# Fourier transform problem

1. Mar 4, 2015

### joshmccraney

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!

2. Mar 6, 2015

### joshmccraney

nevermind, i got it! when using convolution on sine/cosine we use one of each!