How do we prove that $$\int_{-\infty}^{\infty}\bigg|\dfrac{\sin t}{\pi t}\bigg|dt\to \infty$$ This comes in the context of stability of LTI system with impulse response $h(t) = \dfrac{\sin t}{\pi t}$.

Thanks in advance.

  • $\begingroup$ This feels like homework; even if it isn't, please discuss the approach you've got so far! $\endgroup$ Commented Jun 5, 2018 at 11:29
  • $\begingroup$ Hint: Try to find an upper bound the function you are integrating and show that the integral of that upper bound diverges. $\endgroup$
    – Atul Ingle
    Commented Jun 5, 2018 at 18:03
  • $\begingroup$ @AtulIngle that'd be necessary, but not sufficient! But I think I know what you're going for. $\endgroup$ Commented Jun 5, 2018 at 20:53
  • $\begingroup$ oh I should've said lower bound! not upper bound. $\endgroup$
    – Atul Ingle
    Commented Jun 5, 2018 at 21:10
  • $\begingroup$ @AtulIngle Since the sin function has value $0$ at periodically spaced points along the axis, any lower bond on the integrand must also have value $0$ at these points. Did you have any specific function in mind? $\endgroup$ Commented Jun 5, 2018 at 23:39

1 Answer 1


The question has an answer on Math SE. The argument goes as follows:

\begin{eqnarray} \int_{-\infty}^\infty \left| \frac{\sin t}{\pi t} \right|dt &=& \frac{1}{\pi}\sum_{n=-\infty}^\infty \int_{n\pi}^{(n+1)\pi} \left| \frac{\sin t}{t} \right| dt \\ &>& \frac{1}{\pi} \sum_{n=-\infty}^\infty \frac{1}{|n+1|\pi}\int_{n\pi}^{(n+1)\pi} |\sin t|dt \\ &=& \frac{2}{\pi^2} \sum_{n=-\infty}^\infty \frac{1}{|n|}\\ &\rightarrow& \infty \end{eqnarray} where in each interval $[n\pi, (n+1)\pi]$ we lower bound $1/t$ with $1/(n+1) \pi$.

  • 1
    $\begingroup$ nice and elegant :) $\endgroup$ Commented Jun 7, 2018 at 22:10
  • $\begingroup$ yes, and I take no credit for it. Please upvote the linked Math.SE answer! $\endgroup$
    – Atul Ingle
    Commented Jun 8, 2018 at 15:46
  • $\begingroup$ An honest, good idea. Did that! $\endgroup$ Commented Jun 8, 2018 at 15:48

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