# Reversing Fourier series

1. Jul 4, 2006

### gonzo

Anyone have any clues what function f(x) satisfies the following integral (for an integer n > 0):

$\int^{\pi}_{0} f(x)sin(nx)dx=(-1)^{n+1}$

Last edited: Jul 4, 2006
2. Jul 4, 2006

### J77

So an alternating function at +1 and -1?

3. Jul 4, 2006

### gonzo

Yes. In other words, the function whose Fourier series would be:

sin(x)-sin(2x)+sin(3x)...

4. Jul 4, 2006

### gonzo

Nevermind, figured it out.

5. Jul 6, 2006

### eljose

If you define:

$$\int_{0}^{\infty}dxW(x)f(x)sin(nx)=(-1)^{n+1}$$

where x is W(x)=0 iff x>2pi, and W8x)=1 iff x<2pi, the integral above is just a fourier sine transform with inverse:

$$W(x)f(x)=-\frac{2}{\pi}\int_{0}^{\infty}dne^{n\pi i}sin(nx)$$

which is equal to $$f(x)W(x)= 2i(\delta (x+i \pi)-\delta (x-i \pi ))$$

which is real...:bigrin:
: :bigrin:

Last edited: Jul 6, 2006