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$\text{y is a 2$\pi$-periodic function defined as:}$

$y(x) = \begin{cases} \dfrac{\sin(x)}{x} & \text{, $0\le|x|\le \pi$} \\ 1 & \text{, $x=0$} \end{cases}$

$\text{I want to show that the fourier coefficients of y are equal to:}$ $$c_{n}=\frac{1}{2\pi}\int_{(n-1)\pi}^{(n+1)\pi} \frac{\sin x}{x} dx $$

My attempt (using Euler's formula):

$$c_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\sin x}{x}e^{-inx} dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{e^{ix(1-n)}-e^{-ix(1+n)}}{2xi} dx$$

Not sure how to continue though, also am I allowed to integrate from $-\pi$ to $\pi$? Since y is defined differently at $x=0$?

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    We can integrate from $-\pi$ to $\pi$. The value of the integral is the same for all functions [almost everywhere](https://en.wikipedia.org/wiki/Almost_everywhere) equal. In particular, the value of $y$ at a single point (here $x=0$) has absolutely no incidence on the value of the integral. When one is only interested in integrating functions (for example if you study the [$L^p$ spaces](https://en.wikipedia.org/wiki/Lp_space)) it is natural to identify all functions that are equal a.e. If one wants to know if the FS of $y$ converges to $y$ at $x=0$ then the value $y(0)$ is important. – 2017-02-28
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    Okay but regarding the proof...It almost looks like a change of variables can solve it? $u=x(1-n), du=(1-n) dx$ gives $\frac{1}{2\pi} \int_{(n-1)\pi}^{(1-n)\pi} \frac{e^{iu}-e^{-iu(1+n)/(1-n)} }{2iu} du $ You have any idea? But the upper integration limit isnt the same as they asked for, and this isnt equal to eulers formula for sin(u)..but on its way.. – 2017-02-28
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    Interestingly this exercise shows that $\displaystyle\int_{-\infty}^{\infty}\frac{\sin x}{x}\,dx=\pi$ – 2017-02-28
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    Oh, how do you conclude that connection? – 2017-02-28
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    Because $y$ is differentiable at $0$ hence its Fourier series at $0$, which equals $\displaystyle\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\sin x}{x}\,dx$, converges to $y(0)=1$. That is, $$\int_{-\infty}^{\infty}\frac{\sin x}{x}\,dx=\pi$$ – 2017-02-28
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    But isnt the fourier series y(x)= $\sum_{-\infty}^{\infty} \frac{1}{2\pi}\int_{(n-1)\pi}^{(n+1)\pi}\frac{sin(u)}{u} du *e^{inx} $ ? How do you get $\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{sinx}{x} dx$ ? – 2017-02-28
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    When $x=0$, we have $e^{inx}=e^{0}=1$. When summing $n\in\mathbb{Z}$ we cover twice $\mathbb{R}$. Indeed, when $n=0$ the interval of integration is $[-\pi,\pi]$ and you get all translations by integral multiples of $\pi$. For example when $n=1$ the interval is $[0,2\pi]$ and you see that we covered $[0,\pi]$ twice. Hence the Fourier series of $y$ at $0$ is $$\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}\int_{(n-1)\pi}^{(n+1)\pi}\frac{\sin x}{x}\,dx = \frac{1}{2\pi}\cdot2\int_{-\infty}^{\infty}\frac{\sin x}{x}\,dx$$ and as I said this equals $y(0)=1$. – 2017-02-28
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    Im just having trouble seeing how the integration limits equal to $-\infty \to \infty$. Is there maybe a step more before the equality that could help me understand it? – 2017-02-28
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    It's just the definition $$\int_{-\infty}^{\infty}f=\lim_{m,n\to\infty}\int_{-m}^{n}f$$ By definition the series $$\sum_{n=-\infty}^{\infty}\int_{(n-1)\pi}^{(n+1)\pi}f$$ gives exactly (twice) that limit... – 2017-02-28
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    And the extra factor of 2 comes out because we are summing twice? – 2017-02-28
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    Yes. The intervals of integration are not disjoint. If we had $$\sum_{n=-\infty}^{\infty}\int_{n}^{n+1}f$$ we would get $\displaystyle\int_{-\infty}^{\infty}f$. But here $$\int_{(n-1)\pi}^{(n+1)\pi}f=\int_{(n-1)\pi}^{n\pi}f+\int_{n\pi}^{(n+1)\pi}f$$ where $((n-1)\pi,n\pi)$ ($n\in\mathbb{Z}$) are disjoint intervals and same for $(n\pi,(n+1)\pi)$ ($n\in\mathbb{Z}$). Hence we are really summing twice as you said. Anyway, I was just making an unimportant comment. – 2017-02-28

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We have \begin{align} \int_{-\pi}^{\pi} \frac{\sin x}{x}e^{-inx} \,dx &= \int_{-\pi}^{\pi} \frac{\sin x}{x}\cos(nx) \,dx -i \underbrace{\int_{-\pi}^{\pi} \frac{\sin x}{x}\sin(nx) \,dx}_{0 \text{ (odd integrand)}} \\ &= \int_{0}^{\pi} \frac{\sin((n+1)x)}{x} \,dx + \int_{0}^{\pi} \frac{\sin((1-n)x)}{x} \,dx \\ &= \int_0^{(n+1)\pi} \frac{\sin x}{x} \,dx - \int_{0}^{(n-1)\pi} \frac{\sin x}{x} \,dx \\ &= \int_{(n-1)\pi}^{(n+1)\pi} \frac{\sin x}{x} \,dx \end{align}

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    Where do you get the minus sign infront of the second integral in the third line? And how do you got the upper limit of the second integral to be $(n-1)\pi$? Shouldnt it be $(1-n)\pi$ ? – 2017-02-28
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    @fejz1234 Substitution $u=-(1-n)x$ and $\sin(-u)=-\sin(u)$. – 2017-02-28