The Fourier transform of a product of $f$ with $t^n$

Question

There is a proof in my book I don't quite follow. We are supposed to prove that the Fourier transform of a product of $f$ with $t^n$ is given by

$$\mathcal{F}[t^n f(t)] (\lambda) = i^n \frac{d^n}{d \lambda^n} \{\mathcal{F} [f] (\lambda)\}$$

I will show the proof up until the point I don't get. The proof is:

For the Fourier transform of a product of $f$ with $t^n$, we have

$$\mathcal{F}[t^n f(t)] (\lambda) = \frac{1}{2 \pi} \int_{- \infty}^{\infty} t^n f(t) e^{- i \lambda t} dt$$

Using

$$t^n f(t) e^{-i \lambda t} = (i)^n \frac{d^n}{d \lambda^n} \{f(t) e^{-i \lambda t} \}$$,

we obtain. . .

Now the rest of the proof I understand. However, I don't see where the book gets the expression

$$t^n f(t) e^{-i \lambda t} = (i)^n \frac{d^n}{d \lambda^n} \{f(t) e^{-i \lambda t} \}$$

from. If anyone can explain this to me, I would be very grateful!

Have you tried figuring out what $\frac{\mathrm d^n}{\mathrm d\lambda}\{f(t)e^{-i\lambda t}\} = f(t)\frac{\mathrm d^n}{\mathrm d\lambda}\{e^{-i\lambda t}\}$ works out to be? Surely you remember $\frac{\mathrm d}{\mathrm d\lambda}\{e^{-i\lambda t}\}=(-it)e^{-i\lambda t}$ from your calculus class?
Thanks! Sorry, I actually tried your suggestion before posting the question, but didn't get the correct answer. Now I see it's because I made a dumb arithmetic mistake. Appreciate it!
@Kristian I would encourage you to answer your own question and then, after an appropriate amount of time, accept it.

Kristian · Accepted Answer · 2013-01-12 16:00:39Z

We have:

$$\frac{d}{d \lambda} \{f(t) e^{-i \lambda t} \} = (-it) f(t) e^{- i \lambda t}$$

Differentiating again gives:

$$\frac{d^2}{d \lambda ^2} \{ f(t) e^{- i \lambda t} \} = (-it)^2 f(t) e^{- i \lambda t}$$

So in general:

$$\frac{d^n}{d \lambda ^n} \{ f(t) e^{- i \lambda t} \} = (-it)^n f(t) e^{- i \lambda t}$$

It is now an easy task to check that the two sides of the equation stated in the problem are equal.

1 Answer