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 going to make use of the series $\displaystyle \sum_{n=0}^\infty \frac 1{n+1}$.

and that $\displaystyle \int_0^\infty \frac{ \sin^2 x} x \, dx = \sum_{n=0}^\infty\int_{n\pi}^{(n+1)\pi} \frac{ \sin^2 x} x \,dx$

If I use variable substitution $t=x-n\pi$ it gives

$$\tag 1 \sum_{n=0}^\infty\int_0^\pi \frac{\sin^2 t }{n\pi+t}dt $$

gives

$$\tag 2\frac 1 \pi\sum_{n=0}^\infty\int_0^\pi \frac{\sin^2 t }{n+1} \, dt$$

$$\tag 3 \frac 1 \pi \int_{n=0}^\pi \sin^2 t \;dt\cdot \sum_{n=0}^\infty \frac 1 {1+n}$$

I don't really know how to explain this or what i have done. If someone knows how to solve this.

share|improve this question
    
You have a $n+1$ when you should have a $n+\dfrac t\pi$ in $(2)$ –  Pedro Tamaroff Nov 4 '12 at 16:37
    
You might be also interested in this:math.stackexchange.com/q/67198/9464 –  Jack Nov 4 '12 at 17:37

1 Answer 1

First note that in the interval $[n \pi, (n+1) \pi)$, where $n > 0$, we have that $\dfrac{\sin^2(x)}x \geq \dfrac{\sin^2(x)}{(n+1) \pi}$.

This is because $\dfrac1x$ is a decreasing function and hence in the interval $[n \pi, (n+1) \pi)$, we have $$\dfrac1x > \dfrac1{(n+1) \pi}$$

Hence, $$\int_{n \pi}^{(n+1) \pi} \dfrac{\sin^2(x)}x dx > \int_{n \pi}^{(n+1) \pi} \dfrac{\sin^2(x)}{(n+1) \pi} dx = \dfrac1{n \pi} \dfrac{\pi}2 = \dfrac1{2(n+1)}$$

Hence, $$\int_{0}^{\infty} \dfrac{\sin^2(x)}x dx = \int_{0}^{\pi} \dfrac{\sin^2(x)}x dx + \int_{\pi}^{2\pi} \dfrac{\sin^2(x)}x dx + \int_{2\pi}^{3\pi} \dfrac{\sin^2(x)}x dx + \cdots$$ Hence, $$\int_{0}^{\infty} \dfrac{\sin^2(x)}x dx = \lim_{k \to \infty} \sum_{n=0}^{k} \int_{n \pi}^{(n+1) \pi} \dfrac{\sin^2(x)}x dx$$ We showed at the beginning of the post that $$\int_{n \pi}^{(n+1) \pi} \dfrac{\sin^2(x)}x dx > \dfrac1{2(n+1)}$$ Hence, putting these two together, we get that $$\int_{0}^{\infty} \dfrac{\sin^2(x)}x dx > \dfrac12\lim_{k \to \infty} \sum_{n=0}^{k} \dfrac1{n+1}$$ The series on the right side is called the harmonic series and look here to figure out why it diverges.

share|improve this answer
    
can you explain? –  stuck Nov 4 '12 at 16:48
    
You mean $>$ in the last one, yes? –  Pedro Tamaroff Nov 4 '12 at 17:05
    
@PeterTamaroff Yes. Updated. Thanks. –  user17762 Nov 4 '12 at 17:09

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.