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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!

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# Fourier transform problem

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