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vela
#9
Feb26-12, 10:11 PM
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I didn't even notice the k's. Since the original problem uses y(x), let's use k to be the variable conjugate to x:
\begin{align*}
f(x) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F(k)e^{ikx}\, dk \\
F(k) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x)e^{-ikx}\,dx
\end{align*}
So consider F'(k).