# Absolute Integrable Sinc function

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 $h(t) = \dfrac{\sin t}{\pi t}$.

• This feels like homework; even if it isn't, please discuss the approach you've got so far! – Marcus Müller Jun 5 '18 at 11:29
• Hint: Try to find an upper bound the function you are integrating and show that the integral of that upper bound diverges. – Atul Ingle Jun 5 '18 at 18:03
• @AtulIngle that'd be necessary, but not sufficient! But I think I know what you're going for. – Marcus Müller Jun 5 '18 at 20:53
• oh I should've said lower bound! not upper bound. – Atul Ingle Jun 5 '18 at 21:10
• @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? – Dilip Sarwate Jun 5 '18 at 23:39

\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 $1/(n+1) \pi$.