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## Main Question or Discussion Point

Hey Physics Forums,

Grading an assignment, the current topic is continuous Fourier Transforms. They're trying to prove the convenient property:

[tex] \mathcal{F} \left[ \frac{d^n}{dx^n} f(x) \right] = (i \omega)^n \mathcal{F} \left[ f(x) \right] [/tex]

So there's a simple way to get it:

Let [itex]f(x)[/itex] be represented by it's Inverse Fourier Transform [itex]f(x) = \mathcal{F}^{-1} \left[ g( \omega ) \right] = \frac{1}{2 \pi} \int_{-\infty}^{\infty} g( \omega )e^{i \omega x} d \omega [/itex]

Then,

[tex] \frac{d^n}{d x^n} f(x) = \frac{d^n}{d x^n} \frac{1}{2 \pi} \int_{-\infty}^{\infty} g( \omega )e^{i \omega x} d \omega = \frac{1}{2 \pi} \int_{-\infty}^{\infty} g( \omega ) \frac{d^n}{d x^n}e^{i \omega x} d \omega = (i \omega)^n \frac{1}{2 \pi} \int_{-\infty}^{\infty} g( \omega ) e^{i \omega x} d \omega = (i \omega)^n \mathcal{F} \left[ f(x) \right] [/tex]

However, I have some people with a different method. They directly transform the entire expression, and apply integration by parts recursively until they obtain:

[tex] \left. (i \omega)^0 \frac{d^{n-1}}{dx^{n-1}}[f(x)]e^{-i \omega x} + (i \omega)^1 \frac{d^{n-2}}{dx^{n-2}}[f(x)]e^{-i \omega x} + \ldots \hspace{10px} \right|^{\infty}_{-\infty} + (i \omega)^n \int f(x) e^{-i \omega x}dx [/tex]

The claim is that when the delimiter is applied, [itex]\lim_{x \to \pm \infty} e^{-i \omega x} = 0[/itex]. I'm not sure if this is true. [itex]e^{-i \omega x} = \cos(\omega x) - i \sin(\omega x)[/itex] which clearly oscillates. Can anyone else confirm this?

Grading an assignment, the current topic is continuous Fourier Transforms. They're trying to prove the convenient property:

[tex] \mathcal{F} \left[ \frac{d^n}{dx^n} f(x) \right] = (i \omega)^n \mathcal{F} \left[ f(x) \right] [/tex]

So there's a simple way to get it:

Let [itex]f(x)[/itex] be represented by it's Inverse Fourier Transform [itex]f(x) = \mathcal{F}^{-1} \left[ g( \omega ) \right] = \frac{1}{2 \pi} \int_{-\infty}^{\infty} g( \omega )e^{i \omega x} d \omega [/itex]

Then,

[tex] \frac{d^n}{d x^n} f(x) = \frac{d^n}{d x^n} \frac{1}{2 \pi} \int_{-\infty}^{\infty} g( \omega )e^{i \omega x} d \omega = \frac{1}{2 \pi} \int_{-\infty}^{\infty} g( \omega ) \frac{d^n}{d x^n}e^{i \omega x} d \omega = (i \omega)^n \frac{1}{2 \pi} \int_{-\infty}^{\infty} g( \omega ) e^{i \omega x} d \omega = (i \omega)^n \mathcal{F} \left[ f(x) \right] [/tex]

However, I have some people with a different method. They directly transform the entire expression, and apply integration by parts recursively until they obtain:

[tex] \left. (i \omega)^0 \frac{d^{n-1}}{dx^{n-1}}[f(x)]e^{-i \omega x} + (i \omega)^1 \frac{d^{n-2}}{dx^{n-2}}[f(x)]e^{-i \omega x} + \ldots \hspace{10px} \right|^{\infty}_{-\infty} + (i \omega)^n \int f(x) e^{-i \omega x}dx [/tex]

The claim is that when the delimiter is applied, [itex]\lim_{x \to \pm \infty} e^{-i \omega x} = 0[/itex]. I'm not sure if this is true. [itex]e^{-i \omega x} = \cos(\omega x) - i \sin(\omega x)[/itex] which clearly oscillates. Can anyone else confirm this?