Integral of sinc function using Fourier series

[psie]

Homework Statement

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

Relevant Equations

I'm not sure.

Showing that the (complex) Fourier coefficients of $u(x)$ are as specified is a simple exercise, which I've managed to do, but how do I then go about evaluating $\int_0^\infty \frac{\sin x} x dx$? The coefficients do not have an explicit formula, right? Note, the Fourier transform has not been introduced yet. I thought this has something to do with Riemann-Lebesgue's lemma or even Parseval's identity, but probably I'm mistaken.

 

[pasmith]

By evenness of $\sin x/x$, $$\begin{split} \int_0^\infty \frac{\sin x}x\,dx &= \frac12 \int_{-\infty}^{\infty} \frac{\sin x}x\,dx \end{split}.$$ To evaluate the right hand side, one can either split the real line into intervals of width $2\pi$ centered at even multiples of $\pi$ or into intervals of width $2\pi$ centered at odd multiplies of $\pi$. The results must be the same*, so $$\begin{split}\int_{-\infty}^\infty \frac{\sin x}x\,dx &= \frac12 \left(\sum_{n=-\infty}^\infty \int_{((2n)-1)\pi}^{((2n)+1)\pi} \frac{\sin x}{x}\,dx + \sum_{n=-\infty}^\infty \int_{((2n+1)-1)\pi}^{((2n+1)+1)\pi} \frac{\sin x}{x}\,dx \right) \end{split}$$

*Assuming the integral exists, which I don't believe this method establishes.

 

[psie]

*Assuming the integral exists, which I don't believe this method establishes.

Which integral do you mean?

If I understand you right, this is your argument: $\sum_{n\in\mathbb Z} c_n$ is just the function evaluated at $0$, so we have $$1=\sum_{n\in\mathbb Z} c_{2n}+\sum_{n\in\mathbb Z}c_{2n+1}=\frac{2}{2\pi}\int_{-\infty}^\infty \frac{\sin x} x dx.$$ And by evenness, we conclude that $\int_0^\infty \frac{\sin x} xdx=\frac{\pi}{2}$.

 

[psie]

I assume you mean $$\int_{-\infty}^\infty \frac{\sin x} x dx.$$ In my computation above it seems like we require $\sum_{n\in\mathbb Z} c_n$ to converge absolutely, so that one can rearrange the series and sum the even terms first and then the odd one's, or vice versa.

 

[psie]

I think I take back the requirement of absolute convergence. If $\sum_{k\geq} a_k$ converges, we can always write $$\sum_{k\geq 1}a_k = \sum_{k\geq 1}(a_{2k}+a_{2k-1}),$$ but not $$\sum_{k\geq 1}a_k = \sum_{k\geq 1}a_{2k}+\sum_{k\geq 1}a_{2k-1}.$$ Here, however, we can write the series in the latter form, since, assuming that $\int_{-\infty}^\infty \frac{\sin x} x dx$ converges, we have that $\sum_{n\in\mathbb Z} c_{2n}$ and $\sum_{n\in\mathbb Z}c_{2n+1}$ both converge.

 

[psie]

By evenness of $\sin x/x$, $$\begin{split} \int_0^\infty \frac{\sin x}x\,dx &= \frac12 \int_{-\infty}^{\infty} \frac{\sin x}x\,dx \end{split}.$$ To evaluate the right hand side, one can either split the real line into intervals of width $2\pi$ centered at even multiples of $\pi$ or into intervals of width $2\pi$ centered at odd multiplies of $\pi$. The results must be the same*, so $$\begin{split}\int_{-\infty}^\infty \frac{\sin x}x\,dx &= \frac12 \left(\sum_{n=-\infty}^\infty \int_{((2n)-1)\pi}^{((2n)+1)\pi} \frac{\sin x}{x}\,dx + \sum_{n=-\infty}^\infty \int_{((2n+1)-1)\pi}^{((2n+1)+1)\pi} \frac{\sin x}{x}\,dx \right) \end{split}$$

*Assuming the integral exists, which I don't believe this method establishes.

I'm confused. Could you clarify what you are suggesting on how to evaluate $\int_{-\infty}^\infty \frac{\sin x} x dx$? I don't see how this is possible unless we have absolute convergence of $\sum_{n\in\mathbb Z} c_n$.

 

[pasmith]

It is easier to justify if one starts from $$\begin{split} \sum_{n=-N}^N c_n &= \sum_{n=-N}^N \frac{1}{2\pi} \int_{(n-1)\pi}^{(n+1)\pi} \frac{\sin x}x\,dx \\ &= \frac{1}{2\pi}\int_{-(N+1)\pi}^{-N\pi} \frac{\sin x}{x}\,dx + \frac{1}{\pi}\int_{-N\pi}^{N\pi} \frac{\sin x}x\,dx + \frac{1}{2\pi}\int_{N\pi}^{(N+1)\pi} \frac{\sin x}x\,dx \\ &= \frac{2}{\pi} \int_0^{N\pi} \frac{\sin x}x\,dx + \frac1\pi \int_{N\pi}^{(N+1)\pi} \frac{\sin x}x\,dx\end{split}$$ so that $$\left| \sum_{n=-N}^N c_n - \frac2\pi \int_0^{N\pi} \frac{\sin x}x\,dx \right| = \left|\frac 1\pi \int_{N\pi}^{(N+1)\pi} \frac{\sin x}{x}\,dx \right| \leq \frac 1\pi \int_{N\pi}^{(N+1)\pi} \frac 1x\,dx = \frac1\pi \log\left(1 + \frac 1N \right).$$