Why is $\int_{-\infty}^\infty |\text{sinc}(t)|\ dt=\infty$?

Question

Can you help me with that:

Why is $\int_{-\infty}^\infty |\text{sinc}(t)|\ dt=\infty$ ?

I saw this in Signal Processing course and I can’t understand why this is true.

Reference: https://dsp.stackexchange.com/a/1033/49921

Didn’t find any relevant material unfortunately.

score 3 · Accepted Answer · edited Apr 18 '20 at 18:39

3

$$ \int_0^{+\infty}\frac{|\sin(t)|}{t}dt=\sum_{n=0}^{+\infty}\int_{n\pi}^{(n+1)\pi}\frac{|\sin(t)|}{t}dt=\sum_{n=0}^{+\infty}\int_0^{\pi}\frac{|\sin u|}{u+n\pi}du \\ \geqslant \int_0^{\pi}|\sin u|du\sum_{n=0}^{+\infty}\frac{1}{(n+1)\pi}=\frac{2}{\pi}\sum_{n=1}^{+\infty}\frac{1}{n}=+\infty $$

edited Apr 18 '20 at 18:39

Gary

31,845

answered Apr 18 '20 at 18:19

Tuvasbien

8,907

It’s $\infty*0$ is it $\infty$? – Vitali Pom Apr 18 '20 at 18:25
1

It depends, but $\int_0^{\pi}|\sin(u)|du=2$ here – Tuvasbien Apr 18 '20 at 18:26
Wait why is the sum then is $\infty$? – Vitali Pom Apr 18 '20 at 18:28
1

Because of Chasles relation – Tuvasbien Apr 18 '20 at 18:30
I don’t see info on Chasels relation I can understand and relate to it. Is it 100% true? Sorry I nudge you. – Vitali Pom Apr 18 '20 at 18:35
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Yes, I just used $$ \int_a^b f+\int_b^c f=\int_a^c f $$This is Chasles relation. SInce the integration goes to $+\infty$, the sum goes to $+\infty$ aswell – Tuvasbien Apr 18 '20 at 18:36
Let us continue this discussion in chat. – Vitali Pom Apr 18 '20 at 18:37
1

The sum is just the harmonic series which is known to be divergent. – Gary Apr 18 '20 at 18:38
But https://en.m.wikipedia.org/wiki/Convergence_tests here it says it shall converge @Gary What do I miss? – Vitali Pom Apr 18 '20 at 18:43
I apologies I drive you crazy :) – Vitali Pom Apr 18 '20 at 18:45
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@VitaliPom : Nothing on the convergence tests page claims that the harmonic series converges. And for good reason, it does not. Use the integral test from the page you cite to show that it does not. – Eric Towers Apr 18 '20 at 18:49
I see Leibnitz Criterion says it converges. What am I doing wrong? – Vitali Pom Apr 18 '20 at 18:52
Sorry I’m confused ((((( – Vitali Pom Apr 18 '20 at 18:53
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It says that $\sum\frac{(-1)^n}{n}$ converges, not $\sum\frac{1}{n}$ – Tuvasbien Apr 18 '20 at 18:53
Ok. Reset. I’m checking now. Thanks! Tuvasbien and all the others. – Vitali Pom Apr 18 '20 at 18:56
The harmonic series is the classical example of a series whose terms tend to $0$ yet the series diverges. – Gary Apr 18 '20 at 18:59
https://www.quora.com/Why-does-1-n-diverge Mark Regan explained that on Quora. Many thanks gang for the patience! You’re the best. – Vitali Pom Apr 18 '20 at 22:58

Why is $\int_{-\infty}^\infty |\text{sinc}(t)|\ dt=\infty$?

1 Answers1