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}<br /> \int_0^\infty \frac{\sin x}x\,dx &amp;= \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 <br /> \begin{split}\int_{-\infty}^\infty \frac{\sin x}x\,dx &amp;= \frac12 \left(\sum_{n=-\infty}^\infty \int_{((2n)-1)\pi}^{((2n)+1)\pi} \frac{\sin x}{x}\,dx<br /> + \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}<br /> \int_0^\infty \frac{\sin x}x\,dx &amp;= \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 <br /> \begin{split}\int_{-\infty}^\infty \frac{\sin x}x\,dx &amp;= \frac12 \left(\sum_{n=-\infty}^\infty \int_{((2n)-1)\pi}^{((2n)+1)\pi} \frac{\sin x}{x}\,dx<br /> + \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}<br /> \sum_{n=-N}^N c_n &amp;= \sum_{n=-N}^N \frac{1}{2\pi} \int_{(n-1)\pi}^{(n+1)\pi} \frac{\sin x}x\,dx \\<br /> &amp;= \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 \\<br /> &amp;= \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 <br /> \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).