- #1
member 428835
hi pf!
My book presents a problem and has it boiled down to $$S(u) = -S(f(x)) \exp(- \omega y) / \omega$$ where ##S(u)## is the sine Fourier transform of the function ##u##. However, we cannot directly take the transform back since the singularity at ##\omega = 0##. Thus the book then takes $$\frac{\partial}{\partial y} S(u) = S(f(x)) \exp(- \omega y)$$ and now performs an inverse COSINE transform on the exponential (also they use convolution). My question is, why are they using a cosine transform instead of a sine transform?
Thanks so much for your help!
My book presents a problem and has it boiled down to $$S(u) = -S(f(x)) \exp(- \omega y) / \omega$$ where ##S(u)## is the sine Fourier transform of the function ##u##. However, we cannot directly take the transform back since the singularity at ##\omega = 0##. Thus the book then takes $$\frac{\partial}{\partial y} S(u) = S(f(x)) \exp(- \omega y)$$ and now performs an inverse COSINE transform on the exponential (also they use convolution). My question is, why are they using a cosine transform instead of a sine transform?
Thanks so much for your help!
