \right], \qquad J = e^{tx}\left[ \frac{\sin x-t\cos x}{1+t^2} \right]. $$

    Thus $\int_0^\infty e^{-tx}\sin(x) \, dx = \frac{1}{1+t^2}$. Taking limit $t \to 0$

    $$ \lim_{t\to 0}\int_{0}^{\infty} e^{-tx}\sin(x) \, dx = \int_{0}^{\infty} \sin(x) \, dx = 1. $$

Is the integral $1$ or undefined?

skyking
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varang rai
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  • What is the question? – Angina Seng Oct 04 '17 at 06:21
  • @LordSharktheUnknown: is the integral 1 or undefined. – varang rai Oct 04 '17 at 06:22
  • @varangrai It is undefined. – jonsno Oct 04 '17 at 06:25
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    There is no question (not even "where is the mistake?", because the last line is crying "here!"). –  Oct 04 '17 at 06:26
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    It does not converge in improper integral sense, but it has value $1$ in Abel summability sense. It is not surprising that your methods yield different answers as they represent different summability results. – Sangchul Lee Oct 04 '17 at 06:34
  • $\begin{array}{l}t\rightarrow0;can;be;replaced;by;t=0;if;the;function;is;defined;is;defined;\at;t=0,;which;in;this;case;is;as;f(t)=\int_0^\infty(e^{-tx}\sin x;dx);=;\frac1{1+t^2}\end{array}$ – varang rai Oct 04 '17 at 06:53
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    Your comment is true, but the caveat is that while $\frac{1}{1+t^2}$ is defined on all of $t\in\mathbb{R}$, the integral $\int_{0}^{\infty}e^{-tx} \sin x , dx$ is defined only for $t > 0$. So they are different functions which happen to coincide on the set $(0, \infty)$ of positive reals. This is in a similar spirit as the geometric series $$ \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} $$ where the left-hand side converges only for $|x| < 1$ but the right-hand side is defined for all $x$ except $x = 1$. You cannot simply plug $x = -1$ to the sum to claim $1-1+1-1+\cdots = \frac{1}{2}$. – Sangchul Lee Oct 04 '17 at 07:19
  • As a side note, your second method is basically an Abel sum of the integral, which has the analogous:$${\small A}\sum_{n=0}^\infty a_n=\lim_{x\to0^+}\sum_{n=0}^\infty a_ne^{-nx}$$ – Simply Beautiful Art Oct 04 '17 at 19:39
  • Oops, the above was already mentioned by @SangchulLee, but I'm going to leave it since it has references and displays the general definition. – Simply Beautiful Art Oct 04 '17 at 20:29
  • It seems that my question already has an answer here https://math.stackexchange.com/questions/253696/can-a-limit-of-an-integral-be-moved-inside-the-integral. Thanks for your efforts.;) – varang rai Oct 05 '17 at 04:20
  • @SangchulLee $1-1+1-1+\ldots =\frac12$ in the sense of Cesáro mean – Hagen von Eitzen Oct 05 '17 at 11:53
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    @HagenvonEitzen, Of course that is true. But I can't see how it is related to my discussion as I was talking about the ordinary summability defined as the limit of partial sums. My point was that we should not mix different summability methods. – Sangchul Lee Oct 05 '17 at 11:57
  • Taking the Cesáro mean for integrals, you are basically looking at:$$\lim_{n\to\infty}\frac1n\int_0^n\int_0^t\sin(x)~\mathrm dx~\mathrm dt$$which happens to be $1$ @HagenvonEitzen – Simply Beautiful Art Oct 06 '17 at 00:08
  • From this we have (https://math.stackexchange.com/q/794038) $$ \int_0^\infty \frac{\sin x}{x^s} dx = \Gamma(1-s) \cos(\frac\pi 2 s) ~~~~0<s<2$$

    By letting $s\to 0$ we get; $$ \int_0^\infty \sin x dx =1$$

     – Guy Fsone Nov 27 '17 at 15:46

3 Answers3

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In your opinion, is it true that $$\lim_{x\to +\infty}\left(\lim_{t\to 0^+} e^{-tx}\sin(x)\right)= \lim_{t\to 0^+}\left( \lim_{x\to +\infty}e^{-tx}\sin(x)\right)\quad ?$$ Switching the order of limits could be dangerous...

See your last line. Are you sure that $$\lim_{t\to 0^+}\left(\lim_{r\to +\infty}\int_0^r e^{-tx}\sin(x)dx\right)=\lim_{r\to +\infty}\left(\lim_{t\to 0^+}\int_0^r e^{-tx}\sin(x)dx\right)\quad ?$$

Robert Z
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In your last line what you actually make is

$$ \lim_{t\to 0}\int_{0}^{\infty} e^{-tx}\sin(x) \, dx = \int_{0}^{\infty} \lim_{t\rightarrow 0}e^{-tx}\sin(x) \, dx$$

This last step is only allowed if the convergence is uniform. In a sloppy language this means that the "size" of $\sin(x)-e^{-tx} \sin(x)$ is "independent" of x. Which is not the case, that's why your result is wrong.

whoisit
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Jürg W. Spaak
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    "This last step is only allowed if the convergence is uniform." False, it can hold true if the convergence is pointwise. Uniform convergence is a sufficient but unnecessary condition for bringing a limit into an integral. – Simply Beautiful Art Oct 06 '17 at 00:09
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The phenomenon is related to re-summation methods which is a outside my competences, but here is a go:

The first integral $M(u) = \int_0^u \sin x \; dx = 1 -\cos (u)$ indeed does not converge but you may take a Cesàro mean to get a limiting average value: $C(t) = \frac{1}{t} \int_0^T M(u)\; du \rightarrow 1$, as $t\rightarrow \infty$. Unwinding the double integral involed you have: $$ C(t) = \int_0^t \left( 1 - \frac{x}{t}\right) \sin(x) \; dx = \int_0^\infty \left[ (1-x/t) {\bf 1}_{[0,t]}(x) \right] \;\sin(x) \; dx$$ The factor $(1-x/t) {\bf 1}_{[0,t]}(x)$ goes weakly to one as $t\rightarrow \infty$.

This is similar to the second method in which you look at the average of $f(x)=\sin x$ weighted by $e^{-xt}$ which also goes weakly to $1$ as $t\rightarrow 0$. You may certainly make other choices that would give a different limit so one question may be if there is a natural family of weights for which the limiting average is unique? (I don't have an answer).

H. H. Rugh
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