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Is it possible to solve the following integral:

$$\lim\limits_{c \to +\infty} \displaystyle\int_{1}^{1/c} \dfrac{\sin u}{u} \, du$$

using "elementary" methods? By "elementary", I mean those methods that do not involve Complex analysis, Lebesgue Integration, etc (basically, anything beyond an elementary first course in Real Analysis, say, from the first six chapters of Baby Rudin).

I've seen many solutions to this integral (seemingly with different bounds, including the Dirichlet integral), but all of them seem to use methods that would generally not be accessible to someone with just a basic real analysis course.

Arctic Char
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Ricky_Nelson
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    Are any of these particularly useful? -- https://math.stackexchange.com/questions/5248/evaluating-the-integral-int-0-infty-frac-sin-x-x-mathrm-dx-frac-pi/ – PrincessEev Aug 10 '20 at 04:02
  • @EeveeTrainer Unfortunately, no. None of the answers in that link seem to be based just on elementary methods. – Ricky_Nelson Aug 10 '20 at 04:09
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    Did you intend for the lower bound to be $0$, or perhaps $1/c$ if you want to avoid the hole at $0$? – Brian Moehring Aug 10 '20 at 04:13
  • @BrianMoehring There was a typo, which is now fixed. – Ricky_Nelson Aug 10 '20 at 04:24
  • That broke the problem even worse... Since you seem to have expressed interest in $1/c$, I've rewritten it with a lower bound of $1/c$ (though it fairly simply can be replaced with a lower bound of $0$) – Brian Moehring Aug 10 '20 at 04:26
  • @BrianMoehring Perhaps I was wrong it calling the integral I am trying to solve "Dirichlet" integral. But, the bounds of the integral so not have a typo now. It is the integral I am trying to solve. – Ricky_Nelson Aug 10 '20 at 04:29
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    The integral has been changed many times. When I left it, it would evaluate to $\pi/2$. At this point, it would evaluate to $-\text{Si}(1)$ which has no way (as far as I know) to be written in terms of more elementary functions and/or constants. If the definition of $\text{Si}(x)$ is allowed, then this is very elementary. Otherwise, it's not, but we're so far afield from where we started, I have no clue what your question actually is... – Brian Moehring Aug 10 '20 at 04:47
  • @BrianMoehring I have not been editing the integral. The current bounds on the integral seem to be correct, though. I think the question is also resolved as well since I think the Si function is elementary enough. – Ricky_Nelson Aug 10 '20 at 04:58
  • Isn't the method of solving it by Feynman's technique quite elementary? – V.G Aug 10 '20 at 06:03
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    As it stands right now, it is trivial. [$\lim_{c\to+\infty}f(1/c)=f(0)$ for a continuous $f$, which is surely our case.] – metamorphy Aug 10 '20 at 13:41
  • @Ricky_Nelson For $\int\limits_0^\infty \frac{\sin x} x ,\mathrm dx = \frac \pi 2$, Robin Chapman's solution is elementary, in the link given by Eevee Trainer. – River Li Aug 10 '20 at 15:35

2 Answers2

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Note

\begin{align} \int^{\infty}_0 \dfrac{\sin u}{u} \, du &= \int^{\infty}_0 {\sin u}\left(\int_0^\infty e^{-ut}dt\right) du \\ &= \int^{\infty}_0 \left(\int_0^\infty \sin u e^{-ut} du\right)dt\\ & = \int^{\infty}_0 \frac1{1+t^2} dt =\dfrac{\pi}{2} \end{align}

Quanto
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  • @BrianMoehring nevermind, I just saw that OP equated it to $\pi/2$, therefore OP most likely made a typo. – Aniruddha Deb Aug 10 '20 at 04:24
  • You are using Fubini's theorem which is from multivariable calculus, but I can't think of any example simpler than this. – Otomeram Aug 10 '20 at 04:31
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    How do you justify swapping the integrals here? The form of Fubini's I'm aware of would require $\int_0^\infty \frac{|\sin x|}{x},dx$ to be finite, but it's not. – Brian Moehring Aug 10 '20 at 04:33
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    @BrianMoehring the integral can be taken up to some finite $R$, and then afterwards justify the limit to infinity. – Ninad Munshi Aug 10 '20 at 04:56
  • @NinadMunshi Thanks. The additional work looks fairly ugly at the moment, so I would probably spring for dominated convergence or monotone convergence to make it easier, but I guess there might be an "elementary" way to deal with it. – Brian Moehring Aug 10 '20 at 05:07
  • @BrianMoehring: This works (+1). You can justify swapping these integrals with this. – RRL Aug 17 '20 at 23:34
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$$\lim_{c\rightarrow\infty^+}\int_1^\frac{1}{c}\frac{\sin(u)}{u}du = \lim_{t\rightarrow0^+}\int_1^t\frac{\sin(u)}{u}du = -\int_0^1\frac{\sin(u)}{u}du = -Si(1)$$

This is the sine integral

Moko19
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