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Let $f(x)=\mathbb{I}_{[-t,t]}$ where $t\in (0,\pi)$. Using Parseval's theorem to this function we get: $$\sum \limits_{n=1}^{\infty}\dfrac{\sin ^2(nt)}{n^2t}=\dfrac{\pi-t}{2}.$$ Prove that $$\int \limits_{0}^{\infty}\left(\frac{\sin x}{x}\right)^2dx=\lim \limits_{t\to 0}\sum \limits_{n=1}^{\infty}\dfrac{\sin ^2(nt)}{n^2t}=\frac{\pi}{2}.$$ How to prove the first equality strictly using $\varepsilon-\delta$? Unfortunately I have not any ideas. I know that this sum is a Riemann-integral sum but I can't prove it rigorously.

I would be very grateful to anyone who'll post full solution because I can't find it's proof.

RFZ
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2 Answers2

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the Fourier transform of $\mathbb{I}_{[-1/2,1/2]}$ is $$\int_{-\infty}^\infty \mathbb{I}_{[-1/2,1/2]}(x) e^{-2 i \pi f x} dx = \frac{e^{- i \pi f} - e^{ i \pi f}}{-2 i \pi f} = \frac{\sin(\pi f)}{\pi f}$$

so the Parseval theorem for Fourier transforms let us know that

$$\int_{-\infty}^\infty |\mathbb{I}_{[-1/2,1/2]}(x)|^2 dx = 1 = \int_{-\infty}^\infty \frac{\sin(\pi f)^2}{\pi^2 f^2} df$$

now if you want to prove it with the Fourier series, consider the function $f_T(x) = \mathbb{I}_{[-1/2,1/2]}(x)$ for $x \in [-T/2;T/2]$ and $f_T(x) = f_T(x+T)$ :

$$c_T(n) = \frac{1}{\sqrt{T}} \int_{-T/2}^{T/2} f_T(x) e^{-2 i \pi n x / T}dx = \frac{\sin(\pi n/T)}{\pi n / \sqrt{T}}$$

$$f_T(x) = \sum_{n=-\infty}^\infty \frac{c_T(n)}{\sqrt{T}} e^{2 i \pi n x /T}$$ and the Parseval theorem for the Fourier series let us know that : $$\sum_{n=-\infty}^\infty |c_T(n)|^2 = \int_{-T/2}^{T/2} f_T(x)^2 dx = 1$$ so that :

$$\lim_{T \to \infty} \frac{1}{T}\sum_{n=-\infty}^\infty \frac{\sin(\pi n/T)^2}{\pi^2 (n/T)^2 } = \int_{-\infty}^\infty \frac{\sin(\pi x)^2}{\pi^2 x^2} dx= 1$$

reuns
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  • +1! Nice solution but it uses some topics from functional analysis which is little hard to me. But I think that I can understand it soon in future. – RFZ Jan 15 '16 at 14:58
  • it uses only the Parseval theorem, it's just that you made some mistakes calculating the Fourier series and applying the Parseval theorem to it – reuns Jan 15 '16 at 15:04
  • Where I made mistakes? I thought that my calculations are right and they coincides with Rudin's book. – RFZ Jan 15 '16 at 15:11
  • ok, so it's the same solution as in your book. if you want to understand the fourier series, look at the discrete fourier transform first (which is nothing more than a unitary matrix) – reuns Jan 15 '16 at 15:16
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Given $\epsilon\gt0$, select $M > 1/\epsilon$, such that: $$ \left|\int_0^\infty \left(\frac{\sin x}{x}\right)^2 dx - \int_0^M \left(\frac{\sin x}{x}\right)^2 dx\right|\lt \epsilon \tag{1} $$ Then, select $N_0$ so that $N\gt N_0$ implies: $$ \left|\int_0^M \left(\frac{\sin x}{x}\right)^2 dx - \sum_{n=1}^N{\frac{\sin^2\left(n\frac{M}{N}\right)}{n^2\left(\frac{M}{N}\right)}}\right|\lt \epsilon \tag{2} $$ This is possible, since: $$\sum_{n=1}^N{\frac{\sin^2\left(n\frac{M}{N}\right)}{n^2\left(\frac{M}{N}\right)}} = \sum_{n=1}^N{\left(\frac{\sin\left(n\frac{M}{N}\right)}{n\frac{M}{N}}\right)^2\cdot\frac{M}{N}} \longrightarrow \int_0^M \left(\frac{\sin x}{x}\right)^2 dx$$ as $N\rightarrow\infty$, as the expression on the left is a riemann sum of the integral on the right.

Finally, we note that: $$ \left|\sum_{n=1}^N{\frac{\sin^2\left(n\frac{M}{N}\right)}{n^2\left(\frac{M}{N}\right)}} - \frac{\pi-\frac{M}{N}}{2}\right|=\sum_{n=N+1}^\infty{\frac{\sin^2\left(n\frac{M}{N}\right)}{n^2\left(\frac{M}{N}\right)}}\le\frac{N}{M}\sum_{n=N+1}^\infty{\frac{1}{n^2}}\le \frac{1}{M}\lt\epsilon \tag{3} $$ Combining $(1)$, $(2)$, and $(3)$, we get: $$ \left|\int_0^\infty \left(\frac{\sin x}{x}\right)^2 dx - \frac{\pi-\frac{M}{N}}{2}\right|\lt3\epsilon $$ when $N>N_0$. In particular, taking $N\rightarrow\infty$, we get: $$ \left|\int_0^\infty \left(\frac{\sin x}{x}\right)^2 dx - \frac{\pi}{2}\right|\le3\epsilon $$

SomeStrangeUser
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