2

I'm aware of the many answers to evaluating the integral of $\frac{\sin x}{x}$ over $[0,\infty)$ given here. No answer seems to mention the approach below, and neither have I found anything elsewhere, and hence I'm doubting if it's possible.

I stumbled upon this in connection with a problem I was working on.

Consider $$u(x)=\begin{cases} \frac{\sin x}{x}, & 0<|x|\leq\pi, \\ 1, & x=0.\end{cases}$$ [and $2\pi$ periodically extended]. Show its Fourier coefficients are $$c_n=\frac{1}{2\pi}\int_{(n-1)\pi}^{(n+1)\pi}\frac{\sin x}{x}dx.$$ Use this to evaluate $\int_0^\infty \frac{\sin x} x dx$.

The formula for the Fourier coefficients are derived here.

Here is my attempt at evaluating $\int_0^\infty \frac{\sin x} x dx$, which we assume converges. Moreover, since $u(x)$ is at least piecewise $C^1$, we also assume the Fourier series converges pointwise to the function. Now, $\sum_{n\in\mathbb Z} c_n$ is just the function evaluated at $0$, so we have $$1=\sum_{n\in\mathbb Z}c_n=\sum_{n\in\mathbb Z} c_{2n}+\sum_{n\in\mathbb Z}c_{2n+1}=\frac{1}{2\pi}\int_{-\infty}^\infty \frac{\sin x} x dx+\frac{1}{2\pi}\int_{-\infty}^\infty \frac{\sin x} x dx.$$ And by evenness, we conclude that $\int_0^\infty \frac{\sin x} xdx=\frac{\pi}{2}$.

However, are we justified in splitting up the sum into even and odd terms? I can not verify that $\sum_{n\in\mathbb Z} |c_n|$ converges.

psie
  • 765
  • 1
    If $s(x)$ is continuous and the derivative is square integrable then the fourier series will converge absolutely and uniformly to $s(x)$. Can you take it from here? – CyclotomicField Feb 17 '24 at 19:44
  • @CyclotomicField Could you elaborate a bit? In this case, how do we know that the function $u(x)$ and its derivative are in $L^2$? – psie Feb 18 '24 at 02:07
  • Related to this (and in particular to the comment above): https://math.stackexchange.com/questions/1049305/l2-and-uniform-norm-of-textsinc-x-and-its-derivatives (evaluating the $L^2$ norms of $\operatorname{sinc}$ and its derivatives) – leslie townes Feb 18 '24 at 06:12

1 Answers1

0

The series $\sum_{k\in\mathbb Z} |c_k|$ does converge. We have \begin{align}\sum_{k=1}^\infty|c_{2k-1}|&=\frac1{2\pi}\int_0^\infty\frac{\sin(x)}x\,dx , \\ \sum_{k=1}^\infty|c_{2k}|&=-\frac1{2\pi}\int_\pi^\infty\frac{\sin(x)}x\,dx.\end{align} Adding these two convergent series, we get $$\sum_{k=1}^\infty|c_k|=\frac1{2\pi}\int_0^\pi\frac{\sin(x)}x\,dx.$$ Since $c_k=c_{-k}$ and $$c_0=\frac1{2\pi}\int_{-\pi}^\pi\frac{\sin(x)}xdx,$$we obtain finally that $$\sum_{k\in\mathbb Z} |c_k|=\frac2\pi\int_0^\pi\frac{\sin(x)}{x}\,dx.$$

psie
  • 765