> Yes integration is a 1/f Fourier multiplier. Can you explain this? I don’t see the connection. I can see how the zero-frequency value would be equal to the i

> Yes integration is a 1/f Fourier multiplier. Can you explain this? I don’t see the connection. I can see how the zero-frequency value would be equal to the integral (well, the average). Edit: figured it out. Derivative operator multiplies each basis function by its index. D exp(inx) = inexp(inx). Apply the inverse operation (divide by index) to get the integral.