Is it 0 times infinity?

Calculus Level 4

A = lim ⁡ n → 0 ∫ − ∞ + ∞ sin ⁡ ( n x ) x d x A = \displaystyle \lim_{n \to 0} \int_{-\infty}^{+\infty} \dfrac{\sin(nx)}{x} \, dx

Calculate the limit of the Riemann integral , A A .

π \pi 0 − π - \pi A A does not exist

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Hobart Pao by Hobart Pao , 23, USA

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1 solution

We "know" (for instance, with complex analysis) that ∫ − ∞ ∞ sin ⁡ ( x ) x d x = π ⇒ \int_{-\infty}^{\infty} \frac{\sin(x)}{x} \, dx = \pi \Rightarrow lim ⁡ n → 0 + ∫ − ∞ ∞ sin ⁡ ( n x ) x d x = \displaystyle \lim_{n \to 0^{+}} \int_{-\infty}^{\infty} \frac{\sin(nx)}{x} \, dx = Suppose now n > 0 n > 0 a real number, fixed but arbitrary, applying u-substitution n x = t , → d t = n d x nx = t, \space \rightarrow dt = n dx lim ⁡ n → 0 + ∫ − ∞ ∞ sin ⁡ ( t ) t d t = π \displaystyle \lim_{n \to 0^{+}} \int_{-\infty}^{\infty} \frac{\sin(t)}{t} \, dt = \pi . In the same way, lim ⁡ n → 0 − ∫ − ∞ ∞ sin ⁡ ( n x ) x d x = \displaystyle \lim_{n \to 0^{-}} \int_{-\infty}^{\infty} \frac{\sin(nx)}{x} \, dx = lim ⁡ n → 0 − ∫ ∞ − ∞ sin ⁡ ( t ) t d t = − π \displaystyle \lim_{n \to 0^{-}} \int_{\infty}^{-\infty} \frac{\sin(t)}{t} \, dt = -\pi . Therefore, two-sided limits are not equal and this implies that A A doesn't exist.

Note.- If someone wishes a proof (with complex analysis) of ∫ − ∞ ∞ sin ⁡ ( x ) x d x = π \int_{-\infty}^{\infty} \frac{\sin(x)}{x} \, dx = \pi , he/she can ask to me. The proof is based on a corollary of theorem of residues.

5 Helpful 0 Interesting 0 Brilliant 1 Confused

Excellent! That's exactly what I saw. Just fix your LaTeX, little bugs here and there but the idea is there.

Hobart Pao - 4 years, 7 months ago

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Thank you, I'll try to "fix" and get better my LaTex... :) If someone has some doubt or question, please tell me...

Guillermo Templado - 4 years, 7 months ago

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Yeah, for trig functions, use \sin x rather than sin x. It will look nicer. And I think you meant to put an arrow somewhere and there are some unnecessary lines. But I'm just being ridiculously picky.

Hobart Pao - 4 years, 7 months ago

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@Hobart Pao – ok, I'm going to do \sin x...

Guillermo Templado - 4 years, 7 months ago

I don't think there's a need to use heavy machinery like complex analysis for showing that ∫ − ∞ ∞ sin ⁡ x x d x = π \int\limits_{-\infty}^\infty\frac{\sin x}x\,\mathrm dx=\pi . We can just use Laplace Transform along with the fact that the integral is an even function. Denoting by L \mathfrak L the Laplace Transform, we have,

∫ − ∞ ∞ sin ⁡ x x d x = 2 ∫ 0 ∞ sin ⁡ x x d x = 2 L { sin ⁡ x x ; 0 } = 2 ∫ 0 ∞ L { sin ⁡ x ; p } d p = 2 ∫ 0 ∞ 1 p 2 + 1 d p = 2 arctan ⁡ p ∣ 0 ∞ = 2 ( π / 2 − 0 ) = π \begin{aligned}\int\limits_{-\infty}^\infty\frac{\sin x}{x}\,\mathrm dx&=2\int\limits_0^\infty\frac{\sin x}{x}\,\mathrm dx\\&=2\mathfrak L\left\{\frac{\sin x}x;0\right\}\\&=2\int\limits_0^\infty\mathfrak L\{\sin x;p\}\,\mathrm dp\\&=2\int\limits_0^\infty\frac{1}{p^2+1}\,\mathrm dp=2\arctan p|_0^\infty=2(\pi/2-0)=\pi\end{aligned}

Prasun Biswas - 4 years, 6 months ago

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There are a lot of methods for proving this.... Anyway, I think Laplace transform is not light machinery either...

Guillermo Templado - 4 years, 6 months ago

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