Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  taninamdar Mar 8 at 12:09
    
compute it from $0$ to $\pi$. –  OBDA Mar 8 at 12:19

2 Answers 2

up vote 6 down vote accepted

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

share|improve this answer
    
Wow, nice! But can you explain to me please, why you are can simply factor out the 1/((n+1)pi) from the integral? –  Just_a_fool Mar 8 at 12:28
1  
Because $\frac{1}{(n+1)\pi}$ is independent of $x$. –  Priyatham Mar 8 at 12:29
    
Haha, yes! Thank you. –  Just_a_fool Mar 8 at 12:30
    
Great. Thanks so much. –  le duc quang Mar 8 at 12:32

\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}

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.