If $f(x)$ and $g(\alpha)$ is a pair of Fourier transforms, then how can we show that $df/dx$ and $i\alpha g(\alpha)$ is a pair of Fourier transforms?
Conjugate pairs in Fourier transforms but with Fourier coefficients
0
$\begingroup$
real-analysis
fourier-analysis
-
2If you'd like others to take the time to help you, it only makes sense to take the time to e$x$pand on what you have tried and to typeset your question so it's a bit more readable. You've asked 31 question here; by now you should know how to ask a question on this site. – 2012-11-13
1 Answers
1
While I agree that you should type your question more carefully, here's the solution anyway:
- Write down the definition for the inverse Fourier transform
- Take the derivative.
$ f(x)=\int_{-\infty}^\infty g(\alpha)e^{i\alpha x}d\alpha \\ \frac{d}{dx}f(x)=\int_{-\infty}^\infty g(\alpha)\frac{d}{dx}e^{i\alpha x}d\alpha\\ =\int_{-\infty}^\infty i\alpha g(\alpha)e^{i\alpha x}d\alpha $
Since we now have the function $i\alpha g(\alpha)$ in the inverse Fourier transform formula, it must be the Fourier transform of $\frac{df}{dx}$.
-
0Can you show how the inverse FT connects to taking the derivative? they are 2 disjoint steps and I dont see how to proceed with this – 2012-11-14