Reversing Fourier series

[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}

[J77]

So an alternating function at +1 and -1?

[gonzo]

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

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

[gonzo]

Nevermind, figured it out.

[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: