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I'm trying to solve this problem from Real Analysis of Folland but can't find any solution for it. Can anyone help me ?. Thanks so much.

$$\mbox{Show that}\quad \int_{0}^{\infty}\left\vert\,{\sin\left(x\right) \over x}\,\right\vert\,{\rm d}x =\infty $$

And also, can we calculate the similar integral $\int_{0}^{\infty}{\sin\left(x\right) \over x}\,{\rm d}x$ ?. Please help me clarify this. I really appreciate.

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  • $\begingroup$ I don't know about the main question, but the other integral $\int\limits_0^\infty \dfrac{\sin x}{x}dx$ that you mentioned can be solved using various methods and the value of that is $\dfrac{\pi}{2}$. See Sine Integral. $\endgroup$
    – taninamdar
    Mar 8, 2014 at 12:09
  • $\begingroup$ compute it from $0$ to $\pi$. $\endgroup$
    – OBDA
    Mar 8, 2014 at 12:19

2 Answers 2

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$$ \int\limits_0^\infty \left|\frac{\sin x}{x} \right| \mathrm{d}x \\ =\sum\limits_{n = 0}^\infty \int\limits_{n\pi}^{(n+1)\pi} \left|\frac{\sin x}{x} \right| \mathrm{d}x \\ \geq \sum\limits_{n = 0}^\infty \int\limits_{n\pi}^{(n+1)\pi} \left|\frac{\sin x}{(n+1)\pi} \right| \mathrm{d}x \\ = \sum\limits_{n = 0}^\infty \frac{1}{(n+1)\pi}\int\limits_{n\pi}^{(n+1)\pi} \left|\sin x \right| \mathrm{d}x \\ = \sum\limits_{n = 0}^\infty \frac{2}{(n+1)\pi}\\ = \frac{2}{\pi}\sum\limits_{n = 0}^\infty \frac{1}{n+1}\\ = \frac{2}{\pi}\left(1+\frac{1}{2}+\frac{1}{3}+ \dots\right) = \infty $$

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  • $\begingroup$ Wow, nice! But can you explain to me please, why you are can simply factor out the 1/((n+1)pi) from the integral? $\endgroup$ Mar 8, 2014 at 12:28
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    $\begingroup$ Because $\frac{1}{(n+1)\pi}$ is independent of $x$. $\endgroup$
    – Priyatham
    Mar 8, 2014 at 12:29
  • $\begingroup$ Haha, yes! Thank you. $\endgroup$ Mar 8, 2014 at 12:30
  • $\begingroup$ Great. Thanks so much. $\endgroup$ Mar 8, 2014 at 12:32
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\begin{align}\int_0^\infty\left|\frac{\sin x}x\right|dx&=\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}\left|\frac{\sin x}x\right|dx\\&=\sum_{k=0}^\infty\int_{0}^{\pi}\left|\frac{\sin x}{x+k\pi}\right|dx\\&\ge \frac1\pi\sum_{k=0}^\infty\int_{0}^{\pi}\frac{\sin x}{k+1}dx=\frac2\pi\sum_{k=1}^\infty\frac1k=\infty\end{align}

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