# Homework Help: Fourier coefficient

1. May 14, 2005

### Zaare

I'm supposed to show
$$\hat f\left( n \right) = - \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {f\left( {x + {\pi \mathord{\left/ {\vphantom {\pi n}} \right. \kern-\nulldelimiterspace} n}} \right)e^{ - inx} dx}$$
where $$\hat f\left( n \right)$$ is the Fourier coefficient and $$f(x)$$ is a $$2\pi$$-periodic and Riemann integrable on $$[\pi,-\pi]$$.

This is what I've done:
I use the formulae for the Fourier coefficient of a $$2\pi$$-periodic function
$$\hat f\left( n \right) = \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {f\left( x \right)e^{ - inx} dx}$$
and a simple change of variable
$$\hat f\left( n \right) = \frac{1}{{2\pi }}\int\limits_{ - \pi + {\pi \mathord{\left/ {\vphantom {\pi n}} \right. \kern-\nulldelimiterspace} n}}^{\pi + {\pi \mathord{\left/ {\vphantom {\pi n}} \right. \kern-\nulldelimiterspace} n}} {f\left( {x + {\pi \mathord{\left/ {\vphantom {\pi n}} \right. \kern-\nulldelimiterspace} n}} \right)e^{ - in\left( {x + {\pi \mathord{\left/ {\vphantom {\pi n}} \right. \kern-\nulldelimiterspace} n}} \right)} dx} = \frac{1}{{2\pi }}\int\limits_{ - \pi + {\pi \mathord{\left/ {\vphantom {\pi n}} \right. \kern-\nulldelimiterspace} n}}^{\pi + {\pi \mathord{\left/ {\vphantom {\pi n}} \right. \kern-\nulldelimiterspace} n}} {f\left( {x + {\pi \mathord{\left/ {\vphantom {\pi n}} \right. \kern-\nulldelimiterspace} n}} \right)e^{ - inx} e^{ - i\pi } dx} = - \frac{1}{{2\pi }}\int\limits_{ - \pi + {\pi \mathord{\left/ {\vphantom {\pi n}} \right. \kern-\nulldelimiterspace} n}}^{\pi + {\pi \mathord{\left/ {\vphantom {\pi n}} \right. \kern-\nulldelimiterspace} n}} {f\left( {x + {\pi \mathord{\left/ {\vphantom {\pi n}} \right. \kern-\nulldelimiterspace} n}} \right)e^{ - inx} dx}$$.

Everything seems to agree except for the limits of the integral.
1) If I have done some mistake, I'd appreciate it someon would point it out.
2) If I haven't done any mistakes, what's the reasoning behind this? Don't the limits matter as longs as they are $$2\pi$$ apart?

2. May 14, 2005

### dextercioby

You'll have to prove that

$$\int_{-\pi+\frac{\pi}{n}}^{-\pi} f\left(x+\frac{\pi}{n}\right) \ dx =-\int_{\pi}^{\pi+\frac{\pi}{n}} f\left(x+\frac{\pi}{n}\right) \ dx$$

Daniel.

3. May 14, 2005

### Zaare

Ok, here's my reasoning.
The integrand

$$f(x+\frac{\pi}{n})e^{-inx}$$

is $$2\pi$$-periodic. Hence a change from $$x+\frac{\pi}{n}$$ to $$x+\frac{\pi}{n}+2\pi$$ leaves the integrand unchanged and I get

$$\int\limits_{ - \pi }^{ - \pi + {\pi \mathord{\left/ {\vphantom {\pi n}} \right. \kern-\nulldelimiterspace} n}} {f\left( {x + {\pi \mathord{\left/ {\vphantom {\pi n}} \right. \kern-\nulldelimiterspace} n}} \right)e^{ - inx} dx} = \int\limits_\pi ^{\pi + {\pi \mathord{\left/ {\vphantom {\pi n}} \right. \kern-\nulldelimiterspace} n}} {f\left( {x + {\pi \mathord{\left/ {\vphantom {\pi n}} \right. \kern-\nulldelimiterspace} n}} \right)e^{ - inx} dx}$$

which allows me to change the limits of the original integral to $$-\pi$$ and $$\pi$$ without changing its value.
Am I right?

Last edited: May 15, 2005