# Conjugate pairs in Fourier transforms but with Fourier coefficients

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?

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If you'd like others to take the time to help you, it only makes sense to take the time to expand 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. –  amWhy Nov 13 '12 at 1:53

While I agree that you should type your question more carefully, here's the solution anyway:

1. Write down the definition for the inverse Fourier transform
2. 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 $\frac{df}{dx}$.

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I am actually having some trouble with the differentiation. Can you please edit your solution to help? –  Buddy Holly Nov 14 '12 at 2:49
You're learning the Fourier transform and you don't know how to differentiate an exponential function? I find this strange. Well, anyway, $\frac{d}{dx}e^{i\alpha x}=i\alpha e^{i\alpha x}$. You just use the chain rule. –  icurays1 Nov 14 '12 at 4:38
Can 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 –  Buddy Holly Nov 14 '12 at 5:51